{"loaderData":{"0-0-56":{"bookData":{"_id":"680e6593797791ed81d778ae","category_id":"66ceced8c3d3f3f05eac01a2","author_id":[],"book_name":"Physics Problems and Questions Mechanics","book_image":"/books/GoldfarbMechanicsProblemspng-1745858661063.png","book_description":"

N.I. Goldfarb’s Physics Problems and Questions: Mechanics is a classic problem book featuring 444 carefully selected exercises covering fundamental topics like kinematics, Newton’s laws, energy, rotational motion, and gravitation. Designed for high school and introductory college students, it progresses from basic to challenging problems, emphasizing conceptual clarity and problem-solving skills without requiring calculus. It helps students build a strong foundation while preparing for exams like AP Physics, Olympiads, or university entrance tests. Whether used for independent study or classroom practice, this book remains a valuable resource for mastering mechanics through structured problem-solving.

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1. From points $A$ and $B$, separated by a distance $l$, two bodies start moving simultaneously towards each other: the first with velocity $v_1$, the second with $v_2$. Determine the time until they meet and the distance from point $A$ to the meeting point. Solve the problem graphically as well.

Answer

$t_c =\\dfrac{l}{v_1 + v_2}$; $x_c = \\dfrac{lv_1}{v_1 + v_2}$.

Solution

The dependence of the coordinates of the bodies on time is given by the formulas: $x_1 = v_1 t; \\quad x_2 = l - v_2 t.$ The meeting ($x_1 = x_2$) will occur after time $t = l/(v_1 + v_2)$ from the start of motion at a point located at a distance $lv_1/(v_1 + v_2)$ from point A. The graphs of the dependence of the coordinates of the bodies on time are shown below ($\\tan \\alpha = v_1$; $\\tan \\beta = -v_2 < 0$, since $v_2$ is directed in the opposite direction to the one taken as positive). The moment of meeting corresponds to point C of intersection of the graphs.
$\"Description\"$

2. After how much time and where would the bodies (see See Problem 1) meet if they moved in the same direction $A \\rightarrow B$, with the body from point $B$ starting to move $t_0$ seconds after the body from point $A$?

Answer

$t_c = \\dfrac{l - v_2 t_0}{v_1 - v_2}$; $x_c = \\dfrac{(l - v_2 t_0) v_1}{v_1 - v_2}$.

Hint

Use the graphs of the dependence of the coordinates of the bodies on time (the following figure). $\"Description\"$

3. A motorboat covers the distance between two points $A$ and $B$ downstream in time $t_1 = 3\\, \\text{h}$, while a raft takes $t = 12\\, \\text{h}$. How much time $t_2$ will the motorboat take for the return journey?

Answer

$t_2 = 6$ h.

Solution

Let $u$ be the speed of the current, and $v$ be the speed of the boat relative to the water, we can write: $t_1 = \\dfrac{s}{v+u}; \\tag{1}$ $t = \\dfrac{s}{u}. \\tag{2}$ To determine the time $t_2 = s/(v-u)$ taken by the boat for the return trip, it is convenient to rewrite equations (1) and (2) as follows: $\\dfrac{1}{t_1} = \\dfrac{v}{s} + \\dfrac{u}{s}; \\tag{3}$ $\\dfrac{1}{t} = \\dfrac{u}{s}. \\tag{4}$ Subtracting twice equation (4) from equation (3), we get $\\dfrac{1}{t_1} - \\dfrac{2}{t} = \\dfrac{v}{s} - \\dfrac{u}{s} = \\dfrac{1}{t_2},$ whence $t_2 = \\dfrac{t t_1}{t - 2t_1}.$

4. An escalator in the metro descends a person walking down it in 1 minute. If the person walks twice as fast, they descend in 45 seconds. How much time does it take for a person standing still on the escalator to descend?

Answer

1.5 min.

5. A person runs on an escalator. The first time, they count $n_1 = 50$ steps; the second time, moving in the same direction at three times the speed, they count $n_2 = 75$ steps. How many steps would they count on a stationary escalator?

Answer

$n = 100$.

Solution

If the person’s velocity were directed opposite to the motion of the escalator, then the faster he walked, the fewer steps he would count (but no fewer than $ n$. In our case, the directions of the person’s and the escalator’s motion coincide.

Let $v$ be the speed of the escalator; $l$ its length, and $n$ the number of steps on a stationary escalator. The number of steps per unit length of the escalator is $n / l$.

Therefore, if the person walks with speed $u$ relative to the escalator, the time spent on the escalator is $l / (v + u)$, and the distance he travels (relative to the escalator) is $ul / (v + u)$.

Thus, the number of steps the person counts is:
$n_1 = \\dfrac{ul}{v + u} \\cdot \\dfrac{n}{l} = \\dfrac{u}{v + u} n.$

Similarly, in the second case (if the walking speed is tripled), he counts:
$n_2 = \\dfrac{3ul}{v + 3u} \\cdot \\dfrac{n}{l} = \\dfrac{3u}{v + 3u} n.$
Hence, we obtain the following system of equations:

$ \\begin{array}{lcl} \\left\\{ \\begin{array}{l} \\displaystyle \\dfrac{u}{v + u} n = n_1, \\\\ \\displaystyle \\dfrac{3u}{v + 3u} n = n_2 \\end{array} \\right. & \\text{or} & \\left\\{ \\begin{array}{l} \\displaystyle 1 + \\dfrac{v}{u} = \\dfrac{n}{n_1}, \\\\ \\displaystyle 1 + \\dfrac{1}{3} \\cdot \\dfrac{v}{u} = \\dfrac{n}{n_2} \\end{array} \\right. & \\end{array} $

From this, by eliminating the ratio $v/u$, we find:
$n = \\dfrac{2n_1 n_2}{3n_1 - n_2} = 100.$

6. Between two points on a river, located $s = 100\\, \\text{km}$ apart, a boat shuttles. Going downstream, it covers this distance in $t_1 = 4\\, \\text{h}$, and upstream in $t_2 = 10\\, \\text{h}$. Determine the river's current speed and the boat's speed $v$ relative to the water.

Answer

$u = 7.5$ km/h; $v = 17.5$ km/h.

7. A raft passes by a dock. At that moment, a motorboat departs downstream to a village located $s_1 = 15\\, \\text{km}$ from the dock. The boat reaches the village in $t = 3/4\\, \\text{h}$ and, turning back, meets the raft at a distance $s_2 = 9\\, \\text{km}$ from the village. What are the river's current speed and the boat's speed relative to the water?

Answer

$u = 4$ km/h; $v = 16$ km/h.

Solution

Let's choose the raft (water) as the reference frame. In this reference frame, the boat moves down and up the river with the same speed. This means that the time for the boat to move away from the raft is equal to the time to approach it; thus, the boat also returned in 3/4 h. During the elapsed 1.5 h, the raft traveled a distance $s_1-s_2 = 6 \\text{ km}.$ Consequently, the speed of the current (speed of the raft relative to the bank) is $u = 6 / 1.5 = 4~ \\text{km/h}.$ The speed of the boat relative to the water $v = \\dfrac{s_1}{t} - u = 16~\\text{km/h}.$

This solution illustrates how important a successful choice of reference frame is in kinematics.

8. A column of troops marches at a speed of $v_1 = 5\\, \\text{km/h}$, stretched along a road for a distance $l = 400\\, \\text{m}$. A commander at the rear sends a cyclist with a message to the lead detachment. The cyclist rides at $v_2 = 25\\, \\text{km/h}$, delivers the message, and immediately returns at the same speed. How much time $t$ after receiving the message does the cyclist return?

Answer

$t = 2$ min.

Hint

The speed of the cyclist in the reference frame associated with the column is $v_2 - v_1$ when moving towards the head of the column, and $v_2 + v_1$ when returning. Therefore $t = \\dfrac{l}{v_2 - v_1} + \\dfrac{l}{v_2 + v_1} = \\dfrac{2lv_2}{v_2^2 - v_1^2}.$

9. A train car of width $d = 2.4\\, \\text{m}$, moving at $v = 15\\, \\text{m/s}$, is pierced by a bullet flying perpendicular to the car's motion. The displacement of the holes in the car's walls relative to each other is $l = 6\\, \\text{cm}$. What is the bullet's speed?

Answer

600 m/s.

10. What is the speed $v_2$ of vertically falling raindrops if a car driver notices that the drops leave no trace on the rear window, tilted forward at an angle $\\alpha = 60^\\circ$ to the horizon, when the car's speed $v_1$ exceeds 30 km/h?

Answer

$v_2 = v_1 \\tan \\alpha \\approx 14.4$ m/s.

Solution

For the raindrops not to leave a trace on the rear window, the direction of the raindrop velocity vector relative to the car must make an angle with the horizon no greater than $60^\\circ$. The relative velocity is composed of the vertical component, equal to the falling speed of the drop $v_2$, and the horizontal component, equal to the speed of the car $-v_1$ and directed in the opposite direction (following figure). From the velocity triangle, it is clear that $v_2 = v_1 \\tan \\alpha$.
$\"Description\"$

11. It is raining outside. In which case will a bucket standing in the back of a truck fill faster with water: when the truck is moving or when it is stationary?

Answer

Identical.

12. At what speed $v$ and course must an airplane fly to cover a path of $s = 300\\, \\text{km}$ due North in time $t = 2\\, \\text{h}$, if a northwest wind blows at an angle $\\alpha = 30^\\circ$ to the meridian with speed $u = 27\\, \\text{km/h}$ during the flight?

Answer

$v = 174$ km/h; northwest at an angle $4^\\circ 27'$ to the meridian.

Hint

Let $v_0 = s/t$ be the speed of the aircraft relative to the Earth. It is directed, according to the condition, along the meridian; then $\\vec{v}_0 = \\vec{v} + \\vec{u}$. By the cosine theorem $v^2 = u^2 + v_0^2 - 2uv_0 \\cos 150^\\circ.$ The angle $\\beta$ between $\\vec{u}$ and $\\vec{v}_0$ is found from the sine theorem: $\\sin \\beta / u = \\sin 150^\\circ / v.$

13. A blackboard moves on a smooth horizontal table at speed $v$. What shape will chalk leave as a trace on the board if it is thrown horizontally at speed $u$ perpendicular to the board's motion, assuming: a) negligible friction between the chalk and the board; b) significant friction?

Answer

a) and b) - a straight line forming an angle $\\alpha = \\arctan(u/v)$ with the direction of motion of the board. In case b), the trace may not reach the edge of the board.

Hint

Consider the motion of the chalk in the coordinate system associated with the board. Since the friction force is directed along the velocity vector of the chalk relative to the board, it cannot change the direction of the chalk's motion and only reduces its speed.

14. A ship departs from point $A$ at speed $v$, making an angle $\\alpha$ with the line $AB$ (following figure). At what angle $\\beta$ to the line $AB$ should a torpedo be launched from point $B$ to hit the ship? The torpedo must be launched when the ship is at point $A$. The torpedo's speed is $u$. $\"Description\"$

Answer

$\\beta = \\arcsin(\\dfrac{v}{u} \\sin \\alpha)$.

Hint

Point C (see figure below) is the meeting place of the ship and the torpedo. $AC = vt$, $CB = ut$, where $t$ is the travel time of the torpedo. According to the sine theorem $\\dfrac{AC}{\\sin \\beta} = \\dfrac{BC}{\\sin \\alpha}$ or $\\dfrac{vt}{\\sin \\beta} = \\dfrac{ut}{\\sin \\alpha}.$ Hence $\\beta = \\arcsin(\\dfrac{v}{u} \\sin \\alpha).$ $\"Description\"$

15. A cord is attached to a slider that can move along a guide rail (see figure below) and threaded through a ring. The cord is pulled at speed $v$. At what speed $u$ does the slider move when the cord makes an angle $\\alpha$ with the guide rail? $\"Description\"$

Answer

$u = v / \\cos \\alpha$.

Solution

For the same small time interval $\\Delta t$, the slider moves by $AB = \\Delta l$, and the cord is pulled by the length $AC = \\Delta l \\cos \\alpha$ (see figure) ($\\angle BCA$ can be considered right due to the smallness of $\\Delta l$). Therefore, we can write $\\Delta l / u = \\Delta l \\cos \\alpha / v$, whence $u = v / \\cos \\alpha$, i.e., the speed of pulling the rope is equal to the projection of the slider's velocity onto the direction of the rope.
$\"Description\"$

16. Workers lifting a load (following figure) pull ropes at the same speed $v$. What speed $u$ does the load have when the angle between the ropes to which it is attached is $2\\alpha$? $\"Description\"$

Answer

$u = v / \\cos \\alpha$.

Hint

The projection of the velocity $v$ of the load onto the direction of each cable must be equal to the speed $u$ of the cable (see problem 15).

17. A rod of length $l = 1\\, \\text{m}$ is connected by hinges to collars $A$ and $B$, which move along two mutually perpendicular rails (see figure). Collar $A$ moves at a constant speed $v_A = 30\\, \\text{cm/s}$. Find the speed $v_B$ of collar $B$ when the angle $OAB = 60^\\circ$. Taking the time origin as when collar $A$ was at point $O$, determine the distance $OB$ and the speed of collar $B$ as a function of time. $\"Description\"$

Answer

$v_B = v_A \\cot \\alpha = 17.3$ cm/s; $OB = \\sqrt{l^2 - v_A^2 t^2}$; $v_B = v_A^2 t / \\sqrt{l^2 - v_A^2 t^2}$.

Hint

See solution to problem 15. At any moment in time, the projections of the velocities $v_A$ and $v_B$ of the ends of the rod (see figure below) onto the axis of the rod are equal to each other, otherwise the rod would have to shorten or lengthen. Thus, $v_A \\cos \\alpha = v_B \\sin \\alpha$, whence $v_B = v_A \\cot \\alpha$. $\"Description\"$

18. A tank moves at $72\\, \\text{km/h}$. At what speed do the following move relative to the Earth: a) the upper part of the track;

b) the lower part of the track;

c) the point on the track moving vertically relative to the tank at that moment?

Answer

a) 40 m/s;

b) 0;

c) $20\\sqrt{2}$ m/s.

Hint

The velocity of any point on the track relative to the Earth at any given moment is composed of the velocity of the tank relative to the Earth and the velocity of the track link relative to the tank.

19. 1. A car travels the first half of the path at $v_1 = 40\\, \\text{km/h}$ and the second half at $v_2 = 60\\, \\text{km/h}$. Find the average speed over the entire path. 2. A car travels half the path at $v_1 = 60\\, \\text{km/h}$, then half the remaining time at $v_2 = 15\\, \\text{km/h}$, and the last segment at $v_3 = 45\\, \\text{km/h}$. Find the car's average speed over the entire path.

Answer

1. $v_{\\text{avg}} = \\dfrac{2v_1 v_2}{v_1 + v_2} = 48$ km/h;

$v_{\\text{avg}} = \\dfrac{2v_1(v_2+v_3)}{2v_1 + v_2 + v_3} = 40$ km/h.

Solution

The average speed $v_{\\text{avg}}$ is numerically equal to the ratio of the total path traveled by the body to the time of motion: $v_{\\text{avg}} = s_{\\text{total}}/t.$ Time $t = t_1 + t_2 = \\dfrac{s/2}{v_1} + \\dfrac{s/2}{v_2} = \\dfrac{s(v_1+v_2)}{2v_1 v_2},$ therefore $v_{\\text{avg}} = s/t = \\dfrac{2v_1 v_2}{v_1 + v_2}.$

2.Hint. The car traveled the second half of the path with an average speed equal to $(v_2+v_3)/2$.

20. A train travels the first half of the path at a speed $n = 1.5$ times greater than the second half. Its average speed over the entire path is $v_{\\text{avg}} = 43.2\\, \\text{km/h}$. What are the train's speeds $v_1$ and $v_2$ for the first and second halves of the path?

Answer

$v_1 = v_{\\text{avg}} \\dfrac{n+1}{2} = 54$ km/h; $v_2 = v_{\\text{avg}} \\dfrac{n+1}{2n} = 36$ km/h.

21. Two balls start moving simultaneously with the same speed along surfaces shaped as shown in the following figure. How will their speeds and travel times differ when they arrive at point $B$? Neglect friction. $\"Description\"$

Answer

The speeds will be the same. The travel time of the second ball is less.

Solution

Sample graphs of the speed of the balls are shown in the following figure. Since the paths traveled by the balls are equal, as seen from the graph (on the graph, the path is numerically equal to the areas of the shaded figures), $t_2 < t_1$. $\"Description\"$

22. An airplane flies from point $A$ to point $B$ and back. Its speed in calm weather is $v$. Find the ratio of the average speeds for the entire flight in two cases: a) when the wind blows along the line $AB$; b) when the wind blows perpendicular to $AB$. The wind speed is $u$.

Answer

$\\sqrt{v^2 - u^2}/v$.

Solution

If the wind blows along the line AB, the total flight time is $t_1 = \\dfrac{l}{v+u} + \\dfrac{l}{v-u} = \\dfrac{2l v}{v^2 - u^2},$ and the average speed $v_{\\text{avg}1} = \\dfrac{2l}{t_1} = \\dfrac{v^2 - u^2}{v}$ where $l$ is the distance between cities. If the wind is perpendicular to the line AB, the speed of the aircraft relative to the wind must be directed at an angle to the line AB, such that the drift is compensated (see figure). The flight speed in this case is constant and $v_{\\text{avg}2} = \\sqrt{v^2 - u^2}$ (the flight time in this case is $t_2 = 2l/\\sqrt{v^2 - u^2}$). $v_{\\text{avg}1}/v_{\\text{avg}2} = (v^2 - u^2)/(v \\sqrt{v^2 - u^2}) = \\sqrt{v^2 - u^2}/v.$
$\"Description\"$

23. A metro train covers the distance $s = 3\\, \\text{km}$ between two stations at an average speed $v_{\\text{avg}} = 54\\, \\text{km/h}$. It spends $t_1 = 20\\, \\text{s}$ accelerating, then moves uniformly for time $t_2$, and takes $t_3 = 10\\, \\text{s}$ decelerating to a full stop. Plot the train's speed graph and determine its maximum speed $v_{\\text{max}}$.

Answer

$v_{\\text{max}} \\approx 16.2$ m/s. The graph of the train's speed is shown in the following figure.
$\"Description\"$

24. The last car detaches from a moving train. The train continues at the same speed $v_0$. What is the ratio of the distances traveled by the train and the car when the car stops? Assume the car decelerates uniformly. Solve the problem graphically as well.

Answer

2:1; see the following figure.

Hint

The path traveled by the train after the car was detached is equal to the area of the rectangle OABC; the path traveled by the car is the area of the triangle OAC. $\"Description\"$

25. When the train starts moving, a well-wisher begins running uniformly alongside it at $v_0 = 3.5\\, \\text{m/s}$. Assuming the train's motion is uniformly accelerated, determine its speed $v$ when the well-wisher catches up.

Answer

$v = 7$ m/s; see the following figure.

Hint

From the graph, it is clear that the paths traveled by the passerby and the train are equal at the moment when $v = 2v_0$, where $v_0$ is the speed of the passerby. $\"Description\"$

26. The graph of a body's velocity versus time is shown in the following figure. Plot the graphs of acceleration, position, and distance traveled versus time. $\"Description\"$

Answer

Figure below.

Hint

The tangent of the angle of inclination of the tangent to the graph of the dependence of the coordinate (or path) of the body on time is numerically equal to the speed of the body. Therefore, the straight line representing the graph $x(t)$ between $t_1$ and $t_2$ at the points corresponding to moments $t_1$ and $t_2$ will be tangent to the parabolas describing the motion of the body before moment $t_1$ and after moment $t_2$, which are the graphs $x(t)$ or $s(t)$ before moment $t_1$ and after moment $t_2$ respectively. At moments $t=0$ and $t=t_3$, the graphs $s(t)$ and $x(t)$ have a horizontal tangent. $\"Description\"$

27. The graph of a body's acceleration versus time is shown in the following figure. Plot the graphs of velocity, displacement, and distance traveled versus time. The initial velocity is zero (acceleration is zero at the discontinuity). $\"Description\"$

Answer

See the following figure.

Hint

The straight line between points $t_1$ and $t_2$ is tangent to the curves $x(t)$ and $s(t)$ describing the motion of the body before moment $t_1$ and after moment $t_2$. $\"Description\"$

28. A body starts moving from point $A$ at speed $v_0$ and reaches point $B$ (see figure) after some time. What distance did the body travel if it moved with uniform acceleration $a$? The distance between $A$ and $B$ is $l$. Find the body's average speed. $\"Description\"$

Answer

$s = \\dfrac{v_0^2}{a} + l$; $v_{\\text{avg}} = \\dfrac{v_0^2 + al}{v_0 + \\sqrt{v_0^2 + 2al}}$.

Solution

Let's introduce a coordinate system with the origin at point A and the x-axis directed along the velocity of the body at point A (see figure). When the body reaches point B, its coordinate will be $-l$. The body reaches point B only if its acceleration $a$ is directed opposite to the velocity $v_0$. Therefore, the dependence of velocity $v$ on time is expressed by the formula $v = v_0 - at$. At the moment $t_1 = v_0/a$, the velocity becomes zero. At this moment, the body is at the maximum distance from point A. For $t > t_1$, $v < 0$. Substituting $t = t_1$ into the kinematic equation of motion: $x = v_0 t - at^2/2$, we find $x_{\\text{max}} = v_0^2 / (2a)$. The path $s$ traveled by the body during the entire movement is $s = 2x_{\\text{max}} + l = 2\\dfrac{v_0^2}{2a} + l = \\dfrac{v_0^2}{a} + l.$ To find the average speed $v_{\\text{avg}}$, you need to know the time of motion of the body from point A to point B. It can be found by substituting the coordinate value of point B into the equation for x, i.e., $x = -l$: $t_2 = (v_0 + \\sqrt{v_0^2 + 2al})/a$ (the second root is negative, so it must be discarded). Then the average speed of the body is $v_{\\text{avg}} = s/t_2 = (v_0^2 + al)/(v_0 + \\sqrt{v_0^2 + 2al}).$ $\"Description\"$

29. The following figure shows the graph of a body's position versus time. After $t = t_1$, the curve is a parabola. What motion is depicted? Plot the velocity versus time graph. $\"Description\"$

Answer

See the following; $v_1 = \\tan \\alpha$. Point $t_2$ corresponds to the maximum coordinate of the body. Before moment $t=t_1$, the motion is uniform, for $t > t_1$ - uniformly accelerated with negative acceleration (i.e., acceleration directed opposite to velocity $v$). $\"Description\"$

30. The following figure shows velocity graphs for two points moving along the same line from the same starting position. The times $t_1$ and $t_2$ are known. At what time $t_3$ will the points meet? Plot the motion graphs. $\"Description\"$

Answer

$t_3 = t_2 + \\sqrt{t_2(t_2-t_1)}$.

Solution

Equating the paths traveled by the bodies by the moment $t_3$ of meeting: $a_1 t_3^2 / 2 = a_2 (t_3 - t_1)^2 / 2. \\tag{1}$ From the above figure, it is clear that $a_1/a_2 = (t_2-t_1)/t_2$ ($a_1 = \\tan \\alpha$; $a_2 = \\tan \\beta$). Substituting the ratio $a_1/a_2$ into equation (1) and solving the resulting equation for $t_3$, we find $t_3 = t_2 + \\sqrt{t_2(t_2-t_1)}$. The second root is discarded, as the condition $t_3 > t_2$ must be met. The graphs of motion are shown in the following.
$\"Description\"$

31. In uniformly accelerated motion starting from rest, during which second does the distance traveled become three times the distance traveled in the previous second?

Answer

During the second second.

32. A cart must transport cargo the shortest possible distance $L$. It can only accelerate or decelerate with constant magnitude $a$, then move uniformly or stop. What maximum speed $v$ must the cart reach to meet this requirement?

Answer

$v = \\sqrt{La}$.

Solution

The time to transport the cargo will be minimum if the average speed of movement of the trolley is maximum, which is obvious. The latter, under the conditions of this problem (start and end of motion occur with constant acceleration), can only be in the case where the trolley moves the first half of the path with acceleration $+a$, and the second - with acceleration $-a$ (see figure). Thus, we can write the following relations: $v\\tau/2 \\cdot 2 = L/2$, $\\tau/2 = v/a$, whence $v = \\sqrt{La}$.
$\"Description\"$

33. A jet plane flies at $v_0 = 720\\, \\text{km/h}$. At some point, it accelerates for $t = 10\\, \\text{s}$, covering $s = 295\\, \\text{m}$ in the last second. Determine the acceleration $a$ and final speed $v$.

Answer

$a = 10$ m/s$^2$; $v = 300$ m/s.

34. The first train car passes an observer on the platform in $t_1 = 1\\, \\text{s}$, and the second in $t_2 = 1.5\\, \\text{s}$. Each car is $l = 12\\, \\text{m}$ long. Find the train's acceleration $a$ and initial speed $v_0$. Assume uniformly accelerated motion.

Answer

$a = \\dfrac{2l(t_1-t_2)}{t_1 t_2 (t_1+t_2)} = 3.20$ m/s$^2$; $v_0 = l/t_1 + at_1/2 \\approx 13.6$ m/s.

Hint

The paths traveled by the train by moments $t_1$ and $t_1+t_2$ are respectively equal: $l = v_0 t_1 + at_1^2/2$ and $2l = v_0(t_1+t_2) + a(t_1+t_2)^2/2$.

35. A ball rolled up an inclined plane covers two equal segments of length $l$ each and continues further. The first segment takes $t$ seconds, the second $3t$ seconds. Find the ball's speed $v$ at the end of the first segment.

Answer

$v = \\dfrac{5l}{6t}$.

Hint

The problem is solved most simply if we use the reversibility of motion and choose the boundary between the two segments as the origin of coordinates. Then the ball has an initial velocity - the sought value $v$, while the acceleration in the direction towards the initial point is $a$, and towards the final point is $(-a)$. Writing the equation of motion for the paths traveled by the ball, $l = vt + at^2/2$; $l = v(3t) - a(3t)^2/2$, and eliminating $a$, we find the answer.

36. A board divided into five equal segments starts sliding down an inclined plane. The first segment passes a mark on the plane in $t = 2\\, \\text{s}$. How long will it take for the last segment to pass the mark? Assume uniformly accelerated motion.

Answer

$\\tau_n = 0.48$ s.

Solution

Let the entire length of the board be $5l$, then we can write that the time to pass the first segment is $\\tau = \\sqrt{2l/a}$. The times to pass the beginning and end of the n-th segment past the observer are respectively equal: $t_1 = \\sqrt{\\dfrac{2(n-1)l}{a}} = \\tau \\sqrt{n-1}$ and $t_2 = \\sqrt{\\dfrac{2nl}{a}} = \\tau \\sqrt{n},$ whence $\\tau_n = t_2 - t_1 = \\tau (\\sqrt{n} - \\sqrt{n-1}).$

37. A bullet flying at $400\\, \\text{m/s}$ strikes an earthen embankment and penetrates to a depth of $36\\, \\text{cm}$. How long did it move inside the embankment? What was its acceleration? What was its speed at a depth of $18\\, \\text{cm}$? At what depth did its speed reduce by a factor of three? Assume uniformly decelerated motion. What will its speed be when it has traveled 99% of its path?

Answer

$t = 1.8 \\times 10^{-3}$ s; $a \\approx 2.21 \\times 10^5$ m/s$^2$; $v \\approx 282$ m/s; $s = 32$ cm; $v_1 = 40$ m/s.

Hint

When determining at what depth the bullet's speed decreased by three times, use the formula $s = v_0^2 / (2a)$. Starting from the point where the speed decreased by three times, the remaining path will be $3^2 = 9$ times less than the total path.

38. A ball rolled up an inclined board passes a point $l = 30\\, \\text{cm}$ from the start twice: at $t_1 = 1\\, \\text{s}$ and $t_2 = 2\\, \\text{s}$ after starting. Determine the ball's initial speed $v_0$ and acceleration $a$, assuming it is constant.

Answer

$v_0 = 45$ cm/s; $a = 30$ cm/s$^2$.

Solution

1st method. The dependence of the coordinate of the body along the inclined plane on time is expressed by the formula $x = v_0 t - at^2/2$. Hence $t^2 - \\dfrac{2v_0}{a} t + \\dfrac{2x}{a} = 0.$ Since $t_1$ and $t_2$ are the roots of this equation for $x=l$, then, according to Vieta's theorem, $t_1 + t_2 = \\dfrac{2v_0}{a}\\quad\\text{and}\\quad t_1 t_2 = \\dfrac{2l}{a}.$ From the obtained system of equations, it is easy to find $v_0$ and $a$.

2nd method. The dependence of the ball's speed on time is expressed by the formula $v = v_0 - at$. At time moments $t=t_1$ and $t=t_2$, the ball had speeds equal in magnitude and opposite in direction: $v_1 = -v_2$. But $v_1 = v_0 - at_1\\quad\\text{and}\\quad v_2 = v_0 - at_2,$ therefore
$v_0 - at_1 = -v_0 + at_2,\\quad\\text{or}\\quad 2v_0 = a(t_1+t_2).\\tag{1}$
Since the ball moves with uniform acceleration, its average speed over time $t_1$ is $v_{\\text{avg}1} = \\dfrac{v_0+v_1}{2} = \\dfrac{2v_0 - at_1}{2}.$ Therefore, $l = v_{avg1} t_1 = (2v_0 - at_1)\\dfrac{t_1}{2},\\quad\\text{or}\\quad 2v_0 - at_1 = \\dfrac{2l}{t_1}.\\tag{2}$
Solving equations (1) and (2) together, we find $a = \\dfrac{2l}{t_1 t_2} (\\dfrac{1}{t_1} + \\dfrac{1}{t_2}).$

39. A body falls from a height of $100\\, \\text{m}$ without initial speed. How long does it take to cover the first and last meters of its path? What distance does it cover in the first and last seconds of its motion?

Answer

$t_1 \\approx 0.45$ s; $t_2 \\approx 0.023$ s; $s_1 \\approx 4.9$ m; $s_2 \\approx 40$ m.

40. Determine the open time $\\tau$ of a photographic shutter if, when photographing a ball falling vertically along a centimeter scale starting from zero with no initial speed, the negative shows a streak extending from $n_1$ to $n_2$ scale divisions.

Answer

$\\tau = \\sqrt{2l/g}(\\sqrt{n_2} - \\sqrt{n_1})$.

41. A freely falling body covers the last $30\\, \\text{m}$ in $0.5\\, \\text{s}$. Find the height of the fall.

Answer

$h \\approx 195$ m.

42. A freely falling body covers $1/3$ of its path in the last second. Find the fall time and height.

Answer

$t \\approx 5.45$ s; $h \\approx 145$ m.

Hint

Let $h$ be the height from which the body fell, and $t$ be the time of its fall from this height, we can write the following equations: $gt^2/2 = h$; $g(t-1)^2/2 = (2/3)h$. Solving them together, we get $t$ and $h$. The problem can be solved more shortly if we use the formula for the path traveled by the body in the t-th second: $s_t = \\dfrac{g}{2}(2t-1);$ $\\dfrac{g}{2}(2t-1) = \\dfrac{1}{3} \\dfrac{gt^2}{2}.$

43. With what initial speed $v_0$ must a ball be thrown downward from height $h$ to bounce up to height $2h$? Neglect air resistance and energy losses.

Answer

$v_0 = \\sqrt{2gh}$.

44. Two drops fall from a roof eaves. Two seconds after the second drop starts falling, the distance between them is $25\\, \\text{m}$. What is the time interval between their departures? Neglect air resistance.

Answer

$\\tau \\approx 1$ s.

Hint

Let: $\\tau$ be the time interval between the detachment of the 1st and 2nd drops, $t$ be the time from the moment of detachment of the 2nd drop, $d$ be the distance between the drops after time $t$ (from the moment of detachment of the 2nd drop) and considering that the drops move relative to each other uniformly, we can write the equation to determine $\\tau$: $d = \\dfrac{g\\tau^2}{2} + g\\tau t.$

45. A body is thrown vertically upward. An observer notes a time interval $t_0$ between two moments when the body passes point $B$ at height $h$. Find the initial speed $v_0$ and total motion time $t$.

Answer

$v_0 = \\dfrac{g}{2} t_0 + \\dfrac{8h}{t_0}$; $t = \\sqrt{t_0^2 + \\dfrac{8h}{g}}$.

Hint

See problem 38. Solve the problem using another method.

46. From points $A$ and $B$, vertically aligned ($A$ above) and $l = 100\\, \\text{m}$ apart, two bodies are thrown simultaneously at $10\\, \\text{m/s}$: from $A$ downward, from $B$ upward. When and where will they meet?

Answer

5 s; 75 m below point B.

Hint

The relative speed of the bodies is constant and equal to $2v_0$. Both bodies approach each other with this speed. Therefore, the time to meeting is $l/(2v_0) = 5$ s.

47. A body is thrown upward with initial speed $v_0$. At its peak, a second body is thrown from the same point at the same speed. At what height $h$ will they meet?

Answer

$h = (3/4)h_{\\text{max}} = (3/4)v_0^2/(2g)$.

Hint

The problem can be solved beautifully if you realize that the moment of meeting divides the ascent time, as well as the fall time of the bodies, into equal parts. The paths traveled by a body falling without initial velocity in successive equal time intervals relate as 1:3. Therefore, the meeting will occur at 3/4 of the height.

48. Two bodies are thrown upward from the same point at $v_0 = 19.6\\, \\text{m/s}$, with a time interval $\\tau = 0.5\\, \\text{s}$. After what time $t$ and at what height $h$ will they meet?

Answer

$t = \\dfrac{v_0}{g} - \\dfrac{\\tau}{2} = 1.75$ s; $h = \\dfrac{v_0^2}{2g} - \\dfrac{g\\tau^2}{8} \\approx 19.3$ m.

Hint

See problem 38. Solve the problem graphically as well.

49. A balloon rises vertically from Earth at $a = 2\\, \\text{m/s}^2$. After $\\tau = 5\\, \\text{s}$, an object falls out. How long $t$ until it hits the ground?

Answer

$t \\approx 3.4$ s.

Solution

Let's take the coordinate axis directed vertically upwards, with the origin on the Earth's surface. Then the kinematic equation of motion of the object has the form $y = y_0 + v_0 t - g\\dfrac{t^2}{2},$ where $y_0 = a\\tau^2/2$ and $v_0 = a\\tau.$ Substituting the coordinate value of the object at the moment of fall $y=0$ into this equation, we find $t$.

50. A balloon descending at speed $u$ throws an object upward at speed $v_0$ relative to Earth. What will be the distance $l$ between them when the object reaches its peak? What is the maximum distance $l_{\\text{max}}$? How long until they are level again?

Answer

$l = v_0^2 + 2uv_0 / (2g)$; $h_{\\text{max}} = (u+v_0)^2 / (2g)$; $t = 2(v_0+u)/g$.

Hint

The kinematic equation of motion of the body is conveniently written in the coordinate system associated with the moving aerostat. Here $v_0+u$ is the initial velocity of the body; $g$ is the acceleration; $h_{\\text{max}}$ is the coordinate of the body when the velocity of the body relative to the aerostat is zero; $t$ is its coordinate at the moment $t_1$, when its velocity $v$ relative to the Earth is zero. Since $v = v_0 - gt$, then $v=0$ at $t_1 = v_0/g$. At $t = \\tau$, the coordinate of the body is zero.

51. A body at point $B$, $H = 45\\, \\text{m}$ above Earth, starts falling freely. Simultaneously, another body is thrown upward from point $A$, $h = 21\\, \\text{m}$ below $B$, at speed $v_0$. Find $v_0$ if both bodies hit the ground simultaneously. Neglect air resistance. Take $g = 10\\, \\text{m/s}^2$.

Answer

$v_0 = 7$ m/s.

Solution

1st method. Taking the Earth's surface as the origin, let's write the equations of motion for both bodies:
$x_1 = H - \\dfrac{gt^2}{2}; \\tag{1}$
$x_2 = H - h + v_0 t - \\dfrac{gt^2}{2}.\\tag{2}$
At the moment of landing $x_1 = x_2 = 0$. Substituting the values $x_1$ and $x_2$ into equations (1) and (2) and solving them for $v_0$, we find $v_0 = h\\sqrt{\\dfrac{g}{2H}}.$

2nd method. Both bodies approach each other, moving uniformly relative to each other. The initial distance between them is $h$. Therefore, the velocity $v_0$ can be found from the equation of uniform motion $v_0 = h/t$, where $t$ is the fall time of the first body from height H. Since $t = \\sqrt{2H/g}$. We find $v_0 = h/\\sqrt{2H/g} = h\\sqrt{g/(2H)}$.

52. A body falls freely from height $h$. Simultaneously, another body is thrown downward from height $H (H > h)$. Both hit the ground simultaneously. Find the initial speed $v_0$ of the second body. Verify with $h = 10\\, \\text{m}$, $H = 20\\, \\text{m}$, and $g = 10\\, \\text{m/s}^2$.

Answer

$v_0 = \\dfrac{H-h}{2h} \\sqrt{2gh} \\approx 7$ m/s.

53. A stone is thrown horizontally from a mountain slope at angle $\\alpha$. What speed $v_0$ is needed for the stone to land on the mountain at distance $L$ from the peak?

Answer

$v_0 = \\sqrt{\\dfrac{Lg \\cos^2 \\alpha }{2 \\sin \\alpha}}$.

Solution

Writing the coordinate equations for the stone (see figure): $x = v_0 t;\\quad y =\\dfrac{gt^2}{2}$ and considering that at the moment of fall of the stone $x = L \\cos \\alpha,\\quad y = L \\sin \\alpha,$ we get after eliminating $t$, that $v_0 = \\sqrt{\\dfrac{Lg \\cos^2 \\alpha}{2 \\sin \\alpha}}$.
$\"Description\"$

54. Two people play catch with a ball. What is the maximum height the ball reaches if it takes 2 seconds to travel from one person to the other?

Answer

4.9 m.

Hint

The ascent time of the ball is 1 s.

55. An airplane flies at constant altitude $h$ in a straight line at speed $v$. The pilot must drop a bomb on a target ahead. At what angle to the vertical must the pilot see the target at the moment of release? What is the horizontal distance from the target to the point directly below the airplane? Neglect air resistance.

Answer

$\\tan \\alpha = v \\sqrt{2/(gh)}$; $s = v \\sqrt{2h/g}$.

Hint

$\\tan \\alpha = s/h$ (see figure). $\"Description\"$

56. Two bodies fall from the same height. One encounters a platform at $45^\\circ$ and rebounds elastically. How do their fall times and speeds differ?

Answer

The fall time of the first body, which did not experience a collision, is less than the fall time of the second body. The speeds of the bodies at the moments when they were at the same height are identical.

Hint

During an elastic collision, the magnitude of the velocity is conserved, only its direction changes.

57. An elevator ascends at $2\\, \\text{m/s}^2$. When its speed reaches $2.4\\, \\text{m/s}$, a bolt falls from the ceiling. The elevator is $2.47\\, \\text{m}$ tall. Find the bolt's fall time and distance traveled relative to the shaft.

Answer

0.64 s; 0.52 m.

Hint

In the coordinate system associated with the lift, the acceleration of the bolt is $9.8 \\text{ m/s}^2 + 2 \\text{ m/s}^2 = 11.8 \\text{ m/s}^2$. Relative to the shaft, the initial velocity of the bolt is 2.4 m/s, and the acceleration is $9.8 \\text{ m/s}^2$.

58. Two bodies are thrown simultaneously from the same height at $20\\, \\text{m/s}$, one downward and one upward, both at $45^\\circ$ to the vertical. Find their height difference $\\Delta h$ after 2 seconds. How do they move relative to each other?

Answer

$\\Delta h \\approx 56.4$ m; the bodies move apart with constant velocity.

Hint

In the vertical direction, both bodies move away from each other with a velocity equal to $v_0 \\cos \\alpha - (-v_0 \\cos \\alpha) = 2v_0 \\cos \\alpha$.

59. Prove that for freely moving bodies near Earth's surface, their relative speed is constant.

Answer

Solution

If the initial velocities of the bodies are $\\vec{v}_0$ and $\\vec{u}_0$, then after time $t$ their velocities are respectively $\\vec{v} = \\vec{v}_0 + \\vec{g}t$ and $\\vec{u}=\\vec{u}_0+\\vec{g}t$ ($\\vec{g}$ is the acceleration vector of the bodies). The velocity of the second body relative to the first is $\\vec{v}_{\\text{rel}} = \\vec{u} - \\vec{v} = \\vec{u}_0 - \\vec{v}_0.$

60. A body falls freely from point $A$. Simultaneously, another body is thrown from point $B$ at angle $\\alpha$ to the horizon so they collide mid-air (see figure). Show that $\\alpha$ is independent of the initial speed $v_0$ and determine $\\alpha$ if $H/l = \\sqrt{3}$. Neglect air resistance. $\"Description\"$

Answer

$\\tan \\alpha = \\sqrt{3}$; $\\alpha = 60^\\circ$.

Solution

1st method. Both bodies can meet on the line AO (see figure above) at point C (it can also lie below the line BO).

Let's resolve the initial velocity $v_0$ of the body thrown from point B into horizontal $v_{0x}$ and vertical $v_{0y}$ components: $v_{0x} = v_0 \\cos \\alpha,\\quad v_{0y} = v_0 \\sin \\alpha.$ From the start of motion until the moment of meeting, the elapsed time is: $t = l/v_{0x} = l/(v_0 \\cos \\alpha).\\tag{1}$ During this time, body A will descend by $H-h = gt^2/2, \\tag{2}$ and body B will rise to height $h = v_{0y} t - gt^2/2 = v_0 \\sin \\alpha t - gt^2/2. \\tag{3}$ Solving equations (2) and (3) together, we find $H = t v_0 \\sin \\alpha. \\tag{4}$ Substituting the value of $t$ from expression (1) into equation (4), we get $\\tan \\alpha = \\dfrac{H}{l},$ i.e., the throwing angle $\\alpha$ does not depend on the initial velocity, only the location of the meeting point C on the line AO depends on its value.

2nd method. Let's switch to the coordinate system associated with the body falling from point A. Here the velocity of the body thrown from point B is constant (see the previous problem). Therefore, it is clear that the bodies will meet if the velocity vector of the body is directed towards point A, i.e., $\\tan \\alpha = {l}. \\tag{4}$

61. A body is thrown at angle $\\alpha$ to the horizon with speed $v_0$. Find its speed $v$ at height $h$. Does this speed depend on $\\alpha$? Neglect air resistance.

Answer

$v = \\sqrt{v_0^2 - 2gh}$ and does not depend on the throwing angle.

62. A body is thrown at $\\alpha = 60^\\circ$ with $v = 20\\, \\text{m/s}$. After how long $t$ will it move at $\\beta = 45^\\circ$ to the horizon? Neglect friction.

Answer

$t = v (\\sin \\alpha - \\cos \\alpha \\tan \\beta)/g \\approx 0.75$ s.

63. Water jets from three pipes on the ground have the same speed but angles $60^\\circ$, $45^\\circ$, and $30^\\circ$ to the horizon. Find the ratios of their maximum heights $h$ and ranges $l$. Neglect air resistance.

Answer

$h_1:h_2:h_3 = 3:2:1$; $l_1:l_2:l_3 = \\sqrt{3}:\\sqrt{2}:1$.

Hint

Maximum ascent height $h = v_0^2 \\sin^2 \\alpha / (2g)$ and range of water flight $l = t v_0 \\cos \\alpha$, where $t$ is the time of water movement, equal to twice the ascent time to height $h$; $t = 2v_0 \\sin \\alpha / g$, whence $l = v_0^2 \\sin 2\\alpha / g$.

64. From a point at the top of a vertical diameter $d$ of a circle, weights slide without friction along chords of various inclinations (see figure below). Find the time $t$ for them to reach the circle. How does $t$ depend on the chord's angle to the vertical? $\"Description\"$

Answer

$t = \\sqrt{2d/g}$; does not depend on the angle.

Solution

Along the grooves, the weights will move under the action of the component of gravity $mg \\cos \\alpha$ and, consequently, with acceleration $a = g \\cos \\alpha$; the length of the groove $l = d/\\cos \\alpha$ (see the above figure). The travel time $t$ of the weight is $t = \\sqrt{\\dfrac{2l}{a}} = \\sqrt{\\dfrac{2d}{g}}.$ From the formula, it is clear that $t$ does not depend on the angle.

65. A stone's initial speed is $v_0 = 10\\, \\text{m/s}$. After $t = 0.5\\, \\text{s}$, its speed is $v = 7\\, \\text{m/s}$. What is its maximum height above the initial level?

Answer

$H_{\\text{max}} \\approx 2.8$ m.

Solution

Obviously, the stone is thrown upwards at an angle, because if it were thrown vertically, its speed after 0.5 s would be $v_0 - gt = 5.1$ m/s. The height to which the stone rises, $H_{\\text{max}} =\\dfrac{v_{0y}^2}{2g}, \\tag{1}$ where $v_{0y}$ is the vertical component of $v_0$; $v_{0y}$ can be determined from the following system of equations: $v_0^2 = v_x^2 + v_{0y}^2, \\tag{2}$ $v^2 = v_x^2 + (v_{0y} - gt)^2, \\tag{3}$ where $v_x$ is the horizontal projection of $v_0$. Subtracting equation (2) from equation (3), we get $v_{0y} = \\dfrac{v_0^2 - v^2 + g^2 t^2}{2gt}.\\tag{4}$ Substituting expression (4) into equation (1), we get $H_{\\text{max}} = \\dfrac{(v_0^2 - v^2 + g^2 t^2)^2}{8g^3 t^2}.$

66. Balls are thrown simultaneously from the same point at the same speed in all directions. What is the geometric locus of their positions at any time? Neglect air resistance.

Answer

A sphere with radius $v_0 t$, whose center lies below the initial point by $gt^2/2$.

Hint

As the reference body, choose a body that starts falling down from the same point at the moment the balls are thrown. Relative to it, the balls will move (disperse in possible directions) uniformly with speed $v_0$.

67. A target on a hill is visible from a cannon at angle $\\alpha$ to the horizon. The horizontal distance is $L$. The cannon fires at elevation angle $\\beta$ (see figure below). Find the initial speed $v_0$ to hit the target. Neglect air resistance. For what angle $\\beta_0$ is the range along the slope maximized? $\"Description\"$

Answer

$v_0 = \\sqrt{\\dfrac{gL \\cos \\alpha}{2 \\cos \\beta \\sin(\\beta - \\alpha)}}$; $\\beta_0 = \\dfrac{\\pi}{4} + \\dfrac{\\alpha}{2}$.

Solution

Let's choose the coordinate $xy$ system, whose origin is at the location of the gun (see the above figure). Then the kinematic equations of motion of the projectile in coordinates x, y will be expressed as follows: $x = v_0 t \\cos \\beta,\\quad y = v_0 t \\sin \\beta - gt^2/2.$ Substituting the coordinates of the target $x=L$; $y=L \\tan \\alpha$ into these equations, eliminating $t$, we get $v_0 = \\sqrt{\\dfrac{gL \\cos \\alpha}{2 \\cos \\beta \\sin(\\beta - \\alpha)}}.$ The range $l$ of the projectile along the slope is $l = L/\\cos \\alpha$. Therefore, the formula we obtained can be written as: $v_0 = \\sqrt{\\dfrac{gl \\cos^2 \\alpha}{2 \\cos \\beta \\sin(\\beta - \\alpha)}}.$ Hence $l = \\dfrac{2 v_0^2 \\cos \\beta \\sin(\\beta - \\alpha)}{g \\cos^2 \\alpha}.$ This expression is maximum at the maximum product $\\cos \\beta_0 \\sin(\\beta_0 - \\alpha) = \\dfrac{1}{2} [\\sin(2\\beta_0 - \\alpha) - \\sin \\alpha].$ Therefore, $l$ is maximum at the maximum value $\\sin(2\\beta_0 - \\alpha) = 1$ or $\\beta_0 = \\dfrac{\\pi}{4} + \\dfrac{\\alpha}{2}$ (for $\\alpha = 0$ we get the answer $\\beta_0 = \\pi/4 = 45^\\circ$).

68. An elastic body falls from height $h$ onto an inclined plane. How long $t$ after bouncing will it hit the plane again? How does $t$ depend on the plane's angle?

Answer

$t = 2\\sqrt{2gh}/g$; does not depend on the angle.

Solution

Let's take the coordinate system xOy, whose origin is located at the point of the first contact of the body with the inclined plane, the x-axis is directed along the inclined plane, and the $y$-axis is perpendicular to it (see figure below). With this choice of coordinate system, the solution of the problem is significantly simplified. The initial velocity $v_0$ of the body after the collision will be $v_0 = \\sqrt{2gh}$ and forms an angle $\\alpha$ with the $y$-axis. The equation of motion of the body along the $y$-axis will be expressed as follows: $y = v_{0y} t + a_y t^2 / 2,$ where $v_{0y} = v_0 \\cos \\alpha\\text{ and } a_y = -g \\cos \\alpha$ are the projections of the velocity $v_0$ and acceleration $g$ onto the $y$-axis. At the moment the body falls onto the inclined plane $y=0$. Substituting $y=0$ into the equation written above, we get $t = 2v_{0y}/a_y = 2v_0/g = 2\\sqrt{2gh}/g.$ The fall time does not depend on the angle of inclination of the plane. This time will remain unchanged for subsequent falls after the first reflection onto the inclined plane. $\"Description\"$

69. A ball falls freely from height $H$ onto a $45^\\circ$ inclined plane and bounces elastically at the same speed. Find the distances between successive bounces. Solve for any angle $\\alpha$.

Answer

$4H\\sqrt{2}$; $s_1 = 8H \\sin \\alpha$; $s_1:s_2:s_3 = 1:2:3$.

Solution

See the previous problem. The distance from the place of the first impact to the second will be determined from the relation $x = v_{0x} t + a_x t^2 / 2$, where $v_{0x} = v_0 \\sin \\alpha$ and $a_x = g \\sin \\alpha$ (see figure). Substituting the value $t$ obtained in the previous problem, we find that $ \\begin{aligned} s_1 &= v_0 \\dfrac{2v_0}{g} \\sin \\alpha + \\dfrac{g}{2} (\\dfrac{2v_0}{g})^2 \\sin \\alpha \\\\ &= \\dfrac{4v_0^2}{g} \\sin \\alpha \\\\ &= 8H \\sin \\alpha; \\end{aligned} $ and for the components of the velocity of the body $v_{2x}$ and $v_{2y}$ at the moment of the second collision, respectively: $v_{2y} = v_{0y} - a_y t = v_0 \\cos \\alpha - g \\dfrac{2v_0}{g} \\cos \\alpha = -v_{0y};$ $ \\begin{aligned} v_{2x} &= v_{0x} + a_x t \\\\ &=v_0 \\sin \\alpha + g \\dfrac{2v_0}{g} \\sin \\alpha \\\\ &= 3v_0 \\sin \\alpha, \\end{aligned} $ and at the moment of the n-th collision: $v_{ny} = -v_{0y},$ $v_{nx} = (2n-1)v_0 \\sin \\alpha.$ The average velocities along the $x$-axis between impacts are respectively equal: $2v_0 \\sin \\alpha,\\quad 4v_0 \\sin \\alpha,\\quad 6v_0 \\sin \\alpha,$ i.e., they relate as 1:2:3, and since the time of motion between impacts is the same, then $s_1:s_2:s_3... = 1:2:3.$ $\"Description\"$

70. A mountain's distance is determined by the time between a gunshot and its echo. What error $\\tau$ is acceptable in timing if the distance is at least 1 km and must be determined within 3%? Sound speed $c = 330\\, \\text{m/s}$.

Answer

$\\tau \\le 0.09$ s.

Solution

The error in determining the time between the shot and the arrival of the echo is $2\\tau$ seconds. This leads to an error in determining the distance traveled by sound, equal to $2\\tau c$ meters. The path traveled by sound is equal to twice the distance to the mountain $2x$, therefore the relative error in determining this distance is $2\\tau c / (2x) = \\tau c / x$. This is also the relative error in determining the distance to the mountain. Since $\\tau c / x \\le 0.03$, then $\\tau \\le 0.03 (x/c)$. Considering that $x \\ge 1000$ m, we find that the error in determining the moments of the shot and the arrival of the echo should not exceed 0.09 s.

71. A well's depth is measured by dropping a stone and timing the splash sound. For what minimum time $\\tau$ must sound travel time be accounted for? Sound speed $c = 330\\, \\text{m/s}$.

Answer

$\\tau \\approx 1.77$ s.

Solution

If we do not take into account the time of sound propagation, we get that the depth of the well is $h_1 = g\\tau^2/2$. Taking into account the time of sound propagation $\\tau = t_1 + t_2$, where $t_1 = \\sqrt{2h/g}$ is the time of the stone's fall, and $t_2 = h/c$ is the time of sound propagation ($h$ is the true depth of the well): $\\tau = \\dfrac{h}{c} + \\sqrt{\\dfrac{2h}{g}}.\\tag{1}$ The error in measuring the depth of the well is $\\dfrac{\\Delta h}{h} =\\dfrac{h_1 - h}{h} = 0.05.$ Hence $h = 0.95 h_1 = 0.95 \\dfrac{g\\tau^2}{2}.$ Substituting this expression into equation (1) and solving the resulting equation, we find $\\tau = \\dfrac{2(1 - 1/\\sqrt{0.95})c}{0.95 g} \\approx 1.77\\text{ s}.$

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1. Two train cars—one loaded, one empty—roll down a sorting hill. Which will travel farther on the straight track after descending? Assume the drag force is proportional to the load on the wheels and independent of speed. Neglect air resistance.

Answer

Both wagons will travel the same distance.

2. What mass of ballast $m$ must be dropped from a uniformly descending balloon to make it ascend uniformly at the same speed? The balloon's total mass is $M = 1200\\, \\text{kg}$, its lift force is constant at $F = 8000\\, \\text{N}$, and drag is the same for ascent and descent.

Answer

$m = 2(\\frac{F}{a} - M) \\approx 800$ kg.

Hint

The conditions for uniform motion of the aerostat during descent and ascent are expressed respectively: $Mg = F + F_{\\text{res}}$; $(M-m)g + F_{\\text{res}} = F$.

3. A mine cage of mass $M = 3 \\cdot 10^3\\, \\text{kg}$ starts ascending at $a = 0.49\\, \\text{m/s}^2$. Find:

a) the cable tension;

b) the tension when descending at the same acceleration;

c) the tension when moving up or down at constant speed.

Answer

a) $F_N = M(g+a) \\approx 3.1 \\times 10^4$ H;

b) $F_N = M(g-a) \\approx 2.8 \\times 10^4$ H;

c) $F_N = Mg \\approx 2.95 \\times 10^4$ H.

4. A person stands on a spring scale in an elevator. How do the scale readings change as the elevator moves up and down, considering changes in motion?

Answer

When the elevator moves with acceleration directed upwards (this will be at the beginning of the ascent and at the end of the descent), the scale reading is greater than $mg$ ($m$ is the mass of the person). If the acceleration is directed downwards, which will be at the beginning of the descent and at the end of the ascent, the scale reading is less than $mg$. If the elevator moves uniformly, the scale reading is equal to $Mg$.

Hint

The equation of motion for the person is $ma = mg - N$ (equation written in projections onto the direction of $g$). Here $a$ is the acceleration of the elevator (algebraic quantity) and $N$ is the reaction force of the scales. In accordance with Newton's third law, the force $N'$ acting on the scales from the person is equal in magnitude to $N$: $N' = N = m(g-a)$.

5. A person stands on a board and suddenly crouches. Does the board's deflection initially increase or decrease? What happens if the person was squatting and suddenly stands?

Answer

When squatting, it will decrease; when straightening up, it will increase.

Hint

When squatting, the acceleration of the person's center of mass is directed downwards, and when straightening up - upwards.

6. A box filled with balls is thrown upward. How do the forces of the balls on the box's bottom, sides, and each other change during flight? How does the answer change if the box is thrown at an angle? Neglect air resistance.

Answer

During the free movement of the box (movement in the field of gravity in the absence of other forces), the pressure force of the balls on the bottom and walls of the box, as well as on each other, is zero.

7. A heavy object is suspended from a spring attached to an elevator ceiling. How will the object move relative to the elevator if the elevator starts free-falling?

Answer

The body will oscillate on the spring, the amplitude of which is equal to the initial extension of the spring.

Hint

The problem is conveniently solved in a non-inertial coordinate system associated with the moving elevator (see 'Elementary Physics Textbook' edited by Acad. G. S. Landsberg, vol. I, ch. 6). In this reference frame, three forces act on the mass: the force of gravity $mg$, the inertial force $-ma$ balancing it, and also the spring tension force, equal at the initial moment to the force of gravity $mg$ acting on the mass.

8. A body on a stand is suspended from a ceiling by a spring (initially unstretched). The stand is lowered at acceleration $a$. After what time $\\Delta t$ will the body lift off? The spring constant is $k$, and the body's mass is $m$.

Answer

$\\Delta t = \\sqrt{2m(g-a)/(ka)}$.

Solution

Until the moment the body detaches from the support, it moves with acceleration $a$ under the action of three forces: the force of gravity $F_1 = mg$, the spring tension force $F_2 = kx$, and the support reaction force $N$. Therefore $mg - kx - N = ma$. At the moment the body detaches from the support, the support reaction force becomes zero. Considering this and substituting the expression for $x = a(\\Delta t)^2/2$ into the equation of motion, we find $\\Delta t = \\sqrt{2m(g-a)/(ka)}$.

9. Masses $m_1 = 50\\, \\text{g}$ and $m_2 = 75\\, \\text{g}$ are hung from a cord over a pulley. Neglecting friction, cord and pulley mass, and cord stretch, find the accelerations, cord tension, and dynamometer reading supporting the pulley.

Answer

$a \\approx 1.96$ m/s$^2$; $T \\approx 0.6$ H; $F \\approx 1.2$ H.

Solution

The forces acting on the weights are shown in the following figure. Let's write the equations of motion for the weights: $m_1 a_1 = m_1 g - T_1,\\tag{1}$ $m_2 a_2 = T_2 - m_2 g. \\tag{2}$ The accelerations $a_1$ and $a_2$ of the weights are numerically equal, since the string is inextensible and the displacements of the weights are always the same (the relationships between the accelerations of the bodies of the system are determined by their kinematic connection, i.e., the connection between their displacements). In the general case, the forces $T_1$ and $T_2$ applied to the ends of the string, according to Newton's third law, are numerically equal to the forces $T'_1$ and $T'_2$, are not equal to each other, i.e., the difference between these forces imparts acceleration to the string and angular acceleration to the pulley. If the mass of the string and pulley can be neglected, then $T_1$ and $T_2$ and the corresponding forces $T'_1$ and $T'_2$ applied to the weights can be considered equal to each other. Projecting the vectors included in equations (1) and (2) onto the directions $x_1$ and $x_2$, coinciding with the accelerations of each of the weights (conventionally assume $m_1 > m_2$), we obtain the following system of equations: $m_1 a = m_1 g - T$; $m_2 a = T - m_2 g$. Solving this system for the unknowns $a$ and $T$, we get $a = \\frac{m_1 - m_2}{m_1 + m_2} g$; $T = \\frac{2m_1 m_2}{m_1 + m_2} g$. (If $m_1 < m_2$, we get $a < 0$, i.e., the weights move in the direction opposite to the assumed one.) The dynamometer reading is obviously equal to the sum of the string tension forces: $F = 2T = \\frac{4m_1 m_2}{m_1 + m_2} g.$ $\"Description\"$

10. Two masses $m_1 = 7\\, \\text{kg}$ and $m_2 = 11\\, \\text{kg}$ hang from a cord over a pulley. Initially at the same height, how long $t$ until the lighter mass is 10 cm above the heavier one? Neglect pulley and cord mass and friction.

Answer

$t = 0.21$ s.

Hint

See the previous problem. Each weight will travel a path equal to 5 cm.

11. Two identical masses $M$ are hung from a cord over a pulley. A small mass $m$ is placed on one. Find the force $f$ of $m$ on $M$ and the force $F$ on the pulley's axle.

Answer

$F = 4Mg (\\dfrac{m+M}{m+2M})$; $f = 2 \\dfrac{Mm}{m+2M} g$.

Solution

In accordance with Newton's second law, let's write the equations of motion for each of the bodies (see figure): for the first body $Ma = T - Mg;$ for the second body $Ma = Mg + f - T;$ for the load $ma = mg - f',$ where $T$ is the string tension force; $f$ is the pressure force of the load on the body, $a$ is the acceleration of the bodies; $f'$ is the reaction force of the support. According to Newton's third law, $f'$ and $f$, $T$ and $T'$ are numerically equal. The tensions on the right and left are equal, since friction and the mass of the pulley are neglected. Solving the obtained system of equations, we find $a = \\frac{mg}{m+2M};$ $f = 2 \\frac{Mm}{m+2M} g,$ $T = M(\\frac{mg}{m+2M} + g) = 2M (\\frac{m+M}{m+2M}) g.$ The force acting on the axis of the pulley, $F = 2T = 4M (\\frac{m+M}{m+2M}) g.$ $\"Description\"$

12. A horizontal axis passes through the middle of a rod, about which it can rotate. Masses with $m_1 = 2\\,\\text{kg}$ and $m_2 = 8\\,\\text{kg}$ are attached to the ends of the rod. The rod is brought to a horizontal position and released without a push.

Find the force exerted by the rod on the axis at the initial moment after its release. Neglect the mass of the rod and friction at the axis; consider the masses as material points.
Solve the problem for the case when the axis does not pass through the center of the rod but is located at a distance of $1/3$ of its length from the smaller mass.

Answer

1. $F = \\dfrac{4m_1 m_2}{m_1 + m_2} g \\approx 64$ N.

$F = \\frac{9m_1 m_2}{4m_1 + m_2} g \\approx 42.3$ N.

Solution

1. The force $F'$, acting from the axis on the rod and according to Newton's third law numerically equal to the pressure force $F$ on the axis (see figure below, a), can be replaced by two forces $F_1$ and $F_2$, applied to each of the weights and equal to $F/2$. Considering that the rod is massless, we can, using Newton's second law, write: $m_1 a = F/2 - m_1 g$, $m_2 a = m_2 g - F/2$, whence $F = \\frac{4m_1 m_2}{m_1 + m_2} g$.

The force $F'$ must be replaced (see figure below, b) by forces $F_1 = (2/3)F'$ and $F_2 = (1/3)F'$, applied to weights $m_1$ and $m_2$ respectively. The accelerations of the weights are related by $a_1/l_1 = a_2/l_2$, i.e., $a_1 = a_2/2$. $\"Description\"$

13. Masses $m_1$ and $m_2$ hang from a cord over a pulley on a prism with angles $\\alpha$ and $\\beta$ (see figure). Find their accelerations and cord tension. Neglect friction. $\"Description\"$

Answer

$a = \\dfrac{m_1 \\sin \\alpha - m_2 \\sin \\beta}{m_1 + m_2} g$; $T = \\dfrac{m_1 m_2 g (\\sin \\alpha + \\sin \\beta)}{m_1 + m_2}$.

Solution

Let's write the equations of Newton's second law for each of the weights, projecting the forces acting on the weights (see figure) onto the direction of possible motion of the corresponding weights, i.e., along the faces of the prism (assume that weight $m_1$ descends): $m_1 a = m_1 g \\sin \\alpha - T,$ $m_2 a = T - m_2 g \\sin \\beta.$ Solving this system, we find: $a = \\frac{m_1 \\sin \\alpha - m_2 \\sin \\beta}{m_1 + m_2} g;$ $T = \\frac{m_1 m_2 (\\sin \\alpha + \\sin \\beta)}{m_1 + m_2} g.$ If, upon substitution of the numerical values of masses and angles, it turns out that $a < 0$, it means that the entire system moves with acceleration $a$ in the direction opposite to the chosen one. The latter follows from the fact that in the absence of friction, changing the chosen direction of motion of the weights will only change the signs of the right sides of the equations derived above. $\"Description\"$

14. Two masses $m = 0.2 \\, \\text{kg}$ and $M = 4 \\, \\text{kg}$ are connected by a string and lie on a smooth table (friction is neglected). A force $F_1 = 0.2 \\, \\text{N}$ is applied to the first mass along the direction of the string, and a force $F_2 = 0.5 \\, \\text{N}$ is applied to the second mass in the opposite direction. With what acceleration $a$ will the masses move, and what is the tension force $T$ in the connecting string? Solve the problem in general form and draw a conclusion about the tension force in the string when $m \\ll M$.

Answer

$a = 0.12$ m/s$^2$; $T = 0.22$ H; $T = \\dfrac{F_1 M + F_2 m}{M+m} \\approx \\dfrac{F_1 + F_2 \\frac{m}{M}}{1 + \\frac{m}{M}} \\approx F_1$, when $m \\ll M$.

15. Find the accelerations $a_1$ and $a_2$ of masses $m_1$ and $m_2$ and cord tension $T$ in the system shown below. Neglect pulley and cord mass and friction. $\"Description\"$

Answer

$a_1 = 2 \\dfrac{2m_1 - m_2}{4m_1 + m_2} g$; $a_2 = \\dfrac{2m_1 - m_2}{4m_1 + m_2} g$; $T = \\dfrac{3m_1 m_2}{4m_1 + m_2} g$.

Hint

From the kinematic connection, it follows that the acceleration of weight $m_1$ is twice the acceleration of weight $m_2$ (when weight $m_1$ moves by distance $h$, weight $m_2$ moves by distance $h/2$). A force equal to $2T$ is applied to weight $m_2$ from the strings.

16. A cart with a liquid container moves horizontally at acceleration $a$. Find the angle of the liquid's surface to the horizon in steady state.

Answer

$\\alpha = \\arctan(a/g)$.

Solution

Let's conditionally isolate a small part of the liquid inside. Two forces act on this element of liquid with mass $m$: the force of gravity $mg$ and the buoyant (Archimedean) force $N$ from the side of the liquid. The resultant $F$ of these forces must be directed towards the acceleration of the vessel and equal to $ma$ (see figure). Therefore, the force $N$ must form an angle $\\alpha = \\arctan(a/g)$ with the vertical. The direction of the buoyant force $N$ acting on any isolated volume inside the liquid is perpendicular to the planes of equal pressure ($p$=const) in the liquid, is determined by them and, in particular, is perpendicular to its surface ($p=0$). The requirement of perpendicularity uniquely relates to the condition of rest of the liquid relative to the vessel. Any violation of perpendicularity is 'automatically' eliminated due to the fluidity of the liquid. $\"Description\"$

17. Find the angle of a liquid's surface in a container sliding without friction on an inclined plane.

Answer

The surface of the liquid is parallel to the inclined plane.

1st method. See solution to problem 16.

2nd method. Let's solve the problem in the reference frame associated with the moving vessel (in a non-inertial system). In this case, forces $mg$, $N$, and the inertial force $|-m\\vec{a}| = ma$ act on any element of the liquid, directed opposite to the motion of the vessel (see figure). The resultafignt of these forces must be zero (since the liquid is stationary relative to the vessel). This is possible when the force $N$ is perpendicular to the inclined plane. The surface of the liquid is parallel to it.
$\"Description\"$

18. A bucket of water with a floating body is in an elevator. Does the body's submersion depth change if the elevator accelerates upward or downward?

Answer

Will not change.

Solution

When the elevator moves with acceleration $\\vec{a}$, Newton's second law for a floating body is written as: $m\\vec{a} = m\\vec{g} + \\vec{F},\\tag{1}$ where $\\vec{F}$ is the buoyant force; $m$ is the mass of the body.

If the volume displaced by the body were occupied by liquid, a force, also equal to $\\vec{F}$, would act on it from the rest of the liquid. According to Newton's second law
$m_l \\vec{a} = m_l \\vec{g} + \\vec{F}; \\tag{2}$
here $m_l = \\rho_l V$, where $V$ is the volume of the submerged part of the body. Substituting the expression for $\\vec{F}$ found from equality (1) into equation (2), we get
$m(\\vec{a} - \\vec{g}) = \\rho_l V (\\vec{a} - \\vec{g}),$
hence it follows that $V = m/\\rho_l$ does not depend on the acceleration of the elevator, i.e., the immersion depth of the body does not change. Note. The solution to the problem becomes very simple in the reference frame associated with the elevator (see problem 17).

19. A board of mass $M$ can slide without friction on an incline at angle $\\alpha$. A dog of mass $m$ runs on the board. What direction and acceleration must the dog have to prevent the board from sliding? What minimum friction coefficient $k$ between the dog and board is needed?

Answer

$a = g \\sin \\alpha (1 + M/m)$; $k > (M/m) \\tan \\alpha$.

Solution

The dog must run with such acceleration, directed downwards, so that the force $-F_1$, with which the dog pushes the board back, balances the resultant $F_1$ of the forces acting on the board: gravity $Mg$ and reaction force $N_1$, i.e., equals $Mg \\sin \\alpha$ (see figure). On the dog, therefore, in the direction of its acceleration, act the force $mg \\sin \\alpha$ and the force $F_1$ from the board, equal, according to Newton's third law, to $Mg \\sin \\alpha$ (the direction of the dog's movement does not matter). The equation of motion for the dog will be $ma = mg \\sin \\alpha + Mg \\sin \\alpha$, whence $a = g \\sin \\alpha (1 + M/m)$. The force $-F_1$, with which the dog pushes the board, by its nature, is a friction force, which cannot exceed the value $kN$. The problem has a solution (the dog stops the board) if $kN \\ge F_1$, i.e., $kmg \\cos \\alpha \\ge Mg \\sin \\alpha$. Hence $k \\ge (M/m) \\tan \\alpha$.

1st method. See solution to problem 16.

20. A cylinder of mass $M$ has a wound thread. When released, the thread is pulled upward so the cylinder's center remains at the same height. Find the thread's tension.

Answer

$Mg$.

Hint

The cylinder is at rest relative to the Earth.

21. A rope over a fixed pulley has a mass $m_1 = 64\\, \\text{kg}$ on one end and a person $m_2 = 65\\, \\text{kg}$ on the other. The person pulls the rope to lift the mass while staying at the same height. How long $t$ until the mass is raised $h = 3\\, \\text{m}$? Neglect rope and pulley mass.

Answer

$t \\approx 6.26$ s.

Solution

Since the person is stationary relative to the Earth, the forces applied to him are balanced. Thus, the tension force is numerically equal to the weight of the person $m_2 g$. Therefore, for the weight, Newton's second law will be written as: $m_1 a = m_2 g - m_1 g = (m_2 - m_1)g$. Hence the acceleration of the weight $a = g \\frac{m_2 - m_1}{m_1}$ and is directed upwards. The time for the weight to rise is determined from the formula $h = at^2/2$: $t = \\sqrt{2h/a} = \\sqrt{\\frac{2hm_1}{g(m_2 - m_1)}}.$

22. 1. Two boys of equal mass on skates $l$ apart pull a rope: one at speed $v$, the other at $2v$. When and where will they meet? Solve also for mass ratio 1:1.5.

Two gymnasts of equal mass pull a rope over a pulley and begin ascend simultaneously: one at speed $v$, the other at $2v$ relative to the rope. How long until each reaches the pulley? The rope's length is $l$, initially centered.

Answer

1. Both boys will reach the middle of the distance between them simultaneously after time $\\tau = l/(3v)$; if the masses of the boys are unequal, the meeting place divides the distance between them inversely proportional to the masses of the boys.

The gymnasts will reach the block simultaneously after time $\\tau = l/(3v)$.

Solution

Since the forces applied to the boys are equal, according to Newton's third law, the accelerations, and consequently, the velocities of the boys relative to the Earth are equal. The length of the rope between the boys decreases with speed $3v$ and the boys will reach the middle of the distance between them after time $\\tau = l/(3v)$. If the masses of the boys are different, the accelerations and paths traveled by them relative to the Earth are inversely proportional to the masses of the boys.

Compare the solution of this problem with the solution of problem item 1.

23. A body of mass $m = 1\\, \\text{kg}$ on a horizontal plane with friction coefficient $k = 0.1$ is acted on by force $F$. Find the friction force for $F = 0.5\\, \\text{N}$ and $F = 2\\, \\text{N}$.

Answer

0.5 N; 1 N.

Hint

The magnitude of the static friction force cannot exceed $F_{fr.max} = kN$, where $N$ is the normal pressure. If the body is at rest, the friction force is equal in magnitude to the sum of the projections of all forces acting on the body onto the direction of possible motion (i.e., onto the direction in which the body would start moving if the friction force were zero). If the force acting on the body in the direction of possible motion exceeds $F_{fr.max}$, the body moves with acceleration, and the friction force is constant and equal to $F_{fr.max}$.

24. A block of mass $m = 2\\, \\text{kg}$ on a horizontal surface has sliding friction coefficient $k = 0.2$. Plot the friction force versus applied pulling force. Neglect static friction.

Answer

The graph is shown below. $\"Description\"$

25. A block is on a plane whose angle $\\alpha$ can vary from $0^\\circ$ to $90^\\circ$. Plot the friction force versus $\\alpha$. Neglect static friction.

Answer

The graph is shown below. $\"Description\"$

Hint

The motion of a body located on an inclined plane becomes possible if $mg \\sin \\alpha \\ge kmg \\cos \\alpha$, i.e., at an angle $\\alpha \\ge \\alpha_0 = \\arctan k$. If $\\alpha < \\alpha_0$, the body is at rest and the friction force is equal to $mg \\sin \\alpha$ (see problem 23). If $\\alpha \\ge \\alpha_0$, then $F_{\\text{fr}} = F_{\\text{fr.max}} = kmg \\cos \\alpha$.

26. Estimate the friction coefficient $k$ of sand on sand if the angle of a sandpile is $\\alpha$.

Answer

$k = \\tan \\alpha$.

27. A cord is thrown over a light, frictionless pulley. At one end of the cord, a mass $m_1$ is attached. On the other end, a ring of mass $m_2$ can slide along the cord (see figure).

With what acceleration $a$ does the ring move if the mass $m_1$ is stationary? What is the friction force $F_{\\text{fr}}$ between the ring and the cord?
The ring slides down with a constant acceleration $a_2$ relative to the cord. Find the acceleration $a_1$ of the mass $m_1$, and the friction force $F_{\\text{fr}}$ between the ring and the cord. The mass of the cord can be neglected; assume the mass $m_1$ is descending.

$\"Description\"$

Answer

1. $a = (1 - m_1/m_2)g$; $F_{\\text{fr}} = m_1 g$.

$a_1 = \\dfrac{m_1 g - m_2(g-a_2)}{m_1+m_2}$; $F_{\\text{fr}} = \\dfrac{m_1 m_2 (2g - a_2)}{m_1+m_2}$.

Hint

1. Since weight $m_2$ is stationary, the tension of the string, and consequently, the friction force of the ring on the string are equal to $m_2 g$. [Note: Error in original text, should be $m_1 g$ based on context].

The friction force is equal to the tension force of the string, and the acceleration of the ring relative to the block is $a_2 - a_1$. The equations of motion for the weights are expressed as follows: $m_1 a_1 = m_1 g - F_{\\text{fr}};$ $m_2(a_2 - a_1) = m_2 g - F_{\\text{fr}}.$

28. A body of mass $M$ moves with acceleration $a$ on a horizontal plane under force $F$ at angle $\\alpha$. Find $F$ if the friction coefficient is $k$.

Answer

$F = \\dfrac{M(kg+a)}{\\cos \\alpha + k \\sin \\alpha}$.

Hint

The normal pressure force is $Mg - F \\sin \\alpha$ (see figure, a) or $Mg + F \\sin \\alpha$ (see figure, b) depending on the direction of action of force $\\vec{F}$. $\"Description\"$

29. Board $A$ moves on a table under tension from a cord attached to board $B$ hanging over a pulley.

Find the tension $T$ if $m_A = 200, \\text{g}$, $m_B = 300, \\text{g}$, and $k = 0.25$. Neglect pulley mass.
How does the answer change if the boards are swapped? 3. Find the force $F$ on the pulley's axle in both cases.

Answer

1. $T = g \\frac{m_1 m_2 (1+k)}{m_1+m_2} = 1.47$ H. 2. Does not change. 3. $F = 2.08$ H.

30. Two masses $m_1$ and $m_2$ (see figure) are in an elevator ascending at $a$. Find the cord tension $T$ if $m_1s friction coefficient is $k$. Will their motion change if the elevator's acceleration reverses? $\"Description\"$

Answer

$T = \\frac{m_1 m_2 (g+a)(1+k)}{m_1+m_2}$ for $km_1 < m_2$; $T = m_2(g+a)$ for $km_1 > m_2$. The state of motion (rest) does not change.

Solution

The accelerations of the weights relative to the table are equal in magnitude (weights are connected by an inextensible string): $a'_1 = a'_2 = a'$. In a stationary reference frame, the acceleration of weight $m_2$ is directed vertically, and $a_2 = a' - a$. The acceleration of weight $m_1$ has vertical and horizontal components: $a_{1v} = a$, $a_{1h} = a'$. Applying Newton's second law, we get the system of equations: $m_2(a'-a) = m_2 g - T$, $m_1 a' = T - kN$, $m_1 a = N - m_1 g$, where $T$ is the string tension; $N$ is the reaction force of the table. Solving this system, we find the answer. Note. This system of equations can be represented as $m_2 a' = m_2(g+a) - T$, $m_1 a' = T - kN$, $N - m_1(g+a) = 0$. Thus, the upward movement of the elevator with acceleration $a$ is equivalent to a reference frame in which the acceleration of free fall is $g+a$. If $a'=0$ (for $km_1 > m_2$), the answer follows from the equation $m_2 a = T - m_2 g$. In this case, the friction force is no longer equal to $kN$ and the second equation is incorrect. From this note, the answer to the second question of the problem immediately follows.

31. Masses $M_1 = 3\\, \\text{kg}$ and $M_2 = 5\\, \\text{kg}$ on a smooth table are connected by a cord that breaks at $T = 24\\, \\text{N}$. What maximum force $F$ can be applied to $M_1$? To $M_2$? How does friction affect the answer?

Answer

$F_1 = 38.4$ H; $F_2 = 64$ H; does not change.

Solution

Let's write the equations of motion for the bodies (see figure): $F_1 - T = M_1 a$, (1) $T = M_2 a$. (2) From equations (1) and (2) we find $F_1 = T \\frac{M_1+M_2}{M_2}$. If the force is applied to the second weight, then $F_2 = T \\frac{M_1+M_2}{M_1}$. If there is friction, the equations of motion for the weights will be: $F_1 - kM_1 g - T = M_1 a$; (3) $T - kM_2 g = M_2 a$. (4) Solving equations (3) and (4) together, we get $F_1 = T \\frac{M_1+M_2}{M_2}$, i.e., the presence of friction does not change the result if the coefficient of friction is the same. $\"Description\"$

32. $n + 1$ masses $m$ are connected by $n$ massless springs (see figure). Force $F$ accelerates the system horizontally. Find $F$ and each spring's elongation if the friction coefficient is $f$ and the spring constant is $k$. $\"Description\"$

Answer

$F = (n+1)m(a+fg)$; $X_1 = m l \\frac{a+fg}{k}$.

33. Two blocks of masses $m_1$ and $m_2$ are connected by a cord on a horizontal plane. Forces $F_1$ and $F_2$ act at angles $\\alpha$ and $\\beta$ (see figure). Find the system's acceleration $a$ and cord tension $T$. The friction coefficient is $k$ for both blocks. The blocks remain in contact with the plane. $\"Description\"$

Answer

$a = \\frac{F_1(\\cos \\alpha + k \\sin \\alpha) - F_2(\\cos \\beta - k \\sin \\beta)}{m_1+m_2} - gk$; $T = \\frac{F_1 m_2 (\\cos \\alpha + k \\sin \\alpha) + F_2 m_1 (\\cos \\beta - k \\sin \\beta)}{m_1+m_2}$.

Hint

See problem 28.

34. 1. Blocks $A$ and $B$ with masses $m_1$ and $m_1$ are placed on a table (figure, a). A force $\\vec{F}$ is applied to block $B$, directed at an angle $\\alpha$ to the horizontal. Find the accelerations of the blocks' motion if the coefficients of friction between the blocks and between the block and the table are equal to $k_1$ and $k_2$ respectively. The friction force between the surfaces is at its maximum value.

A flat plate with mass $m_2$ is placed on an inclined plane with angle $\\alpha$, and on top of it is a block with mass $m_1$. The coefficient of friction between the block and the plate is $k_1$. Determine for which values of the coefficient of friction $k_2$ between the plate and the inclined plane the plate will not move, if it is known that the block slides on the plate (figure, b).

$\"Description\"$

Answer

1. $a_B = \\dfrac{F(\\cos \\alpha + k_1 \\sin \\alpha)}{m_1} - k_1 g$; $a_A = \\dfrac{(k_1-k_2)(m_1 g - F \\sin \\alpha)}{m_2} - k_2 g$.

$k_2 \\ge \\dfrac{m_1 k_1 + m_2 \\tan \\alpha}{m_1+m_2}$.

Hint

1. The pressure forces of block B on block A and block A on the table are respectively $m_1 g - F \\sin \\alpha$ and $(m_1+m_2)g - F \\sin \\alpha.$.

Since block B slides on the plate, two forces act on the plate from block B: the normal pressure force $N_1 = m_1 g \\cos \\alpha$ and the friction force $F_{fr1} = k_1 N_1 = k_1 m_1 g \\cos \\alpha.$ The normal pressure force of the plate on the inclined plane is $N_2 = N_1 + m_2 g \\cos \\alpha = (m_1+m_2)g \\cos \\alpha.$ The plate does not slide on the inclined plane if $m_2 g \\sin \\alpha + F_{fr1} \\le F_{fr2} = k_2 N_2,$ i.e., if $m_2 g \\sin \\alpha + k_1 m_1 g \\cos \\alpha \\le k_2 (m_1+m_2)g \\cos \\alpha$ or
$k_2 \\ge k_1 \\frac{m_1}{m_1+m_2} + \\frac{m_2}{m_1+m_2} \\tan \\alpha = \\frac{m_1 k_1 + m_2 \\tan \\alpha}{m_1+m_2}$.

35. A body is thrown vertically upward. What is its acceleration at the peak? How does the acceleration change during motion? Consider:

no air resistance;
air resistance increases with speed.

Answer

1) The acceleration of the body at all times during its motion is constant and equal to $g$;

In accordance with Newton's second law $a = (mg + F_{\\text{res}})/m$. When moving upwards, the resistance force is directed, like gravity, downwards and decreases as the body ascends (since the speed of the body decreases), during descent, the resistance force is directed upwards and increases. Therefore, the acceleration of the body at the beginning of the motion is maximum (and greater than $g$), decreases during ascent and becomes equal to $g$ at the highest point of the trajectory, then continues to decrease during descent and may even become zero.

36. Two balls fall in air. One has twice the diameter of the other. What is their speed ratio at terminal velocity? Assume drag is proportional to cross-sectional area and velocity squared.

Answer

The speed of the large ball will be $\\sqrt{2}$ times greater than the speed of the smaller ball.

Solution

Under steady motion, the air resistance force $F_{\\text{res}}$ and the gravity force $Mg$ acting on the ball are equal, but $F_{\\text{res}} = kv^2 (\\pi D^2 / 4),$ and $M = \\pi \\rho D^3 / 6,$ where $\\rho$ is the density of the ball; $D$ is its diameter. Equating $F_{\\text{res}}$ and $Mg$, we find that $v \\sim \\sqrt{D}$.

37. A ball of mass $m$ falls in a liquid (density $\\rho$) at terminal speed $v$. What force is needed to pull it upward at $2v$? The ball's volume is $V$, and drag is proportional to speed.

Answer

$3g(m - \\rho V)$.

38. Why do large raindrops fall faster than small ones?

Answer

See solutions to problems 35 and 36.

39. A ball rises at constant speed in a liquid four times denser than the ball. Find the drag force if the ball's mass is 10 g.

Answer

0.29 N.

40. A body lies on a $4^\\circ$ incline. Find:

a) the minimum friction coefficient $k$ for sliding to begin;

b) the acceleration $a$ if $k = 0.03$;

c) the time $t$ to slide 100 m;

d) the speed $v$ at the end.

Answer

a) $k=0.07$;

b) $a \\approx 0.39$ m/s$^2$;

c) $t = 22.7$ s;

d) $v = 8.85$ m/s.

41. How long $t$ does it take for a body to slide down an incline of height $h$ and angle $\\alpha$, if it moves uniformly on an incline of angle $\\beta$?

Answer

$t = \\frac{1}{\\sin \\alpha} \\sqrt{\\frac{2h}{g(1 - \\cot \\alpha \\tan \\beta)}}$.

42. Two blocks of equal mass $m$ are connected by a cord on an incline $\\alpha$. The upper block's friction coefficient is twice the lower's. Find the cord tension $T$ during motion.

Answer

$T = (1/4) kmg \\cos \\alpha$.

Solution

On the upper weight along the inclined plane act the component of gravity $F = mg \\sin \\alpha$, the string tension force $T$, and the friction force $F_{\\text{fr}} = kmg \\cos \\alpha$. According to Newton's second law, we get $ma = mg \\sin \\alpha - kmg \\cos \\alpha + T$. For the lower weight, we have accordingly $ma = mg \\sin \\alpha - (k/2)mg \\cos \\alpha - T$ (the weight moves with the same acceleration, and the coefficient of friction between it and the inclined plane is half as much). Subtracting the second from the first equation, we find the string tension force $T = (1/4) kmg \\cos \\alpha$.

43. A mass $m_1 = 2\\, \\text{kg}$ on an incline $\\alpha = 20^\\circ$ is connected over a pulley to a hanging mass $m_2 = 1\\, \\text{kg}$. The friction coefficient is $k = 0.1$. Find the acceleration $a$ and cord tension $T$. Neglect pulley and cord mass.

Answer

$a = g \\dfrac{m_2 - m_1(\\sin \\alpha + k \\cos \\alpha)}{m_1+m_2} \\approx 0.42$ m/s$^2$; $T = g \\dfrac{m_1 m_2}{m_1+m_2} (1 + \\sin \\alpha + k \\cos \\alpha) \\approx 9.36$ H.

44. A weightless pulley is fixed at the top of two inclined planes forming angles $\\alpha = 30^\\circ$ and $\\beta = 45^\\circ$ with the horizontal. Weights of equal mass $m_1 = m_2 = 1\\,\\text{kg}$ are connected by a thread passing over the pulley. Find:

a) the acceleration $a$ with which the weights move;

b) the tension force $T$ in the thread.

The coefficient of friction between the weights and the inclined planes is $k = 0.1$ (see figure). Neglect friction in the pulley, as well as the mass of the thread and its elasticity.

$\"Description\"$

Answer

a) $a \\approx 0.24$ m/s$^2$;

b) $T = 6$ H.

Hint

According to Newton's second law $m_1 g \\sin \\beta - km_1 g \\cos \\beta - T = m_1 a$; (1) $-m_2 g \\sin \\alpha - km_2 g \\cos \\alpha + T = m_2 a$. (2) Solving equations (1) and (2) together, we find $a$ and $T$.

45. An icy slope has angle $\\alpha = 10^\\circ$. A stone is launched upward, slides back down, and takes twice as long to descend as to ascend. Find the friction coefficient $k$.

Answer

$k = \\tan \\alpha \\frac{n^2-1}{n^2+1} = 0.1$.

Hint

From the equations of motion of the stone up and down, find that its accelerations are respectively equal: $a_1 = g(\\sin \\alpha + k \\cos \\alpha)$; $a_2 = g(\\sin \\alpha - k \\cos \\alpha)$. (1) (2) The time of motion of the stone $t_1 = \\sqrt{2l/a_1}$; this follows from two equations: $l = v_0 t_1 - a_1 t_1^2 / 2$, $v_0 - a_1 t_1 = 0$, and $t_2 = \\sqrt{2l/a_2}$, where $l$ is the distance to the highest point of ascent of the stone. Hence $t_1/t_2 = \\sqrt{a_2/a_1} = 1/n$ (according to the condition). (3) Solving equations (1), (2), and (3) together, find the value of $k$.

46. A block is on a horizontal board with friction coefficient $k = 0.1$. What horizontal acceleration $a$ must the board have for the block to slide off?

Answer

$a > kg = 0.1g$.

Hint

The force acting on the weight from the side of the board cannot be greater than the maximum static friction force.

47. A cylinder (height 20 cm, diameter 2 cm) stands on a sheet of paper. What minimum acceleration must the sheet have to make the cylinder topple? Assume no slipping.

Answer

$a > dg/h = 0.1g$.

Solution

1st method. The acceleration $a$ of the body is provided by the resultant of the force $F$ and the force of gravity $P=mg$ (see figure). The force $F$ is created by the normal pressure force $N$, applied in the limiting case (before tipping) at point C, and the friction force $F_{\\text{fr}}$. If the acceleration is greater than the limiting value, the corresponding force $F$ cannot be provided, since for this the force $N$ (which is the resultant of the distributed pressure forces on the support) would have to be applied outside the support of the body. From the similarity of triangles OAB and COD $ma_{\\text{lim}}/(mg) = d/h$, where $a_{\\text{lim}}$ is the limiting acceleration at which the body does not yet tip over. Hence $a_{\\text{lim}} = dg/h$; for tipping the body $a > a_{\\text{lim}}$, i.e., $a > dg/h$.

2nd method. In the non-inertial reference frame associated with the sheet of paper, the force $F = -ma_{\\text{lim}}$ (see solution to problem 17) and the force of gravity $P=mg$ act on the body at the moment of detachment. At $a_{lim}$, the sum of the moments of these forces relative to point C is zero, i.e., $-ma_{\\text{lim}} h/2 + mg d/2 = 0$, hence $a_{\\text{lim}} = dg/h$; for tipping $a > dg/h$.
$\"Description\"$

48. A cart of mass $M$ moves without friction on a horizontal track. A block of mass $m$ lies at its rear. The friction coefficient between block and cart is $k$. A horizontal force $F$ makes the block slide. How long $t$ until the block falls off if the cart's length is $l$ (see figure)? What is the minimum force $F_0$ to start sliding? $\"Description\"$

Answer

$t = \\sqrt{\\dfrac{2l}{F/m - kg(1+m/M)}}$; $F_0 = kmg(1+m/M)$.

Solution

A force $F$ and a friction force $kmg$ act on the block. According to Newton's second law, its acceleration $a_1$ is determined by the equation $m a_1 = F - kmg$. A force $kmg$ acts on the cart from the side of the block, according to Newton's third law. Its acceleration $a_2$ is found from the equation $M a_2 = kmg$. The block slides relative to the cart with acceleration $a = a_1 - a_2 = F/m - kg(1+m/M)$. It travels the length $l$ of the cart in time $t = \\sqrt{2l/a}$. The minimum value $F_0$ of force $F$ is determined by the condition $a=0$.

49. A block of mass $m$ slides at $v_0$ on a long board of mass $M$ held by a cord (see figure). At $t = 0$, after the block has traveled $l_0$, the cord is cut. Describe the subsequent motion. When $t_1$ does the friction force between them change? What minimum board length $l$ prevents the block from sliding off? $\"Description\"$

Answer

$t_1 = v_0 M / F$; $F_{fr1} = F$, $F_{fr2} = FM/(M+m)$; $l = l_0 + Mv_0^2 / (2F)$.

Solution

From the condition of constant velocity of the block relative to the stationary board, it follows that the friction force between them is maximum and equal to $F$. After cutting the cord, the block does not change its state of motion (sliding and, consequently, the equality to zero of the sum of forces acting on it is preserved). The board, however, begins to move with uniform acceleration under the action of the force $F_{\\text{fr}1} = F$: $a_1 = F/M$ until some moment $t_1 = v_0/a_1 = Mv_0/F$. At moment $t_1$, the speed of the board equals the speed of the block, its sliding stops, and further they will move as a single unit with acceleration $a_2 = F/(M+m)$, and the static friction force $F_{fr2}$ will become equal to $F - ma_2 = FM/(M+m)$. The block travels a path $v_0 t_1 - a_1 t_1^2 / 2 = Mv_0^2 / (2F)$ along the board during time $t_1$. Thus, the minimum required length of the board, at which the block still manages to stop on it (neglecting the size of the block), is $l = l_0 + Mv_0^2 / (2F)$. Plot the velocity graph for both bodies.

50. The last car (mass $m$) detaches from a train (mass $M$) moving at constant speed. The car travels distance $s$ and stops. How far $l$ is the train from the car when it stops? Assume constant engine thrust and drag proportional to weight.

Answer

$l = \\dfrac{M}{M-m} s$.

Solution

After the wagon detaches, the train will move with acceleration imparted to it by a force equal to $F_{thrust} - F_{\\text{res}} = F_{thrust} - (F_{res.init} - kmg) = kmg$ (the initial resistance $F_{res.init}$ was equal to the thrust force of the locomotive); $a_1 = kmg/(M-m)$. The wagon will move with uniform deceleration with acceleration $a_2 = -kg$. Relative to each other, the wagon and the locomotive move with acceleration $a_{rel} = a_1 - a_2 = \\frac{kmg}{M-m} + kg = kg \\frac{M}{M-m}$. The distance $l$ between them will relate to the path $s$ traveled by the wagon until stopping as $a_{rel}/|a_2|$. Another solution see in problem 40 in Chapter: Work, Power, Energy .

","chapter_express_content":"","public_visibility":true,"allow_part_see":false,"book_slug":"physics-problems-and-questions-mechanics","slug":"newtons-laws","regular_authors":"","express_authors":"","created_at":"2025-04-26T12:42:36.598Z","__v":0,"header_data":[]},{"_id":"680e6671797791ed81d77917","book_id":"680e6593797791ed81d778ae","chapter_name":"Momentum. Law of Conservation of Momentum","sequence":3,"chapter_content":"

1. A ball of mass $m = 100$ g, moving at a speed of $v = 1$ m/s, elastically collides with a plane. Determine the change in the ball's momentum if the angle between the velocity and the plane is $\\alpha$, equal to: a) $90°$; b) $30°$.

Answer

a) $\\Delta p = 2mv = 0.2 \\text{ N} \\cdot \\text{s}$ (see figure, a in Solution);

b) $|\\Delta p| = 2mv \\sin \\alpha = 0.1 \\text{ N} \\cdot \\text{s}$.

Solution

b) The impulse of the ball can be resolved into two components: perpendicular to the plane and parallel to it (see figure, b). The latter does not change during the impact of the ball (since no forces act on the ball in this direction), while the normal component changes its direction to the opposite, maintaining its numerical value. Therefore, the ball reflects from the plane at the same angle at which it falls onto it. The change in the ball's impulse, as seen from the figure, is $2mv \\sin \\alpha$. $\"Description\"$

2. A ball of mass $m = 10$ g falls onto a horizontal plane from a height of $h_1 = 27$ cm. Find the average impact force $F_{cp}$ in the following cases:

a) the ball is made of clay (perfectly inelastic collision);

b) the ball and the plane are made of steel (perfectly elastic collision);

c) the ball is made of plastic and rises to a height of $h_2 = 12$ cm after the impact.

Consider the first two cases of the ball hitting a plane inclined at an angle of $\\alpha = 30°$. Assume in all cases that the contact between the ball and the plane lasts (the duration of the impact) 0.03 s.

Answer

a) $F_{avg} = F'_{avg} = m\\sqrt{2gh_1}/\\Delta t + mg \\approx 0.87$ H;

b) $F_{avg} = (2m\\sqrt{2gh_1}/\\Delta t + mg) \\cos \\alpha \\approx 1.43$ H;

c) $F_{avg} = m\\sqrt{2g}(\\sqrt{h_1} + \\sqrt{h_2})/\\Delta t + mg \\approx 1.38$ H.

Hint

Use Newton's second law in the form $\\Delta(m\\vec{v}) = \\vec{f} \\Delta t,$ where $\\vec{f} = \\vec{f}_p + m\\vec{g}.$ Remember that impulse is a vector quantity, so in the third case $|\\Delta(m\\vec{v})| = mv_1 + mv_2.$ The change in the ball's impulse is directed upwards. According to Newton's third law, a force equal to $-\\Delta(m\\vec{v})/t$ and directed downwards will act on the plane during the impact.

3. Two particles with masses $m$ and $2m$ move in mutually perpendicular directions with velocities $2v$ and $v$. The same force begins to act on the particles. Determine the magnitude and direction of the velocity of the particle with mass $2m$ at the moment when the velocity of the particle with mass $m$ becomes as shown by the dashed line:

a) in figure, a;

b) in figure, b.

$\"Description\"$

Answer

a) $v\\sqrt{5}$ at an angle $\\alpha = \\arctan 2$ to the initial direction;

b) $v\\sqrt{5}$ at an angle $\\beta = \\arcsin(1/\\sqrt{5})$ to the initial direction.

Solution

The change in the impulses of the particles is the same: the same forces acted on them for the same amount of time. In case a), the magnitude of the impulse change of the first particle is $2mv - (-2mv) = 4mv.$ The vector $\\Delta \\vec{p}$ is directed horizontally (figure, a). Therefore, the impulse of the second particle will be $\\sqrt{(4mv)^2 + (2mv)^2} = mv\\sqrt{20},$ and the speed is $v\\sqrt{5}$, angle $\\alpha = \\arctan(4mv/2mv) = \\arctan 2.$ Similarly, we find that in case b) the magnitude of the impulse change of the first particle is $2mv\\sqrt{2}$ (figure, b). The impulse of the second particle changes similarly. Its magnitude will become $mv\\sqrt{20}$ (this can be easily found using the cosine theorem). Its speed will accordingly be $v\\sqrt{5}$; angle $\\beta = \\arcsin(1/\\sqrt{5})$ (using the sine theorem). $\"Description\"$

4. A water jet with a cross-sectional area of $S = 6$ cm² strikes a wall at an angle of $\\alpha = 60°$ to the normal and bounces off elastically (without loss of speed). Find the force $F$ acting on the wall, if it is known that the water flow speed in the jet is $v = 12$ m/s.

Answer

$F = 2\\rho Sv^2 \\cos \\alpha = 86.4$ N.

Hint

Note that in time $t$, the mass of water hitting the wall is the mass contained in a cylinder of length $l=vt$ and cross-section $S$, i.e., $m = \\rho Svt$ ($\\rho$ is the density of water).

5. A railway platform with a gun mounted on it moves at a speed of $v_1 = 9$ km/h. The total mass is $M = 20$ t. A projectile of mass $m = 25$ kg is fired from the gun at a speed of $v_2 = 700$ m/s relative to the center of mass. Determine the speed of the platform after the shot: a) when the shot is fired in the direction of the platform's movement; b) when the shot is fired in the opposite direction. Neglect the friction of the platform against the rails.

Answer

a) $u = v_1 - \\dfrac{m}{M-m} v_2 \\approx 1.62$ m/s;

b) $u = v_1 + \\dfrac{m}{M-m} v_2 \\approx 3.39$ m/s.

Hint

Since the change in the impulses of the platform and the projectile occurred due to the action of internal forces, the total change in their impulses should be zero.

6. A grenade flying at a speed of 10 m/s explodes into two fragments. The larger fragment, whose mass was 60% of the total mass of the grenade, continued to move in the same direction, but with an increased speed of 25 m/s. Find the speed of the smaller fragment.

Answer

12.5 m/s and will move in the direction opposite to the initial one.

7. A projectile at the highest point of its trajectory at a height of $h = 100$ m explodes into two parts: $m_1 = 1$ kg and $m_2 = 1.5$ kg. The projectile's speed at this point is $v_0 = 100$ m/s. The speed of the larger fragment $v_2$ turned out to be horizontal, coinciding in direction with $v_0$ and equal to 250 m/s. Determine the distance $s$ between the points where both fragments fall. Do not consider air resistance.

Answer

$s \\approx 1695$ m.

Hint

From the law of conservation of momentum $(m_1 + m_2)v_0 = m_1 v_1 + m_2 v_2$, find $v_1 = \\frac{(m_1 + m_2)v_0 - m_2 v_2}{m_1} = -125$ m/s. Consequently, $s = (v_1 + v_2)/2 \\cdot \\sqrt{2h/g} \\approx 1695$ m.

8. A projectile fired from a gun at some angle to the horizon, at the highest point of its parabolic trajectory, explodes into two fragments of equal mass. One fragment returns to the gun along the previous trajectory after the explosion. Where will the second fragment fall? Will both fragments fall to the ground at the same time? Do not consider air resistance.

Answer

The second fragment will fall to Earth twice as far as the projectile would have fallen. They will fall simultaneously.

Hint

The fragment of the projectile that returned to the initial point obviously acquired a velocity equal in magnitude to the velocity of the projectile at the highest point of the trajectory, but opposite in direction, during the explosion. From the law of conservation of momentum, it follows that the second fragment after the explosion will have a velocity equal to three times the velocity of the projectile at the highest point.

9. A projectile explodes at the highest point of its trajectory at a height of $h = 19.6$ m into two identical parts. After a time $\\tau = 1$ s after the explosion, one part falls to the Earth under the place where the explosion occurred. At what distance $s_2$ from the firing point will the second part of the projectile fall, if the first one fell at a distance of $s_1 = 1000$ m? Do not consider air resistance when solving the problem.

Answer

$s_2 = 5000$ m.

Solution

The law of conservation of momentum in this case is conveniently written for the projections of the impulses on the x and y axes (see figure) (since the time of rupture is short, the impulse of the gravity force can be neglected): $2mv_x = mv_{1x} + mv_{2x},\\quad 2mv_y = mv_{1y} + mv_{2y},$ where $m$ is the mass of the fragment; $v_x, v_{1x}, v_{2x}, v_y, v_{1y}, v_{2y}$ are the projections of the velocities of the projectile and fragments on the x and y axes, respectively. Considering that the first fragment fell under the place of the projectile's rupture and therefore $v_{1x} = 0$, and also that the explosion occurred at the highest point of the trajectory and $v_y = 0$, we get: $v_{2x} = 2v_x,\\quad v_{2y} = -v_{1y}.$ Since the projectile ruptured at the highest point of the trajectory $h = 19.6$ m, before the rupture it moved for $t = \\sqrt{2h/g} = 2$ s, which means $v_x = s_1/t = 500$ m/s. Let's now write the equations for the coordinates of the first and second fragments: $x_1 = 0,\\qquad y_1 = v_{1y}t - gt^2/2;$ $x_2 = v_{2x}t,\\qquad y_2 = v_{2y}t - gt^2/2.$ Substituting $t=1$ s, $y_1 = y_2 = -h = -19.6$ m into the second and fourth equations and considering that $v_{2y} = -v_{1y}$, we find $t_2$ and then $x_2$. The distance from the firing point to the landing point of the second fragment $s_2 = s_1 + x_2$. $\"Description\"$

10. From a two-stage rocket with a total mass of $M = 1000$ kg at the moment of reaching a speed of $v_0 = 171$ m/s, its second stage with a mass of $m = 400$ kg separated, the speed of which increased to $v_2 = 185$ m/s. Find with what speed $v_1$ the first stage of the rocket began to move. The speeds are indicated relative to an observer on Earth.

Answer

$v_1 = \\dfrac{Mv_0 - mv_2}{M-m} \\approx 161.7$ m/s.

11. A spacecraft flies at a constant speed in a cloud of stationary micrometeoroid particles, which experience absolutely inelastic collisions with it. How many times do you need to increase the engine thrust so that:

a) the speed of the ship increases by a factor of two?

b) when entering a region of particles with a density three times greater, the speed of the ship does not change?

Answer

a) 4 times; b) 3 times.

Hint

The mass of particles entrained per unit time is $\\dot{m} = m_0 n S v,$ where $m_0$ is the mass of one particle; $n$ is their number per unit volume (density); $S$ is the cross-sectional area of the spacecraft; $v$ is the speed of the spacecraft. Since the collisions are inelastic, the impulse imparted to the entrained mass per unit time is $m_0 n S v^2.$ According to Newton's third law, the force $F$ acting on the spacecraft is equal to the change in impulse of the mass interacting with it per unit time: $F = m_0 n S v^2;$ a) if $v$ increases 2 times, $F$ increases 4 times; b) if $n$ increases 3 times, $F$ increases by the same factor.

12. The third stage of the rocket consists of a carrier rocket with a mass of $m_p = 500$ kg and a head cone $m_k = 10$ kg. A compressed spring is placed between them. During tests on Earth, the spring imparted a speed of $v_{отн} = 5.1$ m/s to the cone relative to the carrier rocket freely suspended in a horizontal position. What will be the speeds of the cone $v_k$ and the rocket $v_p$ if their separation occurs in orbit when moving at a speed of $v = 8000$ m/s?

Answer

$v_k = 8005$ m/s; $v_p = 7999.9$ m/s.

Solution

Let's write the law of conservation of momentum in a reference frame moving with velocity $v$ in the direction of the rocket's flight. In this reference frame, before the cone separation, the rocket was at rest: $m_p \\vec{v}_p' + m_k \\vec{v}_k' = 0$ (here velocities $\\vec{v}_p'$ and $\\vec{v}_k'$ are algebraic quantities). According to the condition $v_k' = v_{rel} + v_p'$. Solving these equations together for $v_p'$, we find $v_p' = -m_k v_{rel} / (m_p + m_k)$. The velocity of the cone relative to the Earth is $v_k = v + v_k'$. The velocity of the rocket will be $v_p = v_k - v_{rel}$.

13. A homogeneous rod of length $l$ touches the smooth horizontal surface with its lower end. The upper end of the rod is suspended on a thread so that the rod forms an angle $\\alpha$ with the horizontal plane. The thread is cut. In which direction and by how much will the lower end of the rod shift when it falls?

Answer

$\\Delta x = l \\sin^2(\\alpha/2)$.

Hint

Since no forces act on the rod in the horizontal direction, its center of mass will move vertically.

14. Two balls with masses $m_1$ and $m_2$ are connected by a non-stretchable weightless horizontal rod. At the initial moment, the balls are given speeds $v_1$ and $v_2$ directed at angles $\\alpha$ and $\\beta$ to the horizon (see figure). What relationship should exist between the angles $\\alpha$ and $\\beta$ so that these speeds can be imparted to the balls? What is the nature of the system's movement? To what height will the center of mass of the system rise? $\"Description\"$

Answer

From the condition of inextensibility of the rod, it follows that $v_1 \\cos \\alpha = v_2 \\cos \\beta$, or $\\cos \\alpha / \\cos \\beta = v_2 / v_1$; the system will perform complex motion: rotate around the center of mass, which will simultaneously move along the trajectory along which a material point thrown at an angle to the horizon with an initial velocity $v_0 = \\frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$ would move; the components of this velocity: horizontal $v_{0x} = v_1 \\cos \\alpha = v_2 \\cos \\beta,$ vertical $v_{0y} = \\frac{m_1 v_1 \\sin \\alpha + m_2 v_2 \\sin \\beta}{m_1 + m_2};$ the height $h$ to which the center of mass of the system will rise is found from the law of conservation of energy $ \\frac{1}{2}mv_{0y}^2 = mgh,$ whence $h = \\frac{(m_1 v_1 \\sin \\alpha + m_2 v_2 \\sin \\beta)^2}{2g(m_1 + m_2)^2}.$

15. A person with a mass of $m = 70$ kg is on the stern of a boat located on a lake. The length of the boat is $l = 5$ m and its mass is $M = 280$ kg. The person moves to the bow of the boat. By what distance will the person move relative to the bottom?

Answer

The person will move a distance $x_1 = \\frac{M}{M+m} l = 4$ m, and the boat a distance $x_2 = \\frac{m}{M+m} l = 1$ m.

Solution

1st method. The impulse of an isolated system is constant, and its center remains at rest or maintains its velocity unchanged. Therefore, the position of the center of mass of the boat-person system $x_{cm}$ in the coordinate system associated (see figure) with the water should not change during the person's movement: $x_{cm} = \\frac{Ma + ml}{M+m} = \\frac{M(a+x_2) + mx_1}{M+m}$ ($x_2$ is the coordinate of the boat's nose after the person moves to the nose; $a$ is the coordinate of the boat's center of mass before the movement; $m$ is the mass of the person; $M$ is the mass of the boat). From this, we find that the boat moves a distance $x_2 = \\frac{m}{M+m} l$ relative to the bottom, and the person a distance $x_1 = l - x_2 = \\frac{M}{M+m} l$ relative to the bottom.

n2nd method. Denoting by $v$ the velocity of the person relative to the boat, and $u$ the velocity of the boat relative to the water, we can write: $m(v-u) = Mu$ [($v-u$) is the velocity of the person relative to the bottom], whence $u = \\frac{m}{M+m} v.$ Considering that the paths traveled by the person and the boat are proportional to their velocities, we find the displacement of the boat relative to the bottom
$x_2 = \\frac{m}{M+m} l,$ and the person $x_1 = \\frac{M}{M+m} l.$
$\"Description\"$

16. A boat is stationary on a lake. Fishermen are sitting on the stern and bow of the boat at a distance of $l = 5$ m from each other. The mass of the boat is $M = 150$ kg, the masses of the fishermen are $m_1 = 90$ kg and $m_2 = 60$ kg. The fishermen change places. How much will the boat move in this case? Neglect water resistance.

Answer

$x = (m_1 - m_2)l / (m_1 + m_2 + M) = 0.5$ m.

Hint

See the solution to problem 3.15, 1st method.

17. Three boats of the same mass $M$ are going in a line (one after the other) at the same speed $v$. From the middle boat, loads of mass $m$ are simultaneously thrown into the front and rear boats at a speed $u$ relative to the boat. What will be the speeds of the boats after the loads are thrown? Neglect water resistance.

Answer

$v_1 = \\dfrac{m(v+u) + Mu}{M+m}$; $v_2 = u$; $v_3 = \\dfrac{m(u-v) + Mu}{M+m}$.

Solution

Let's solve the problem in a reference frame moving with velocity $u$ relative to the shore. Obviously, in this reference frame, the boats before the cargo transfer are stationary. From the law of conservation of momentum, it follows: $mu = (M+m)v'_1,$ therefore the velocity of the front boat $v'_1 = mu/(m+M);\\qquad 0 = (M+m)v'_2,$ for the middle boat $v'_2 = 0;\\quad -mu = (M+m)v'_3,$ for the rear boat $v'_3 = -mu/(m+M).$ The velocity $v_1$ of the first boat relative to the Earth is obtained from the formula $v_1 = v'_1 + u,$ similarly we find the velocities of the 2nd and 3rd boats.

18. A trolley with a mass of $M = 120$ kg moves along rails without friction at a speed of $v = 6$ m/s. A person with a mass of $m = 80$ kg jumps off the trolley at an angle of $\\alpha = 30°$ to the direction of its movement in the horizontal plane. The speed of the trolley decreases to $v' = 5$ m/s. What was the person's speed $u$ relative to the ground during the jump?

Answer

$u = \\frac{(M+m)v - Mv'}{m \\cos \\alpha} \\approx 8.6$ m/s.

Solution

The law of conservation of momentum can only be applied for the components of the impulses along the direction of motion, since external forces do not act on the system in this direction: $(M+m)v = Mv' + mu \\cos \\alpha.$ Hence $u = \\frac{(M+m)v - Mv'}{m \\cos \\alpha}.$

19. Two pipes with cross-sections $S_1$ and $S_2$ are connected to each other, filled with explosive gas, and closed with pistons of masses $m$ and $M$ (see figure). After the explosion, the pistons fly out of the pipes. The first one flew out at a speed of $v_1$. At what speed $v_2$ did the second one fly out, if: a) the pipes are fixed; b) the mass of the pipes is $M_0$ and they are not fixed? What speed will the pipes have in this case? Neglect the friction of the pistons against the pipe walls and the mass of the gas. The time of movement of both pistons inside the pipes is the same. $\"Description\"$

Answer

$v_2 = \\frac{S_2}{S_1} \\frac{m}{M} v_1$; $u = \\frac{mv_1}{MS_1} (S_2 - S_1)$.

Solution

Considering that the forces acting on the pistons (gas pressure is the same) are proportional to the magnitudes of their surfaces $S_1$ and $S_2$ and the action time is the same, we can, using Newton's second law in 'impulse' form, obtain $mv_1 / Mv_2 = S_1 / S_2.$ Hence $v_2 = \\frac{S_2}{S_1} \\frac{m}{M} v_1$ in both cases. The velocity of the tubes in the second case can be easily found from the law of conservation of momentum: $mv_1 + M_0 u = M v_2.$ For $u$ we get: $u = \\frac{m}{M_0} (\\frac{S_2}{S_1} - 1) v_1.$

20. A box with sand of mass $M$ begins to slide without friction down an inclined plane making an angle $\\alpha$ with the horizon. At the moment when the box has traveled a distance $l$, a body of mass $m$ falls into it, the speed of which is directed at an angle $\\beta$ to the horizon. The box stops at this point. With what speed $v$ was the body moving?

Answer

$v = \\dfrac{M\\sqrt{2gl \\sin \\alpha}}{m \\cos(\\alpha \\pm \\beta)}$.

Solution

By the time the box with sand travels path $l$, it will have a speed $u = \\sqrt{2al}$, where $a = g \\sin \\alpha$. For the box to stop, the sum of the projections of the impulses of the box and the body along the inclined plane must be zero: $-mv \\cos(\\alpha \\pm \\beta) + Mu = 0,$ the sign before $\\beta$ is chosen depending on the direction of the body's velocity relative to the horizon (the interaction time of the body and the box is assumed to be small). Substituting the expression for $u$ into this equation, we get the answer.

21. A small body slides down a smooth inclined plane making an angle of $\\alpha = 45°$ with the horizon from a height $h$. How will the body move if, at the end of the inclined plane, it encounters:

a) a perfectly elastic horizontal plane;

b) a horizontal plane that is inelastic but smooth?

Answer

a) Will describe a parabola above the plane, the vertex of which will be at height $h/2$;

b) will slide uniformly along the plane with velocity $v = \\sqrt{gh}$.

Hint

a) The horizontal component of the body's velocity $v_x = \\sqrt{2gh} \\cos \\alpha$ remains unchanged, while the vertical component $v_y = \\sqrt{2gh} \\sin \\alpha$ will change its direction to the opposite upon impact. The trajectory of the body will be segments of parabolas.

b) The vertical component of the body's velocity will become zero, and the body will move uniformly with velocity $v = v_x = \\sqrt{2gh} \\cos \\alpha$.

22. A vessel with water moves along an inclined plane with an angle of inclination $\\alpha$ so that the water level is parallel to this plane (see figure). Water flows out of an opening near the bottom of the vessel at a speed $v$. Determine the coefficient of friction $k$ between the vessel and the plane, if the mass of the vessel with water is $m$, and the area of the opening is $S$. Neglect the change in the mass of water associated with its outflow from the vessel. $\"Description\"$

Answer

$k = \\rho Sv^2 / (mg \\cos \\alpha)$.

Solution

For the container with water, Newton's second law has the form $ma = mg \\sin \\alpha + F_p - kmg \\cos \\alpha.\\tag{1}$ The forces acting on the container are shown below. From the condition that the water level is parallel to the inclined plane, it follows that the container with liquid moves along the inclined plane with acceleration $a = g \\sin \\alpha$ (see problem 2.17). The reaction force (reactive force) $F_p$, arising due to the outflow of the jet from the container, according to Newton's second (in impulse form) and third laws, is numerically equal to the change in the impulse of the water flowing out of the container per unit time: $F_p = \\frac{\\Delta(mv)}{\\Delta t} = \\frac{\\rho S \\Delta l}{\\Delta t}v = \\rho S v^2.$ Substituting the values of $a$ and $F_p$ into equation (1), we get $k = \\frac{\\rho S v^2}{mg \\cos \\alpha}.$ $\"Description\"$

23. What external force causes a car to move?

Answer

The external force causing the car to move is the static friction force of the driving wheels on the road.

24. From a jet engine with a mass of $M = 0.5$ t, initially at rest, two portions of a substance are ejected sequentially in the horizontal direction at a speed of $v_0 = 1000$ m/s relative to the engine. The mass of each portion is $m = 25$ kg. What will be the speed of the engine $v_2$ after the second portion is ejected? There is no friction.

Answer

$v_2 = -mv_0 \\left(\\dfrac{1}{M} + \\dfrac{1}{M-m}\\right) \\approx -102.6$ m/s.

Hint

The law of conservation of momentum when the first and second portions are ejected will be written respectively as: $(M-m)v_1 + m(v_0+v_1) = 0;\\quad (M-2m)v_2' + m(v_0+v_2') = 0,$ where $v_1$ is the velocity acquired by the setup after ejecting the first portion. The second equation is written in a coordinate system moving with velocity $v_1$. (Here all velocities are algebraic quantities). From the recorded equations, find $v_1$ and $v_2'$ and finally $u_2 = v_1 + v_2'.$ (See also the solution to problem 25.)

25. Products of combustion are ejected from a rocket of mass $M$ in portions, the masses of which are $m$, at a speed $u$ relative to the rocket. Neglecting the effect of gravity and air resistance, determine the speed $u_n$ of the rocket after the release of the n-th portion.

Answer

$\\displaystyle u_n = v \\sum_{k=1}^{n} \\frac{m}{M-(k-1)m} = v \\left(\\frac{m}{M} + \\frac{m}{M-m} + ... + \\frac{m}{M-(n-1)m}\\right)$.

Solution

Let's write the law of conservation of momentum for the rocket-gas system when the k-th portion of gas is ejected in a reference frame moving with the velocity equal to the rocket's velocity after the ejection of the (k-1)-th portion: $(M-km)\\Delta u_k = m(v - \\Delta u_k)$, where $\\Delta u_k$ is the velocity acquired by the rocket in this reference frame after the ejection of the k-th portion of gas; $(v - \\Delta u_k)$ is the velocity of the k-th portion at the moment of its separation from the rocket, when their interaction ended and the latter had already acquired velocity $\\Delta u_k$. Obviously, the velocity of the rocket relative to the Earth will be equal to the sum of the changes in its velocity $\\Delta u_k$ associated with the ejections of gas portions from the first to the n-th: $u_n = \\sum_{k=1}^{n} \\Delta u_k = v (\\frac{m}{M} + \\frac{m}{M-m} + \\frac{m}{M-2m} + ... + \\frac{m}{M-(n-1)m}).$ This solution differs from the previous problem in that here all velocities are arithmetic quantities (vector magnitudes), and their assumed directions are taken into account when choosing signs, i.e., all velocity vectors are projected onto a single direction.

26. On a platform of mass $M$, which can move along a horizontal plane without friction, there are $n$ people, each of mass $m$. In which case will the platform be given a greater speed $u$: a) if each of them sequentially runs along the platform with a relative speed $u$ and jumps off onto the ground; b) in the case when all people simultaneously run along the platform and simultaneously jump off it with the same relative speed $u$? Analyze the case when these people are standing on the edge of the platform and jump off either one at a time or simultaneously.

Answer

In both cases a) and b) the answers are the same:

a) $u = v (\\frac{m}{M+nm} + \\frac{m}{M+(n-1)m} + ... + \\frac{m}{M+m})$ (see solution to problem 25);

b) $u = v nm / (M+nm)$.

In cases a), as seen from the answers, the speed of the raft is greater.

27. A rocket of mass $M$ is launched vertically. The exhaust gas velocity from the engine nozzle is $u$. At what fuel consumption rate $μ$ (mass per unit time) will the engine thrust be sufficient to: a) balance the force of gravity acting on the rocket; b) give the rocket an acceleration of $a = 19.6$ m/s².

Answer

a) $\\mu_1 = Mg/v$; b) $\\mu_2 = M(a+g)/v$.

Solution

If the fuel consumption rate is $\\mu$, and the velocity of gas ejection from the nozzle is $v$, then per unit time the rocket is imparted an impulse $\\mu v$, which, according to Newton's laws, is equal to the reactive force. Therefore:

a) $\\mu_1 v = Mg$, $\\mu_1 = Mg/v$;

b) $Ma = \\mu_2 v - Mg$, $\\mu_2 = M(a+g)/v$.

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1. Two people grab a rope and pull it in opposite directions. One of them overpowers the other. Does this mean that the first person applies a greater force to the rope than the second? Compare the work done by the forces applied to the rope.

Answer

The forces are equal in magnitude; the works are also equal in magnitude but opposite in sign, because in one case the direction of the force acting on the rope and its displacement coincide, while in the other case they are opposite.

2. What is the work $A$ required to lift a chain, taken by one end and lying on a plane, to a height equal to its length? The chain has a length $l = 2 \\, \\text{m}$ and a mass $m = 5 \\, \\text{kg}$.

Answer

$A = mg \\frac{l}{2} \\approx 50$ J.

Hint

The work done is equal to the work done in lifting the center of gravity of the chain.

3. A window shade with a mass $M = 1 \\, \\text{kg}$ and length $l = 2 \\, \\text{m}$ rolls up onto a thin roller at the top of the window. What work is done in this process? Neglect friction.

Answer

$A = Mg \\frac{l}{2} \\approx 10$ J.

4. A flexible rubber hose of length $l$ hangs such that one end is $\\frac{1}{3} l$ lower than the other. The hose is filled with the maximum possible amount of water, with density $\\rho$. What is the minimal work needed to pour the water out of the hose by lifting its lower end while keeping the upper end at a constant height? The inner diameter of the hose is $d$. Neglect the mass of the hose. The radius of curvature of the hose bend is much smaller than $l$.

Answer

$A = \\frac{\\rho g l^2 \\pi a^2}{12}$.

Hint

The work goes into lifting the center of mass of the water in the hose from a height of $\\frac{1}{6}l$ to a height of $\\frac{2}{3}l$.

5. A chain of mass $M$ and length $l$ lies at the boundary of two adjoining half-planes made of different materials (see figure). What work must be done to move the chain onto the second half-plane? The coefficients of friction between the chain and the half-planes are $k_1$ and $k_2$, respectively. Solve the problem graphically as well. $\"Description\"$

Answer

$A = \\frac{1}{2}(k_1 + k_2) Mgl$.

Solution

The friction force overcome when moving the chain is: $\\begin{aligned} F_{fr} &= F_1 + F_2 \\\\ &= k_1 \\frac{Mg}{l}(l - x) + k_2 \\frac{Mg}{l} x \\\\ &= k_1 Mg - \\frac{Mg}{l}(k_1 - k_2) x \\end{aligned},$ where $x$ is the distance from the boundary between the semi-planes to the beginning of the chain. The graph of the dependence of the friction force on the displacement of the chain is shown below. The work done is numerically equal to the area of the figure bounded above by the graph of the force, i.e. $A = \\frac{1}{2}(k_1 + k_2) Mgl.$

The problem can also be solved by plotting the graphs of the dependence of the friction force on the displacement of the chain on each of the semi-planes and determining the total area of the resulting figures.
$\"Description\"$

6. An engine with a useful power of 15 kW, installed on a car, can propel it at a speed of 90 km/h on a horizontal road. The same engine, installed on a motorboat, ensures a speed no greater than 15 km/h. Determine the resistance force $F_c$ acting on the car and the motorboat at the given speeds.

Answer

$F_{c1} = 600$ H; $F_{c2} = 3600$ H.

7. A tram of mass $M$ moves uphill at a constant speed on a street inclined at an angle $\\alpha$ to the horizon. On a horizontal section, it can pull an additional car of mass $M_1$ at the same speed. What is the mass $M_1$ if the rolling friction coefficient of the wheels is $k$? The engine power is constant.

Answer

$M_1 = M \\left( \\frac{\\sin \\alpha}{k} + \\cos \\alpha - 1 \\right)$.

Hint

Power is equal to $N = Fv$ (where F is the traction force, v is the speed). Since the speeds are the same in both cases, the traction forces developed by the wagon's engine are also the same: $Mg \\sin \\alpha + kMg \\cos \\alpha = (M + M_1)gk$.

8. A locomotive, operating at constant power, can pull a train of mass $M = 2000$ tonnes uphill on a slope of $\\alpha_1 = 0.005$ at a speed $v_1 = 30$ km/h or on a slope of $\\alpha_2 = 0.0025$ at a speed $v_2 = 40$ km/h. Determine the resistance force $F_c$, assuming it is constant.

Answer

$F_c = \\dfrac{Mg (v_1 \\sin \\alpha_1 - v_2 \\sin \\alpha_2)}{v_2 - v_1} \\approx 9800$ H.

Hint

The power of the locomotive (see previous problem) $N = (Mg \\sin \\alpha_1 + F_c) v_1 = (Mg \\sin \\alpha_2 + F_c) v_2.$

9. A bullet flying at a certain speed penetrates a wall to a depth $l_1 = 10$ cm. To what depth $l_2$ will the bullet penetrate the same wall if its speed is doubled?

Answer

$l_2 = 40$ cm.

10. A bullet flying at speed $v_0$ pierces several identical boards placed at some distance from each other. In which board (counted from the first) will the bullet get stuck if its speed after passing the first board is $v_1 = 0.83 v_0$?

Answer

In the fourth board.

Hint

The kinetic energy losses of the bullet when passing through each board are identical and equal to: $\\Delta E = \\frac{m v_0^2}{2} - \\frac{m v_1^2}{2} = \\frac{m (v_0^2 - v_1^2)}{2}.$ The bullet will lose all its energy after piercing $n = \\frac{mv_0^2/2}{\\Delta E} = \\frac{v_0^2}{v_0^2 - v_1^2} \\approx 3.3\\text{ boards},$ i.e., the bullet will get stuck in the fourth board.

11. What work must be done to make a train of mass $M = 800$ tonnes: a) increase its speed from $v_1 = 36$ km/h to $v_2 = 54$ km/h; b) stop when the initial speed is $v_3 = 72$ km/h? Neglect resistance.

Answer

a) $A_1 = \\frac{M v_1^2}{2} - \\frac{M v_0^2}{2} = 5 \\cdot 10^4$ kJ;

b) $A_2 = mv_0^2/2 = 1.6 \\cdot 10^5$ kJ.

12. A train of mass $M = 2000$ tonnes, starting from rest with an acceleration $a = 0.2 \\, \\text{m/s}^2$, reaches the desired speed in one minute and then moves uniformly. Determine the power of the locomotive during steady motion if the resistance coefficient $k = 0.005$.

Answer

$N = Mgkat \\approx 1.2 \\cdot 10^6$ W.

13. A car of mass $M = 2000$ kg starts from rest and moves uphill on a slope of $\\alpha = 0.02$. After traveling a distance $s = 100$ m, it reaches a speed $v = 32.4$ km/h. The resistance coefficient is $k = 0.05$. Determine the average power developed by the car's engine.

Answer

$N_{avg} = M \\left( \\frac{v^2}{2s} + ga + kg \\right) \\frac{v}{2} \\approx 9900$ W (assuming $\\sin \\alpha \\approx \\alpha, \\cos \\alpha \\approx 1$).

14. A rocket of mass $M$ hovers motionless above Earth with its engine running. The speed of the exhaust gases is $u$. Determine the engine's power.

Answer

$N = Mgu/2$.

Hint

The engine power is $N = \\mu u^2 / 2$, where $\\mu$ is the fuel consumption rate. The reactive force $\\mu u$ balances the force of gravity: $\\mu u = Mg$.

15. In which case must the car's engine do more work: accelerating from rest to a speed of 27 km/h or increasing the speed from 27 km/h to 54 km/h? Assume the resistance force and acceleration time are the same in both cases.

Answer

In the second case: $A_2 / A_1 = 3$.

Hint

Since the accelerations of the car and the resistance forces are the same in both cases, the traction forces $F_T$, equal to $ma + F_{res}$, are also equal. However, the distance traveled by the car in the second case is three times greater.

16. A stone of mass $m = 200$ g is thrown from a horizontal surface at an angle to the horizon and lands back on the surface at a distance $s = 5$ m after $t = 1.2$ s. Find the work done during the throw. Neglect air resistance.

Answer

$A = \\dfrac{m}{2} \\left( \\dfrac{g^2 t^4}{4} + \\dfrac{l^2}{t^2} \\right) \\approx 5.2$ J.

Hint

The work is spent on giving the body kinetic energy.

17. Determine the work required to compress a spring by $x = 10$ cm if a force $F_0 = 100 \\, \\text{N}$ is needed to compress it by $x_0 = 1$ cm.

Answer

$A = \\frac{F_0 x^2}{2 x_0} = 50$ J.

Solution

The work done in compressing the spring can be found by plotting the graph of the dependence of the force $F$ acting on the spring on the magnitude of its deformation $x$, i.e., on the displacement of the end of the spring from its equilibrium position. According to Hooke's law, this force is $F = kx$, where $k$ is the spring constant, which in this case is $k = F_0/x_0 = 10^4$ N/m. The work of compressing the spring will be numerically equal to the area of the triangle OAB (see figure): $A = \\frac{kx \\cdot x}{2} = \\frac{kx^2}{2} = \\frac{F_0}{x_0} \\frac{x^2}{2} = 50 \\text{ J}.$ $\"Description\"$

18. A railcar of mass $M = 2 \\cdot 10^4 \\, \\text{kg}$, moving at a speed $v = 0.5 \\, \\text{m/s}$, collides with two stationary spring buffers. Find the maximum compression $x$ of the buffers if a force of $5 \\cdot 10^4 \\, \\text{N}$ compresses a buffer by 1 cm. Neglect friction.

Answer

$x \\approx 2.2$ cm.

Hint

Upon impact, the kinetic energy of the wagon is converted into the work of compressing the springs, i.e., it transforms into their potential energy: $W_p = 2 kx^2 / 2$, where $k = 5 \\cdot 10^6$ N/m (see problem 17).

19. A constant force $F = 200 \\, \\text{N}$ lifts a load of mass $M = 10 \\, \\text{kg}$ to a height $h = 10 \\, \\text{m}$. What work $A$ does the force $F$ perform? What potential energy $U$ will the lifted load possess?

Answer

$A = Fh = 2 \\cdot 10^3$ J; $\\Pi = mgh = 10^3$ J. The body also possesses kinetic energy $T = A - \\Pi$.

20. An elevator of mass $M = 1000 \\, \\text{kg}$ is accelerated upward by a winch. Over a segment of the path of length $l = 1 \\, \\text{m}$, the elevator moves at an average speed $v_{\\text{avg}} = 5 \\, \\text{m/s}$, and its speed increases by $\\Delta v = 0.5 \\, \\text{m/s}$. What work is done by the force moving the elevator over this segment?

Answer

$A = Mgl + Mv_{avg} \\Delta v = 12.3 \\cdot 10^3$ J.

Hint

$\\frac{v_1^2 - v_2^2}{2} = \\frac{v_1 + v_2}{2} (v_1 - v_2) = v_{avg} \\Delta v.$

21. What work will a force $F = 30 \\, \\text{N}$ do by lifting a load of mass $m = 2 \\, \\text{kg}$ up an inclined plane to a height $h = 2.5 \\, \\text{m}$ with an acceleration $a = 10 \\, \\text{m/s}^2$? The force acts parallel to the inclined plane. Neglect friction.

Answer

$mgh \\dfrac{F}{F - ma} \\approx 147$ J.

Hint

The work goes into increasing the potential energy of the load $mgh$ and imparting kinetic energy $\\frac{mv^2}{2}$. $\\frac{mv^2}{2} = \\frac{mah}{\\sin \\alpha} = \\frac{mah}{h/l} = mal.$ It can also be calculated simply using the formula $A = Fl = F \\frac{h}{\\sin \\alpha}.$ Find $\\sin \\alpha$ from the condition $F - mg \\sin \\alpha = ma$. The magnitude of $\\sin \\alpha$ is found from the condition $F - mg \\sin \\alpha = ma$.

22. A certain force pushes a body of mass $m = 16 \\, \\text{kg}$ up an inclined plane of length $l = 3.1 \\, \\text{m}$ and inclination $\\alpha = 30^\\circ$ to the horizon.

The body's speed at the base of the plane was $v_0 = 0.6 , \\text{m/s}$, and at the top edge, it was $v_1 = 3.1 , \\text{m/s}$. What is the work done by the force? There is no friction.
What is the work done by the same force, and what will be the kinetic energy of the body at the top of the inclined plane if there is friction with a coefficient $k = 0.1$? The force acts along the inclined plane.

Answer

1. $A = mgl \\sin \\alpha + \\frac{m (v_1^2 - v_0^2)}{2} = 317$ J.

The work will be the same, but the kinetic energy is less by the amount $kmgl \\cos \\alpha \\approx 42.5$ J, and at the highest point it is equal to 34.4 J.

23. A truck of mass $M = 6 \\cdot 10^3 \\, \\text{kg}$ drives onto a ferry tied to the shore with two ropes at a speed $v = 18 \\, \\text{km/h}$. After boarding, the truck stops, traveling a distance $s = 10 \\, \\text{m}$ while braking. Determine the total tension force in the ropes.

Answer

$F_{drag} = Mv^2/(2s) = 7500$ H.

24. A car moving at speed $v = 54 \\, \\text{km/h}$ brakes sharply and begins to skid (the locked wheels slide on the road). Determine the acceleration $a$ and the distance $s$ the car will travel if the sliding friction coefficient between the wheels and the asphalt is:

a) $k_1 = 0.3$ in wet weather;

b) $k_2 = 0.7$ in dry weather.

Answer

a) $a_1 \\approx 2.94$ m/s², $s_1 \\approx 38.3$ m;

b) $a_2 \\approx 6.86$ m/s², $s_2 \\approx 16.4$ m.

Hint

The kinetic energy of the car is spent on the work of overcoming friction: $mv^2/2 = kmgs$.

25. A car with fully engaged brakes (wheels locked) can remain stationary on a slope with an inclination of up to $23^\\circ$. What is the braking distance $s$ of the car on a horizontal road when the initial speed is 10 m/s? The coefficient of friction between the wheels and the ground is the same on the slope and the road.

Answer

$s \\approx 12$ m.

Hint

The coefficient of friction between the wheels and the ground is determined from the condition of immobility of the car at the critical angle of inclination of the mountain $\\alpha$ and is equal to $\\tan \\alpha$.

26. A sled with a load of mass $M = 120 \\, \\text{kg}$ slides down a slope inclined at $\\alpha = 14^\\circ$ to the horizon. The length of the descent is $l = 60 \\, \\text{m}$. The sliding friction coefficient of the sled is $k = 0.14$. Determine:

a) the acceleration $a_1$ of the sled during the descent;

b) the speed $v$ at the end of the descent;

c) the time $t_1$ of the descent;

d) the kinetic energy $T_1$ at the end of the descent;

e) the distance $s$ the sled will travel horizontally after the descent;

f) the time $t_2$ of horizontal motion;

g) the acceleration $a_2$ during horizontal motion.

Answer

a) $a_1 = g (\\sin \\alpha - k \\cos \\alpha) = 1.04$ m/s²;

b) $v = \\sqrt{2a_1 l} \\approx 11.2$ m/s;

c) $t_1 = v/a_1 = \\sqrt{2l/a_1} = 10.7$ s;

d) $T_1 = mgl (\\sin \\alpha - k \\cos \\alpha) \\approx 7620$ J;

e) $s = \\frac{l}{k} (\\sin \\alpha - k \\cos \\alpha) = 45.3$ m;

f) $t_2 = 2s/v \\approx 8$ s; g) $a_2 = -kg = -0.14g$.

27. A body slides down an inclined plane. The angle of inclination is $\\alpha = 20^\\circ$, the length is $l = 4 \\, \\text{m}$, and the friction coefficient between the body and the plane is $k = 0.2$. What will be the speed $v$ of the body as it transitions from the inclined plane to the horizontal surface?

Answer

$v = \\sqrt{2gl (\\sin \\alpha - k \\cos \\alpha)} \\approx 3.4$ m/s.

28. A pool of area $S = 100 \\, \\text{m}^2$, filled with water to a depth $h = 1 \\, \\text{m}$, is divided in half by a vertical partition. The partition is slowly moved horizontally so that it divides the pool in a 1:3 ratio. What work must be done for this if no water leaks through the partition?

Answer

$A = \\dfrac{1}{6} \\rho g h^2 S = 1.63 \\cdot 10^5$ J.

Hint

The work is equal to the difference in potential energies of the water before and after moving the partition.

29. Two cars start simultaneously from rest and accelerate uniformly. Their masses are equal. How many times greater is the average power of the first car's engine compared to the second if, in the same time, the first car reaches twice the speed of the second? Neglect resistance.

Answer

By 4 times.

Hint

Each of the cars acquires kinetic energy over time $t$, equal to $N_1 t$ and $N_2 t$ respectively, therefore $N_1 / N_2 = v_1^2 / v_2^2$.

30. An airplane needs a speed $v = 25 \\, \\text{m/s}$ for takeoff. The takeoff run is $s = 100 \\, \\text{m}$. What is the power of the engines if the airplane's mass is $m = 1000 \\, \\text{kg}$ and the resistance coefficient is $k = 0.02$? Assume the airplane's motion during takeoff is uniformly accelerated.

Answer

$N = mv \\left( \\dfrac{v^2}{2s} + kg \\right) \\approx 83.1$ kW.

Hint

The power of the motors during takeoff is $N = F_T v,$ where $F_T$ is the thrust force of the motors, which can be found by writing Newton's second law: $F_T - kmg = ma$ and substituting the acceleration of the aircraft during takeoff $a = v^2 / (2s)$.

31. A train of mass $M = 5 \\cdot 10^5 \\, \\text{kg}$ climbs a slope of 10 m per kilometer at a speed of 30 km/h. The resistance coefficient is $k = 0.002$. Determine the power developed by the locomotive.

Answer

$N \\approx 500$ kW.

32. After accelerating, an ice skater moves uniformly along a horizontal ice track for some time. Then, without pushing off, the skater travels a distance $s = 60 \\, \\text{m}$ in $t = 25 \\, \\text{s}$ before stopping. The skater's mass is $m = 50 \\, \\text{kg}$. Determine: a) the friction coefficient; b) the power expended by the skater during uniform motion.

Answer

a) $k = 2s/(gt^2) \\approx 0.02$; b) $N = 4ms^2/t^3 \\approx 46$ W.

33. A locomotive pulls a train with a total mass $m = 2000$ tonnes. Given that the locomotive's power $N$ is constant and equal to 1800 kW, and the resistance coefficient $k = 0.005$, determine:

a) the train's acceleration $a$ when its speed is $v_1 = 4 , \\text{m/s}$ and when its speed is $v_2 = 12 , \\text{m/s}$;

b) the maximum speed $v_{\\text{max}}$ of the train.

Answer

a) $a = \\frac{N}{mv} - kg$; $a_1 = 0.176$ m/s²; $a_2 = 0.026$ m/s²;

b) $v_{max} = N/(kmg) \\approx 66$ km/h.

34. A pulley of radius $R$ rotates at $n$ revolutions per second, transmitting power $N$ via a belt. Find the tension force $T$ in the belt, assuming no slipping.

Answer

$T = N/(2\\pi Rn)$.

Hint

The power transmitted by the belt is $N = Tv$, where $v = 2\\pi Rn$ is the speed of the belt.

35. Find the power of an air stream with a circular cross-section of diameter $d = 18 \\, \\text{m}$, flowing at a speed $v = 12 \\, \\text{m/s}$. The air density (under normal conditions) is $\\rho = 1.3 \\, \\text{kg/m}^3$.

Answer

$N \\approx 284$ kW.

Hint

The power of the flow is equal to the work that the flow can perform in one second during the complete conversion of its kinetic energy into work: $N = \\frac{mv^2/2}{t}$. Here $m$ is the mass of air passing through the cross-section $S$ of the flow in time $t$. Therefore $N = \\frac{\\pi \\rho d^2 v^3}{8}$ (assuming \$a\$ in OCR corresponds to diameter \$d\$), i.e., the power of the flow is proportional to the cube of its velocity.

36. A mountain stream with a cross-sectional area $S$ forms a waterfall of height $h$. The water flow speed is $v$. Find the power of the waterfall.

Answer

$N = \\rho g S v h + \\frac{\\rho S v^3}{2}$.

Hint

See solution to problem 35. The speed of water at the base of the waterfall $v_1$ can be easily found from the energy conservation law: $m \\frac{v^2}{2} + mgh = m \\frac{v_1^2}{2}.$

37. A road section has a slope of 0.05. Descending with the engine off, a car moves uniformly at $v = 60 \\, \\text{km/h}$. What engine power is required for the car to ascend the same slope at the same speed? The car's mass is $m = 1.5$ tonnes.

Answer

$N = 2mgv \\sin \\alpha = 24.5$ kW, where $\\sin \\alpha = 0.05$.

Solution

The power of the car's engine must be equal to $N = F_T v$, where $F_T$ is the traction force. Since the car moves uniformly, $F_T$ is equal to the sum of the resistance force acting on the moving car and the component of gravity along the inclined plane: $F_T = F_{res} + mg \\sin \\alpha.$ For a car moving downhill, one can write, on the other hand: $mg \\sin \\alpha - F_{res} = 0,$ from which $F_{res} = mg \\sin \\alpha$, and $F_T = 2mg \\sin \\alpha$. Thus, $N = 2mgv \\sin \\alpha$.

38. Trucks equipped with engines of power $N_1$ and $N_2$ achieve speeds $v_1$ and $v_2$, respectively. What will be their speed if they are connected by a tow rope?

Answer

$v = \\dfrac{(N_1 + N_2) v_1 v_2}{N_1 v_2 + N_2 v_1}$.

Hint

For trucks connected by a rope, $N_1 + N_2 = (F_1 + F_2) v$, where $F_1$ and $F_2$ are the traction forces of the trucks, equal to the resistance forces of the motion.

39. A snowmobile moves uphill on a gentle slope at a steady speed $v_1 = 20 \\, \\text{m/s}$. If it moves downhill at the same engine power, its speed becomes $v_2 = 30 \\, \\text{m/s}$. What speed $v$ will it achieve on a horizontal path with the same engine power?

Answer

$v \\approx 2 \\dfrac{v_1 v_2}{v_1 + v_2} \\approx 24$ m/s.

Solution

When the aerosled moves uniformly upwards, the traction force is $F_1 = mg \\sin \\alpha + kmg \\cos \\alpha. \\tag{1}$ When moving downwards $F_2 = kmg \\cos \\alpha - mg \\sin \\alpha, \\tag{2}$ where $k$ is the coefficient of friction; $m$ is the mass of the aerosled. Since the engine power is constant, $N = F_1 v_1 = F_2 v_2. \\tag{3}$ When moving on a horizontal path $N = kmgv$ (4). Adding equations (1) and (2) term by term, we get $F_1 + F_2 = 2kmg \\cos \\alpha. \\tag{5}$ Solving equations (3), (4), and (5) together, we find $v = 2 \\cos \\alpha \\times \\frac{v_1 v_2}{v_1 + v_2}.$ Since the ascent is slight, i.e., $\\alpha$ is small, then $\\cos \\alpha \\approx 1$ and $v = 2 \\frac{v_1 v_2}{v_1 + v_2}.$

40. A train of mass $m = 500$ tonnes moves uniformly on a horizontal track. The last car of mass $m_1 = 20$ tonnes detaches. After traveling $s = 240 \\, \\text{m}$, the driver shuts off the steam supply. How far apart will the detached car and the rest of the train come to rest? Assume the traction force is constant, and the resistance is proportional to the mass.

Answer

250 m.

Solution

Let the speed of the train at the moment the car detaches be $v$. By the time the car stops, its kinetic energy will be completely converted into work against friction: $m_1 v^2 / 2 = km_1 g s_1,$ where $s_1$ is the distance traveled by the car until it stops. During uniform motion of the train, the traction force is balanced by the total resistance force: $F_T = kmg.$ When the rear car detached, the resistance force decreased, but the traction force, according to the problem statement, did not change. A net force $F$ started acting on the train, accelerating its motion: $F = kmg - k(m - m_1)g = km_1 g.$ After the car detached, the train began to move under the action of this force with acceleration, and its kinetic energy over the path $s$ increased by an amount $\\Delta T$, equal to the work of the force $F$ over the path $s$: $\\Delta T = km_1 g s.$ Before the detachment, the kinetic energy of this part of the train was $(m - m_1)/m$ times the energy of the car, i.e., it was equal to $k(m - m_1)gs_1$. Therefore, by the time the steam supply stopped, the train had kinetic energy $T = k(m - m_1)gs_1 + km_1 g s.$ This energy is spent on the work of overcoming the resistance of the train's motion over the path $s_2$: $k(m - m_1)gs_1 + km_1 g s = k(m - m_1)gs_2.$ From this equation, we find $s_2 - s_1 = m_1 s / (m - m_1)$, and the distance between the stopped train and the car is $x = s_2 - s_1 + s = \\frac{ms}{m - m_1}.$ Another solution is in problem 50 in Chapter: Newton's Laws.

41. Find the work required to drag a body of mass $m = 50 \\, \\text{kg}$ up a hill of arbitrary profile along a flat trajectory from point $A$ to point $B$, where the horizontal distance is $l = 10 \\, \\text{m}$ and the vertical height is $h = 10 \\, \\text{m}$. The friction coefficient between the body and the hill is $k = 0.1$. The hill's profile is such that the tangent at any point forms an acute angle with the horizon. The applied force always acts tangentially to the path.

Answer

$A = mg(kl + h) \\approx 5500$ J.

Solution

Let's divide the entire path into small straight sections. Then the work on each section is $\\Delta A = kmg \\Delta l \\cos \\alpha + mg \\Delta l \\sin \\alpha$, where the first term is the work against the friction force, and the second is the part of the work that goes into increasing the potential energy of the body. Noting that $\\Delta l \\cos \\alpha$ and $\\Delta l \\sin \\alpha$ are the projections of the path segment onto the horizontal and vertical directions, respectively, we find that the total work will be equal to $A = kmgl + mgh = mg(kl + h)$.

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1. A thread with a weight suspended from it is deflected by an angle $\\alpha$ and released. At what angle $\\beta$ will the thread with the weight deflect if, during its motion, it is stopped by a pin placed vertically at the midpoint of the thread's length?

Answer

$\\beta = \\arccos(2 \\cos \\alpha - 1)$.

2. 1. A body is thrown vertically upward with a velocity $v_0 = 16 \\, \\text{m/s}$. At what height $h$ will the kinetic energy of the body equal its potential energy?

With what initial velocity must a ball be thrown from a height $h$ so that it bounces to a height $2h$? Assume the collision is elastic and air resistance is negligible.

Answer

1. $h = v_0^2 / (4g) \\approx 6.5$ m.

$v_0 = \\sqrt{2gh}$.

3. A stone is thrown horizontally from a tower of height $H = 25 \\, \\text{m}$ with a velocity $v_0 = 15 \\, \\text{m/s}$. Find the kinetic and potential energies of the stone one second after the motion begins. The mass of the stone is $m = 0.2 \\, \\text{kg}$. Neglect air resistance.

Answer

$T = 32.2$ J; $\\Pi = 39.4$ J.

Hint

The initial values of kinetic and potential energy are respectively: $T = mv_0^2/2$,\\qquad \\Pi = mgH.$ The change in potential energy $\\Delta \\Pi = mg\\Delta h$, where $\\Delta h = gt^2/2$ (taking $t=1$ s) is the change in the body's height over 1 s. The kinetic energy of the body increases by this same amount.

4. Determine the kinetic energy of a body with a mass of 1 kg, thrown horizontally with a velocity of $20 \\, \\text{m/s}$, at the end of the fourth second of its motion. Assume $g = 10 \\, \\text{m/s}^2$.

Answer

$T = 1000$ J.

Solution

At the end of the fourth second, the velocity of the body, thrown horizontally with speed $v_0$ from some height, will be composed of the horizontal velocity $v_0$ and the vertical velocity $v_y = gt$. The velocity of the body $\\vec{v}$ is found using the parallelogram rule (see fgigure). Hence, the kinetic energy of the body at the end of the fourth second is $ T = \\dfrac{mv^2}{2} = \\dfrac{m}{2} [v_0^2 + (gt)^2]. $ (see figure) $\"Description\"$

5. A flexible uniform rope of length $L$ lies on a smooth horizontal table. One end of the rope is at the edge of the table. At some moment, a slight push sets the rope in motion, causing it to slide off the table continuously. How do the acceleration and velocity of the rope depend on the length $x$ of the hanging part? What will be the velocity of the rope when it completely slides off the table?

Answer

$a = xg/L$; $v = x\\sqrt{g/L}$; $v_{final} = \\sqrt{gL}$.

Solution

The entire rope is accelerated by the force of gravity acting on the hanging part $F_g = mxg/L$, where $m$ is the mass of the rope. According to Newton's second law, $mxg/L = ma$. Hence $a = xg/L$. The speed of the rope can be found from the law of conservation of energy. If a part of the rope of length $x$ slides off the table, its center of mass descends by $h = x/2$, therefore $mgx^2/(2L) = mv^2/2$. Hence $v = x\\sqrt{g/L}$ and $v_{final} = \\sqrt{gL}$.

6. A rope of length $L$ is thrown over a peg. Initially, the ends of the rope are at the same level. After a slight push, the rope starts moving. Determine the velocity $v$ of the rope when it slides off the peg. Neglect friction.

Answer

$v = \\sqrt{gL/2}$.

Hint

See the solution to problem 5. Initially, the center of mass of the rope was at a distance $L/4$ from the peg, and at the moment the rope leaves the peg, its center of mass will be at a distance $L/2$ from the peg.

7. A skater, having accelerated to a velocity $v = 27 \\, \\text{km/h}$, enters an icy hill. To what height $H$ from the initial level will the skater ascend if the hill rises $h = 0.5 \\, \\text{m}$ for every $s = 10 \\, \\text{m}$ horizontally? The coefficient of friction between the skates and the ice is $k = 0.02$.

Answer

$H \\approx 2$ m.

Solution

The decrease in the mechanical energy of the skater is equal to the work done against friction: $ \\dfrac{mv^2}{2} - mgH = A_{fr} \\tag{1} $ Here $H$ is the height to which the skater ascends; on the other hand, $ \\begin{aligned} A_{fr} &= kmgl \\cos \\alpha \\\\ &= k \\dfrac{H}{\\sin \\alpha} mg \\cos \\alpha \\\\ &= k \\dfrac{mgH}{s} \\end{aligned}\\tag{2} $ where $l$ is the path traveled by the skater along the inclined plane. Substituting expression (2) into equation (1), we find that $ H = \\dfrac{v^2}{2g(1 + k/s)} $.

8. A body of mass $m = 1.5 \\, \\text{kg}$ is thrown vertically upward from a height $h = 4.9 \\, \\text{m}$ with a velocity $v_0 = 6 \\, \\text{m/s}$ and falls to the ground with a velocity $v = 5 \\, \\text{m/s}$. Determine the work done by air resistance forces.

Answer

$A \\approx -80.2$ J.

Hint

The work done by resistance forces is equal to the change in the mechanical energy of the body: $ A = E_2 - E_1 = \\dfrac{mv_2^2}{2} - \\left( mgh + \\dfrac{mv_1^2}{2} \\right). $

9. A stone with a mass of 50 g is thrown at an angle to the horizontal from a height of 20 m above the ground with a velocity of 18 m/s and lands with a velocity of 24 m/s. Find the work done to overcome air resistance.

Answer

$A \\approx 3.5$ J.

Hint

See problem 8. The work done against resistance forces is positive.

10. An airplane with a mass $m = 10^3 \\, \\text{kg}$ flies horizontally at an altitude $H = 1200 \\, \\text{m}$ with a velocity $v_1 = 50 \\, \\text{m/s}$. The engine is then turned off, and the plane transitions to a gliding flight, reaching the ground with a velocity $v_2 = 25 \\, \\text{m/s}$. Determine the average air resistance force during the descent, assuming the descent path is 8 km long.

Answer

$F_{res} \\approx 1570$ N.

Hint

The work done by resistance forces $A = \\dfrac{mv_2^2}{2} - \\left( \\dfrac{mv_1^2}{2} + mgH \\right)$. On the other hand, $A = -F_{res}l$ ($F_{res}$ is the average resistance force, which is directed opposite to the displacement, $l$ is the length of the slope).

11. A body of mass $m = 1 \\, \\text{kg}$ moves across a table with an initial velocity $v_0 = 2 \\, \\text{m/s}$. Upon reaching the edge of the table, which is $h = 1 \\, \\text{m}$ high, the body falls. The coefficient of friction between the body and the table is $k = 0.1$. Determine the amount of heat released during the inelastic collision with the ground. The path traveled by the body on the table is $s = 2 \\, \\text{m}$.

Answer

$Q \\approx 9.8$ J.

Hint

From the law of conservation of energy, it follows that $ mgh + \\dfrac{mv_0^2}{2} = kmgs + Q, $ where $Q$ is the amount of heat released during the impact.

12. A weight attached to a vertical spring is slowly lowered to its equilibrium position, stretching the spring by a length $x_0$. How much will the spring stretch if the same weight is allowed to fall freely from a position where the spring is unstretched? What maximum velocity $v_{\\text{max}}$ will the weight attain? Describe the nature of the weight's motion. The mass of the weight is $m$. Neglect the mass of the spring.

Answer

$x_{\\text{max}} = 2x_0$; $v_{\\text{max}} = \\sqrt{gx_0}$; oscillatory.

Solution

From the equilibrium condition of the load $mg = kx_0$, we find the spring stiffness $k = mg/x_0$. Now writing the law of conservation of energy $kx^2/2 = mgx$ and substituting the expression for $k$ into the equation, we find $x = 2x_0.$ The speed of the load is maximum when the acceleration is zero and $mg = kx$. Since $mg = kx_0$, this occurs at $x=x_0$. The energy conservation at this point is $\\dfrac{mv_{\\text{max}}^2}{2} + \\dfrac{kx_0^2}{2} = mgx_0.$ Substituting $k=mg/x_0$, we get $\\dfrac{mv_{\\text{max}}^2}{2} + \\dfrac{mgx_0}{2} = mgx_0,$ so $v_{\\text{max}} = \\sqrt{gx_0}$. The load will oscillate around the equilibrium position $x_0$ with amplitude $x_0$ (from $x=0$ to $x=2x_0$).

13. A falling weight of mass $M = 5 \\cdot 10^2 \\, \\text{kg}$ drives a pile into the ground from a height $h = 1.2 \\, \\text{m}$. The pile sinks $s = 2 \\, \\text{cm}$ into the ground upon impact. Determine the average force of impact $F_{cp}$ and its duration $\\tau$, assuming the mass of the pile is much smaller than the mass of the weight.

Answer

$F_{res} = Mg(1 + h/s) \\approx 3 \\times 10^5$ N; $\\tau = 2s/\\sqrt{2gh} \\approx 8 \\times 10^{-3}$ s.

Solution

The work done against the resistance force of the soil is equal to the change in potential energy of the pile: $F_{res}s = Mg(h + s)$. Hence $F_{res} = Mg(1 + h/s) = Mg \\dfrac{h+s}{s} \\approx 3 \\times 10^5$ N. The impact time is determined from the relation $\\tau = s/v_{avg} = 2s/v_0$, where $v_0$ is the speed of the pile at the beginning of the impact (we assume the motion of the pile is uniformly decelerated, so $v_{avg} = v_0/2$). In accordance with the law of conservation of energy $v_0 = \\sqrt{2gh}$. Therefore $\\tau = 2s/\\sqrt{2gh} \\approx 8 \\times 10^{-3}$ s.

14. A sled slides down a hill of height $h = 2 \\, \\text{m}$ and base $b = 5 \\, \\text{m}$, then stops after traveling a horizontal distance $l = 35 \\, \\text{m}$ from the base of the hill. Find the coefficient of friction.

Answer

$k = 0.05$.

Solution

At the end of the path, the sled stops, and, consequently, all the initial potential energy is spent on work against friction forces on the inclined and horizontal sections of the path: $ mgh = kmgs \\cos \\alpha + kmgl, \\tag{1} $ where $s$ is the length of the slope; $\\alpha$ is the angle of inclination of the hill. In this case $s \\cos \\alpha = b$. Therefore, from equation (1) we get $h = k(b + l)$, whence $k = h/(b + l)$.

15. A steel ball of mass $m = 20 \\, \\text{g}$ falls from a height $h_1 = 1 \\, \\text{m}$ onto a steel plate and rebounds to a height $h_2 = 81 \\, \\text{cm}$. Find: a) the impulse of the force acting on the plate during the collision; b) the amount of heat released during the collision.

Answer

a) $p = 0.17$ N·s;

b) $Q = 3.7 \\times 10^{-2}$ J.

Solution

The impulse of the force acting on the ball during the impact with the plate, in accordance with Newton's second law, is equal to: $ p = |\\Delta (m\\vec{v})| = m(v_1 + v_2), $ where $v_1$ and $v_2$ are the speeds of the ball respectively before and after the impact. From the law of conservation of energy, we find: $v_1 = \\sqrt{2gh_1}$ and $v_2 = \\sqrt{2gh_2}$, therefore $p = m\\sqrt{2g}(\\sqrt{h_1} + \\sqrt{h_2})$. According to Newton's third law, the impulse of the force that acted on the plate during the impact is numerically equal to $p$. The amount of heat released during the impact of the ball on the plate is equal to the difference in the ball's energies before and after the impact: $ Q = \\dfrac{mv_1^2}{2} - \\dfrac{mv_2^2}{2} = mg(h_1 - h_2). $

16. A light ball begins to fall freely and, after traveling a distance $l$, collides elastically with a heavy plate moving upward with a velocity $u$. To what height $h$ will the ball rebound after the collision?

Answer

$h = (\\sqrt{2gl} + 2u)^2 / (2g)$.

Solution

It is more convenient to solve the problem by considering the collision in a coordinate system associated with the moving plate. Since the speed of the plate changes negligibly due to the impact, this coordinate system can be considered inertial. In it, the ball before the impact with the plate has speed $v' = v + u$, where $v = \\sqrt{2gl}$ is the speed of the ball in the stationary coordinate system associated with the Earth. The impact of the ball on the plate is perfectly elastic, therefore, after the impact, the ball in the moving system will have speed $v_1' = -v' = -(v + u)$. In the coordinate system associated with the Earth, the speed of the ball will be $v_1 = -(v + u) - u = -(v + 2u)$ (the minus sign means that the ball will have a velocity directed upwards). Therefore, after the impact, the ball will bounce to a height $h = \\frac{v_1^2}{2g} = \\frac{(v + 2u)^2}{2g} = \\frac{(\\sqrt{2gl} + 2u)^2}{2g}.$

17. A balloon held by a rope rises to a certain height. How does the potential energy of the balloon-air-Earth system change?

Answer

The potential energy of the ball-air-Earth system decreased, because as the ball rises, the volume occupied by the ball is replaced by air, which has a greater mass than the ball.

18. A hockey puck with an initial velocity $v_0 = 5 \\, \\text{m/s}$ slides a distance $s = 10 \\, \\text{m}$ before hitting the boards. The collision is perfectly elastic, the coefficient of friction between the puck and the ice is $k = 0.1$, and air resistance is negligible. Determine the distance $l$ the puck will travel after the collision.

Answer

$l \\approx 2.7$ m.

Hint

During the motion of the puck, its kinetic energy is spent on doing work against friction: $mv_0^2/2 = kmg(s + l).$

19. A body slides without friction off a wedge lying on a horizontal plane twice: first, the wedge is fixed; second, the wedge can slide without friction. Will the speed of the body at the end of the slide be the same in both cases if the body starts from the same height each time?

Answer

In the first case, the speed of the body is greater than in the second.

20. Why is it difficult to jump to shore from a light boat near the shore, but easy to do so from a steamship at the same distance?

Solution

Let $v$ denote the minimum speed with which the person must jump to reach the shore. During the jump, the person imparts a velocity $u$ to the boat (or steamship), equal, based on the law of conservation of momentum, to $u = mv/M$ where $m$ is the mass of the person and $M$ is the mass of the boat or steamship.

By jumping, the person performs work $A$, equal to the kinetic energy acquired by the person and the boat (or steamship):
$A = \\dfrac{mv^2}{2} + \\dfrac{Mu^2}{2} = \\dfrac{mv^2}{2} \\left( 1 + \\dfrac{m}{M} \\right). $
This work is smaller, the smaller the ratio of the masses of the person and the boat (steamship). The mass of the steamship is many times greater than the mass of the person ($m/M \\ll 1$), therefore, jumping from it, the person performs less work than in the case when they jump from the boat ($m/M \\approx 1$).

21. A skater of mass $M = 70 \\, \\text{kg}$, standing on ice skates on ice, throws a stone of mass $m = 3 \\, \\text{kg}$ horizontally with a velocity $v = 8 \\, \\text{m/s}$ relative to the Earth. Find the distance $s$ the skater will recoil if the coefficient of friction between the skates and the ice is $k = 0.02$.

Answer

$s = \\left( \\dfrac{m}{M} \\right)^2 \\dfrac{v^2}{2kg} \\approx 0.29$ m.

Solution

Based on the laws of conservation of momentum and energy, we can write the equations: $Mv' = mv\\quad\\text{ and }\\quad M\\frac{(v')^2}{2} = kMgs$. Excluding $v'$ from these equations, we find $s = \\left( \\dfrac{m}{M} \\right)^2 \\dfrac{v^2}{2kg}.$

22. A person stands on a stationary cart and throws a stone of mass $m = 8 \\, \\text{kg}$ horizontally with a velocity $v_1 = 5 \\, \\text{m/s}$ relative to the Earth. Determine the work done by the person if the mass of the cart with the person is $M = 160 \\, \\text{kg}$. Analyze the dependence of the work on the mass $M$. Neglect friction.

Answer

$A \\approx 105$ J.

Solution

The work that the person performs by throwing a stone from the cart is spent on imparting kinetic energy to the stone and the cart with the person: $A = \\dfrac{mv_1^2}{2} + \\dfrac{Mv_2^2}{2}, $ where $v_2$ is the speed of the cart. By the law of conservation of momentum, $mv_1 = Mv_2$, whence $v_2 = mv_1/M$; substituting $v_2$ into the expression for work, we get $ A = \\dfrac{mv_1^2}{2} + \\dfrac{m^2v_1^2}{2M} = \\dfrac{mv_1^2}{2} \\left( 1 + \\dfrac{m}{M} \\right). $ If $M \\gg m$, then $A \\approx mv_1^2/2$, i.e., all the work done by the person goes into imparting kinetic energy to the stone. In the general case, the kinetic energies acquired by the interacting bodies are inversely proportional to their masses.

23. A rifle of mass $M = 3 \\, \\text{kg}$ is suspended horizontally by two parallel threads. When fired, it recoils and deflects upward by $h = 19.6 \\, \\text{cm}$. The mass of the bullet is $m = 10 \\, \\text{g}$. Determine the velocity $v_1$ of the bullet as it exits the barrel. $\"Description\"$

Answer

$v_1 \\approx 590$ m/s.

Solution

Based on the law of conservation of energy, we determine the recoil speed of the rifle after the shot $v$ (figure above): $Mgh = Mv^2/2$,\\qquad v^2 = 2gh.$ At the moment of firing, only the pressure forces of the powder gases act on the rifle and the bullet. These are internal forces. Therefore, the rifle-bullet system can be considered closed and the law of conservation of momentum can be applied to it. Before the shot, the momentum of the system was zero, therefore, after the shot, the total momentum of the system must also be zero: $M\\vec{v} + m\\vec{v}_1 = 0.$ Hence $v_1 = -\\dfrac{Mv}{m} = -\\dfrac{M}{m} \\sqrt{2gh}.$ The '$- sign means that the bullet's velocity is directed opposite to the rifle's recoil velocity.

24. A bullet flying horizontally with a velocity $v = 40 \\, \\text{m/s}$ hits a block suspended by a thread of length $l = 4 \\, \\text{m}$ and becomes embedded in it. Determine the angle $\\alpha$ by which the block will deflect if the mass of the bullet is $m_1 = 20 \\, \\text{g}$ and the mass of the block is $m_2 = 5 \\, \\text{kg}$.

Answer

$\\alpha \\approx 15^{\\circ}$.

Solution

The block, having received velocity $u$ upon impact by the bullet, rises in accordance with the law of conservation of mechanical energy to a height $h = u^2/(2g)$. On the other hand, $h = l(1 - \\cos \\alpha) = 2l \\sin^2 (\\alpha/2)$. Therefore $\\sin^2 (\\alpha/2) = u^2/(4gl)$; $u$ can be determined from the law of conservation of momentum: $m_1v = (m_1 + m_2)u$, $u = \\dfrac{m_1}{m_1 + m_2} v.$ Substituting the expression for $u$ into the equation for the sine of the angle, we find $ \\sin \\dfrac{\\alpha}{2} = \\dfrac{m_1v}{m_1 + m_2} \\dfrac{1}{\\sqrt{4gl}}. $

25. A bullet flying horizontally hits a ball suspended by a very light rigid rod and becomes embedded in it. The mass of the bullet is $n = 1000$ times smaller than the mass of the ball. The distance from the suspension point to the center of the ball is $l = 1 \\, \\text{m}$. Find the velocity $v$ of the bullet if the rod with the ball deflects by an angle $\\alpha = 10^\\circ$ after the collision.

Answer

$v = (1 + n) \\sqrt{4gl} \\sin(\\alpha/2) \\approx 550$ m/s.

Hint

See the solution to problem 24.

26. A bullet of mass $m_1 = 10 \\, \\text{g}$, flying horizontally with a velocity $v_1 = 600 \\, \\text{m/s}$, strikes a freely suspended wooden block of mass $m_2 = 0.5 \\, \\text{kg}$ and penetrates it by $s = 10 \\, \\text{cm}$. Find the resistive force $F_c$ of the wood acting on the bullet. How deep $s_1$ would the bullet penetrate if the block were fixed?

Answer

$F_c = \\dfrac{m_1 m_2 v_1^2}{2(m_1 + m_2)s} \\approx 1.8 \\times 10^4$ H; $s_1 = s \\dfrac{m_1 + m_2}{m_2} \\approx 10.2$ cm.

Solution

Let's write the laws of conservation of energy and momentum for the bullet-block system: $ \\dfrac{m_1 v_1^2}{2} = \\dfrac{(m_1 + m_2)v^2}{2} + F_c s; \\tag{1}$ $ m_1 v_1 = (m_1 + m_2)v, \\tag{2} $ where $v$ is the velocity acquired by the block. Solving the system of equations (1) and (2), we find $F_c$. When the bullet hits a fixed block, the law of conservation of energy is written differently: $m_1 v_1^2 / 2 = F_c s_1$. Substituting the expression for $F_c$ here, we find $s_1 = s \\dfrac{m_1 + m_2}{m_2}.$ Explain what significance the length of the thread suspension has.

27. A bullet of mass $m = 0.01 \\, \\text{kg}$ strikes a stationary ball of mass $M = 1 \\, \\text{kg}$ suspended by a long rigid rod hinged at the top. The angle between the bullet's trajectory and the rod is $\\alpha = 45^\\circ$. The collision is central. After the collision, the bullet embeds itself in the ball, and the ball rises to a height $h = 0.12 \\, \\text{m}$ relative to its initial position. Find the velocity $v$ of the bullet. Neglect the mass of the rod.

Answer

$v = \\dfrac{M+m}{m \\sin \\alpha} \\sqrt{2gh} \\approx 219$ m/s.

Solution

Since the ball at the moment of impact can only move in the horizontal direction (see figure), applying the law of conservation of momentum to the bullet-ball system, we must consider only the horizontal component of the momentum: $ (M+m)v_1 = mv \\sin \\alpha. \\tag{1} $ By the law of conservation of mechanical energy $(M+m)v_1^2/2 = (M+m)gh$, and consequently, $v_1^2 = 2gh$. Substituting this value into equation (1), we get the bullet speed $ v = \\dfrac{M+m}{m \\sin \\alpha} \\sqrt{2gh}. $ (see figure) $\"Description\"$

28. A pendulum consists of a thin rigid rod of length $l = 1.5 \\, \\text{m}$ with a steel ball of mass $M = 1 \\, \\text{kg}$ at its end. A steel ball of mass $m = 20 \\, \\text{g}$ flying horizontally with a velocity $v = 50 \\, \\text{m/s}$ strikes the pendulum ball. Determine the maximum deflection angle of the pendulum, assuming the collision is elastic and central. Neglect the mass of the rod.

Answer

$\\alpha = 2 \\arcsin \\dfrac{mv}{(m+M)\\sqrt{gl}} \\approx 30^{\\circ}$.

29. Two masses $m_1$ and $m_2$ of unequal weights are suspended on a thread thrown over a pulley. Find the acceleration of the center of mass of this system. Solve the problem in two ways: 1) using the law of conservation of energy, and 2) using the law of motion of the center of mass. Neglect the masses of the pulley and the thread.

Answer

$a_{cm} = \\left( \\dfrac{m_1 - m_2}{m_1 + m_2} \\right)^2 g$.

Solution

1st method. Write the law of conservation of energy for the center of mass of the system: $ (m_1 + m_2) g h_{cm} = m_1gh - m_2gh, $ where $m_1$ and $m_2$ are the masses of the weights; $h_{cm}$ is the height to which the center of mass of the system descends (rises); $h$ is the distance by which each of the weights has risen (descended). But $h_{cm} = a_{cm}t^2/2$; $h = at^2/2$, where $a$ is the acceleration of the weights: $ a = \\dfrac{m_1 - m_2}{m_1 + m_2} g. $ Therefore $ a_{cm} = \\dfrac{m_1 - m_2}{m_1 + m_2} a = \\left( \\dfrac{m_1 - m_2}{m_1 + m_2} \\right)^2 g. $

2nd method. According to the law of motion of the center of mass, one can write
$ (m_1 + m_2) a_{cm} = m_1g + m_2g - 2T, \\tag{1} $
where $a_{cm}$ is the acceleration of the center of mass; $T$ is the tension force of the thread. From the equations of motion of the weights, it is easy to find that
$ T = \\dfrac{2m_1m_2}{m_1 + m_2} g. \\tag{2} $
Substitute expression (2) into equation (1):
$ a_{cm} = \\left( \\dfrac{m_1 - m_2}{m_1 + m_2} \\right)^2 g. $

30. A hammer of mass $m = 1.5 \\, \\text{t}$ strikes a hot workpiece lying on an anvil and deforms it. The mass of the anvil with the workpiece is $M = 20 \\, \\text{t}$. Determine the efficiency $\\eta$ of the hammer's strike, assuming the collision is inelastic. Consider the work done in deforming the workpiece as useful.

Answer

$\\eta = M/(M+m) \\approx 93\\%$.

Solution

By definition, efficiency is $ \\eta = A_{useful} / A_{expended}. \\tag{1} $ In this case, the useful work will be the work $A_{useful}$ on the deformation of the blank, and the expended work $A_{expended}$ is the kinetic energy that the hammer possessed before the impact: $A_{expended} = T_{initial} = mv^2/2$. The deformation of the blank ends at the moment when the speed of the hammer equals the speed of the anvil, i.e., at the moment when the impact itself ends. The hammer-blank-anvil system can be considered closed with good accuracy, since the anvil is mounted on a foundation through shock absorbers, damping the vibrations of the anvil, and during the short impact time, a large additional force cannot develop in them due to the small displacement of the anvil during this time. Thus, the momentum that the hammer initially possessed is conserved, and as a result of the inelastic impact, it transfers to the system as a whole: $ mv = (m + M)u, \\tag{2} $ where $u$ is the joint speed of the system at the moment the impact ends. The 'residual' kinetic energy $T_k$ associated with this motion is the unused (lost) part of the initial energy $T_{initial}$. Then the useful work $A_{useful} = T_{initial} - T_k$, and the efficiency is $ \\eta = \\dfrac{A_{useful}}{A_{expended}} = \\dfrac{T_{initial} - T_k}{T_{initial}} = 1 - \\dfrac{T_k}{T_{initial}}. $ Using formula (2), we get $T_k = (M+m)u^2/2 = m^2v^2/[2(M+m)]$, and finally $ \\eta = 1 - \\dfrac{m^2v^2/[2(M+m)]}{mv^2/2} = 1 - \\dfrac{m}{M+m} = \\dfrac{M}{M+m}. $

31. A body of mass $m_1$ makes an inelastic collision with a stationary body of mass $m_2$. Find the fraction $q$ of kinetic energy lost in the collision.

Answer

$q = m_2/(m_1 + m_2)$.

Hint

The fraction of lost energy is $q = \\Delta T / T = (m_1v_1^2 - (m_1 + m_2)u^2) / (m_1v_1^2)$, where $v_1$ is the speed of mass $m_1$ before the impact; $u$ is the speed of both bodies after the impact; $u$ is easily found from the law of conservation of momentum: $m_1v_1 = (m_1 + m_2)u$. This problem is actually identical to the previous one. The lost energy $\\Delta T$ went into the work of deformation of the bodies during the impact.

32. A short block of mass $m$ is dropped from a small height onto the front edge of a platform of mass $M$ moving horizontally without friction at velocity $v$. What is the minimum length $l$ of the platform required for the block not to fall off, given the coefficient of friction $k$ between block and platform? How much heat $Q$ will be generated in this process?

Answer

$l_{min} = \\dfrac{v^2}{2kg(1 + m/M)}$; $Q = \\dfrac{Mv^2}{2} \\dfrac{m}{m+M}$.

Solution

In the horizontal direction, only friction forces act, which are internal forces for the platform-block system. Applying the laws of conservation of momentum and energy, we obtain the shortest solution: $ Mv = (M+m)u; \\tag{1} $ $ \\dfrac{(M+m)u^2}{2} - \\dfrac{Mv^2}{2} = -kmgS, \\tag{2} $ where $S$ is the distance traveled by the block on the platform. The second expression equates the change in kinetic energy of the system (difference between final and initial values) to the work of the friction forces, which serves as a measure of the conversion of kinetic energy into internal energy (heat). Excluding $u$ from (1) and (2), we find $ S = \\dfrac{M^2 v^2}{2kg(M+m)^2}. $ (Note: The original text has a different expression for S. Re-deriving: From (1), $u = Mv/(M+m)$. Substitute into (2): $\\dfrac{(M+m)}{2} \\dfrac{M^2v^2}{(M+m)^2} - \\dfrac{Mv^2}{2} = -kmgS$. $\\dfrac{M^2v^2}{2(M+m)} - \\dfrac{Mv^2(M+m)}{2(M+m)} = -kmgS$. $\\dfrac{Mv^2(M - (M+m))}{2(M+m)} = -kmgS$. $\\dfrac{Mv^2(-m)}{2(M+m)} = -kmgS$. $S = \\dfrac{mMv^2}{2kg(M+m)}$. The condition of the problem is met if $S \\le l$. Therefore, $l_{min} = S = \\dfrac{mMv^2}{2kg(M+m)}$. The amount of heat released $Q = kmgS$ is equal to the decrease in kinetic energy of the system: $Q = \\dfrac{Mv^2}{2} - \\dfrac{(M+m)u^2}{2} = \\dfrac{Mv^2}{2} - \\dfrac{M^2v^2}{2(M+m)} = \\dfrac{Mv^2(M+m) - M^2v^2}{2(M+m)} = \\dfrac{Mmv^2}{2(M+m)}$. Solve the problem another way, determining the accelerations of the block and the platform.

33. A body of mass $m = 1$ kg lying on a long horizontal platform of a moving cart is given velocity $v = 10$ m/s. The coefficient of friction between body and platform is $k = 0.2$. How far will the cart move by the time the body stops on it? How much heat will be released during the body's motion along the platform? The cart rolls on frictionless rails and has mass $M = 100$ kg.

Answer

$s = \\dfrac{mMv^2}{2kg(m+M)^2} \\approx 0.25$ m; $Q = \\dfrac{mMv^2}{2(m+M)} \\approx 50$ J.

Hint

From the law of conservation of momentum $mv = (m+M)u$, we find $u = mv/(M+m)$ - the speed of the cart after the body stops. The path traveled by the cart is $s = u^2/(2a)$, where $a = kmg/M$ is the acceleration of the cart. The amount of heat released is determined from the law of conservation of energy: $ Q = \\dfrac{mv^2}{2} - \\dfrac{(m+M)u^2}{2} = \\dfrac{mMv^2}{2(m+M)}. $

34. Two masses $m_1 = 10$ kg and $m_2 = 15$ kg are suspended on threads of length $l = 2$ m such that they touch each other. The smaller mass is displaced by angle $\\alpha = 60^\\circ$ and released. To what height will both masses rise after collision? Treat the collision as perfectly inelastic. How much heat $Q$ will be released?

Answer

$h \\approx 0.16$ m; $Q \\approx 58.8$ J.

Solution

The height $h$ of the rise of the weights after the impact is found from the law of conservation of mechanical energy: $ (m_1 + m_2)gh = (m_1 + m_2) \\dfrac{u^2}{2}. \\tag{1} $ We find the speed $u$ using the law of conservation of momentum: $ m_1v_1 = (m_1 + m_2)u. \\tag{2} $ From (1), $h = u^2/(2g)$. Substituting $u$ from (2): $h = \\dfrac{1}{2g} \\left( \\dfrac{m_1v_1}{m_1+m_2} \\right)^2$. The amount of heat $Q$ released during the impact is $ Q = \\dfrac{m_1v_1^2}{2} - \\dfrac{(m_1+m_2)u^2}{2} = \\dfrac{m_1v_1^2}{2} - (m_1+m_2)gh = \\dfrac{m_1v_1^2}{2} \\left( 1 - \\dfrac{m_1}{m_1+m_2} \\right) = \\dfrac{m_1m_2v_1^2}{2(m_1+m_2)}. $

35. A ball moves between two very heavy vertical parallel walls, colliding with them elastically. One wall is fixed, the other moves away with constant horizontal velocity $u_x = 0.5$ m/s. Determine the number of collisions $n$ and the final velocity $v_x$ of the ball if its initial velocity was $v_{0x} = 19.5$ m/s.

Answer

$n = 19$ impacts; $v_x = 0.5$ m/s.

Hint

See problem 16. In the coordinate system associated with the moving wall, the speed of the ball before the first collision was $v_{0x} - u_x$. After the impact, it reverses direction, and consequently, in the stationary coordinate system, it will be $-(v_{0x} - u_x) - u_x = v_{0x} - 2u_x$. After $n$ collisions with the moving wall, the speed of the ball is $v_{nx} = v_{0x} - 2nu_x.$ Collisions will continue until the speed of the ball becomes less than or equal to the speed of the wall $u_x$.

36. Two balls are suspended on parallel threads of equal length so they touch. Their masses are $m_1 = 0.2$ kg and $m_2 = 100$ g. The first ball is displaced so its center rises $h = 4.5$ cm and released. Find the heights reached after collision if: a) collision is elastic; b) collision is inelastic.

Answer

a) $h_1 = 5 \\times 10^{-3}$ m, $h_2 = 0.08$ m; b) $H = 0.02$ m.

Solution

a)** Write the conservation laws of mechanical energy and of momentum for the two‐ball system, expressing the post‐collision speeds in terms of the heights $h_1$ and $h_2$ to which they rise: $ m_1 g\\,h \\;=\\; m_2 g\\,h_2 \\;+\\; m_1 g\\,h_1, \\tag{1} $ $ m_1\\sqrt{2gh}\\;=\\;m_2\\sqrt{2gh_2}\\;+\\;m_1\\sqrt{2gh_1}. \\tag{2} $ Transforming $(1)$ and $(2)$ as in Problem 28 and solving them jointly gives $ \\sqrt{h_2} \\;=\\;\\sqrt{h_1} \\;+\\;\\sqrt{h}. \\tag{3} $ Substitute $(3)$ into $(2)$ to find $ h_1 \\;=\\; h\\,\\Bigl(\\frac{m_1 - m_2}{m_1 + m_2}\\Bigr)^{2}. \\tag{4} $ Finally, using $(1)$ together with $(4)$ yields $ h_2 \\;=\\; 4\\,h\\,\\Bigl(\\frac{m_1}{m_1 + m_2}\\Bigr)^{2}. $

b) (See Problem 34.)
$
H ;=; h,\\Bigl(\\frac{m_1}{m_1 + m_2}\\Bigr)^{2}.
$

37. By what factor does a helium atom's speed decrease after a central elastic collision with a stationary hydrogen atom, given hydrogen's mass is four times smaller than helium's?

Answer

$n = 5/3$.

Hint

The speed of the helium atom after the collision $v_1' = v_1 \\dfrac{m_1 - m_2}{m_1 + m_2}$ (see problem 5.28), where $m_1$ is the mass of the helium atom; $m_2$ is the mass of the hydrogen atom; $v_1$ is the speed of the helium atom before the collision; $n = v_1 / |v_1'| = \\dfrac{m_1+m_2}{|m_1-m_2|} = \\dfrac{m_1/m_2 + 1}{|m_1/m_2 - 1|}$. (Note: $m_1 \\approx 4u, m_2 \\approx 1u$, so $n = (4+1)/|4-1| = 5/3$).

38. A ball lying on a smooth horizontal surface is struck by another identical ball moving horizontally. The collision is elastic and central. Plot the fraction of energy transferred versus the mass ratio $\\alpha = m_1/m_2$.

Answer

$E_1 = \\dfrac{4\\alpha}{(1+\\alpha)^2} E_0$ (see figure).

Hint

From the laws of conservation of energy and momentum, it follows that the energy transferred to the initially stationary ball is $ E_1 = \\dfrac{4\\alpha}{(1+\\alpha)^2} E_0, $ where $E_0$ is the total energy of the balls (initial kinetic energy of the first ball); $\\alpha$ is the ratio of the mass of the incident ball to the mass of the initially stationary ball. $E_1$ is maximum when the expression $\\dfrac{(1+\\alpha)^2}{\\alpha} = \\dfrac{1}{\\alpha} + \\alpha + 2$ is minimum. But $\\dfrac{1}{\\alpha} + \\alpha \\ge 2$ (arithmetic mean-geometric mean inequality), and $\\dfrac{1}{\\alpha} + \\alpha = 2$ if $1/\\alpha = \\alpha$, i.e., $\\alpha = 1$ (since $\\alpha > 0$). Therefore, $E_1$ is maximum when $\\alpha = 1$. In this case $E_1 = E_0$. (see figure) $\"Description\"$

39. To obtain slow neutrons, they are passed through hydrogen-containing substances (e.g., paraffin). Find the maximum fraction of kinetic energy a neutron (mass $m_0$) can transfer to: a) a proton (mass $m_0$); b) a lead nucleus (mass $207 m_0$). Maximum transfer corresponds to central elastic collision.

Answer

a) 100%. In an elastic collision of particles with equal mass, they exchange velocities;

b) 1.9%.

Hint

See problems 28, 37 and 38.

40. Two perfectly elastic balls of masses $m_1$ and $m_2$ move along the same line with velocities $v_1$ and $v_2$. During collision they deform, converting some kinetic energy to potential energy of deformation. Find the maximum potential energy of deformation during the collision.

Answer

$\\Pi_{\\text{max}} = \\dfrac{1}{2} \\dfrac{m_1m_2}{m_1+m_2} (v_1 - v_2)^2$.

Solution

At the moment of maximum deformation, the balls move together with the same velocity $u$. For this moment, the laws of conservation of mechanical energy and momentum take the following form: $ \\dfrac{m_1v_1^2}{2} + \\dfrac{m_2v_2^2}{2} = \\dfrac{(m_1+m_2)u^2}{2} + \\Pi_{\\text{max}}; \\tag{1} $ $ m_1v_1 + m_2v_2 = (m_1+m_2)u. \\tag{2} $ From equation (2), $u = \\frac{m_1v_1 + m_2v_2}{m_1+m_2}.$ Substituting this expression into equation (1), we obtain $\\Pi_{\\text{max}} = \\dfrac{1}{2} \\dfrac{m_1m_2}{m_1+m_2} (v_1 - v_2)^2.$

41. A streamlined body of density $\\rho_1$ falls through air from height $h$ onto a liquid surface (density $\\rho_2$, where $\\rho_1 < \\rho_2$). Determine: immersion depth $h_1$, immersion time $t$, and acceleration $a$. Neglect liquid resistance.

Answer

$h_1 = h \\dfrac{\\rho_1}{\\rho_2 - \\rho_1}$;

$t = \\sqrt{\\dfrac{2h}{g}} \\dfrac{\\rho_1}{\\rho_2 - \\rho_1}$;

$a = g \\dfrac{\\rho_2 - \\rho_1}{\\rho_1}$.

Solution

For a body that has fallen from height $h$ above the surface of a liquid and submerged to depth $h_1$, one can write the law of conservation of energy in the form $ mg(h+h_1) = F_{\\text{buoy}} h_1. \\tag{1} $ On the right side of equality (1) is the work of the buoyant force $F_{\\text{buoy}}$ during the submersion of the body. According to Archimedes' principle $ F_{\\text{buoy}} = V_{body} g \\rho_2 = mg \\rho_2 / \\rho_1, \\tag{2} $ where $m = V_{body} \\rho_1$ is the mass of the body. Substituting expression (2) into equation (1), we get $h_1 = h \\rho_1 / (\\rho_2 - \\rho_1)$. The time $t$ can be found from the equality of the change in the body's momentum upon immersion to the impulse of the force: $ (F_{\\text{buoy}} - mg)t = m\\sqrt{2gh}. \\tag{3} $ From relations (3) and (2) we find $t = \\dfrac{m\\sqrt{2gh}}{mg(\\rho_2/\\rho_1 - 1)} = \\sqrt{\\dfrac{2h}{g}} \\dfrac{\\rho_1}{\\rho_2 - \\rho_1}$. The acceleration (according to Newton's second law) is equal to $a = \\dfrac{F_{\\text{buoy}} - mg}{m} = g \\dfrac{\\rho_2 - \\rho_1}{\\rho_1}$.

42. A mass $m$ hangs on a thread of length $l$. Find the minimum height from which it must fall to break the thread, given that the minimum mass $M$ needed to break the thread stretches it by 1% of $l$ at breaking point. Assume Hooke's law holds until breaking.

Answer

$h_{min} = 0.01 Ml / (2m)$.

Solution

The weight $m$ will perform work in stretching the thread, where the tension force is directly proportional to the elongation $\\Delta l$ (according to Hooke's law). This work ends with the breaking of the thread and can be written as (see problem 17 in Chapter: Work, Power, Energy): $ A = F_{\\text{max}} \\Delta l / 2, \\tag{1} $ where $F_{\\text{max}} = Mg$; $\\Delta l = 0.01 l$ (according to the condition). The minimum height $h_{min}$, falling from which the weight will break the thread, is determined using the law of conservation of energy: $ mgh_{min} = A \\tag{2} $ (we neglect the elongation of the thread during its tension here). Substituting expression (1) into equation (2), we get $h_{\\text{min}} = F_{\\text{max}} \\frac{\\Delta l}{2mg} = Mg \\frac{0.01l}{2mg} = 0.01 Ml / (2m).$

43. Find the maximum range $s$ of a water jet from a hose of diameter $d = 4$ cm where the piston is acted upon by force $F = 30$ N. Liquid density $\\rho = 1000$ kg/m³. Neglect air resistance.

Answer

$s \\approx 4.9$ m.

Solution

The range of flight will be maximum at an angle of inclination of the initial velocity of the stream to the horizon equal to $45^{\\circ}$, and is determined by the formula $s = v^2/g$ (see problem 63 in Chapter: Kinematics). The speed $v$ of the stream can be determined by equating the kinetic energy $mv^2/2$ of the liquid ejected from the syringe to the work of the piston $Fl$, where $l$ is its displacement. This can be done because the condition $S_{\\text{hole}} \\ll S_{\\text{piston}}$ allows neglecting the kinetic energy that the liquid possessed while moving inside the wide part of the syringe before the outlet orifice (see problem 44). Since the liquid is practically incompressible, the mass of the ejected liquid is $m = \\rho Sl$, where $S = \\pi d^2/4$ is the area of the piston; $\\rho$ is the density of the liquid. $Fl = \\rho Sl v^2/2$, whence $v = \\sqrt{2Fl/(\\rho Sl)} = \\sqrt{2F/(\\rho S)} = \\sqrt{8F/(\\pi d^2 \\rho)}$. Substituting this expression into the formula for the range of the stream, we find $s = 8F / (\\rho g \\pi d^2)$.

44. A horizontal cylinder of diameter $D$ is filled with water. With what velocity $u$ does the piston move when acted upon by force $F$, while water jets out through a bottom hole of diameter $d$? Neglect friction and gravity. Liquid density $\\rho$.

Answer

$u = \\dfrac{d^2}{D^2} \\sqrt{\\dfrac{8F}{\\pi \\rho (D^4 - d^4)}}$.

Hint

The work done by the piston, moving with speed $u$ during time $\\Delta t$, goes into increasing the kinetic energy of the liquid mass $\\Delta m = \\rho S u \\Delta t$ (where $S = \\pi D^2/4$) flowing out of the opening during this time: $Fu \\Delta t = \\dfrac{\\Delta m v^2}{2} - \\dfrac{\\Delta m u^2}{2}$. The speed $v$ with which water flows out of the opening is related to $u$ by the continuity equation $v/u = S/s = D^2/d^2$ (where $s = \\pi d^2/4$).

45. Two beads of masses $m_1$ and $m_2$ can slide without friction on a smooth horizontal circular wire. Initially connected by a thread with a compressed spring between them, the thread is burned. After motion begins, the spring is removed. Where will the beads collide for the 11th time? Collisions are perfectly elastic. Neglect spring mass.

Answer

$l_1/l_2 = m_2/m_1$, where $l_1$ and $l_2$ are the lengths of the arcs of the ring from the starting point to the point of the 11th collision.

Hint

The initial momenta of the beads are equal $m_1v_{01} = m_2v_{02},$ since equal forces acted on them for the same time. From the laws of conservation of energy and momentum, it follows that after the collision, the momenta of the balls will not change in magnitude, but only reverse their direction. The 11th collision will occur at the same place as the first (like any other odd collision), i.e., at the point dividing the ring together with the starting point in the ratio $m_2/m_1$.

46. A proton (mass $m$) with velocity $v_0$ collides with a stationary atom (mass $M$), then moves oppositely with $0.5v_0$ while the atom becomes excited. Find the atom's velocity $v$ and excitation energy $E$.

Answer

$v = \\dfrac{3}{2} \\dfrac{m}{M} v_0$; $E = \\dfrac{3mv_0^2(M-3m)}{8M}$.

Solution

From the law of conservation of momentum $mv_0 = -\\dfrac{1}{2}mv_0 + Mv.$ We find $v = \\dfrac{3}{2} \\dfrac{m}{M} v_0.$

The excitation energy $E$ follows from energy conservation:
$
\\begin{aligned}
E&=\\frac{m,v_0^2}{2}-\\frac{m,(v_0/2)^2}{2}-\\frac{M,v^2}{2}\\
&=\\frac{3,m,v_0^2,(M-3m)}{8,M}.
\\end{aligned}
$

47. A stationary nucleus decays into three fragments (masses $m_1$, $m_2$, $m_3$) with total kinetic energy $E_0$. Find their velocities if the velocity directions are 120° apart.

Answer

$v_1 = \\sqrt{\\dfrac{2E_0 m_2 m_3}{m_1(m_1m_2 + m_1m_3 + m_2m_3)}}$;

$v_2 = \\sqrt{\\dfrac{2E_0 m_1 m_3}{m_2(m_1m_2 + m_1m_3 + m_2m_3)}}$;

$v_3 = \\sqrt{\\dfrac{2E_0 m_1 m_2}{m_3(m_1m_2 + m_1m_3 + m_2m_3)}}$;

or in general form $v_i = \\dfrac{1}{m_i} \\sqrt{\\dfrac{2E_0 m_1 m_2 m_3}{m_1m_2 + m_1m_3 + m_2m_3}}$.

Hint

Since the initial momentum of the nucleus is zero, the momenta of the three fragments must be equal—only then, with 120° between each pair of their velocity vectors, will their vector sum vanish: $ m_1v_1 = m_2v_2 = m_3v_3. \\tag{1 \\& 2} $ From energy conservation one has $ m_1v_1^2 + m_2v_2^2 + m_3v_3^2 = 2E_0. \\tag{3} $ Solving equations (1)–(3) together yields the required result.

48. One ball strikes another identical stationary ball not along their line of centers. At what angle will they separate if perfectly elastic and smooth?

Answer

At a right angle.

Solution

1st method. Write the law of conservation of momentum in vector form and the law of conservation of kinetic energy: $ m\\vec{v}_1 = m\\vec{v}_1' + m\\vec{v}_2', \\tag{1} $ $ \\dfrac{1}{2}mv_1^2 = \\dfrac{1}{2}mv_1'^2 + \\dfrac{1}{2}mv_2'^2, \\tag{2} $ where $\\vec{v}_1$ is the initial velocity of the ball; $\\vec{v}_1'$ and $\\vec{v}_2'$ are the velocities of the balls after the impact. Canceling the mass in both parts of equations (1) and (2), we get $ \\vec{v}_1 = \\vec{v}_1' + \\vec{v}_2', $ $ v_1^2 = v_1'^2 + v_2'^2. $ The simultaneous fulfillment of these equalities is possible if the vectors $\\vec{v}_1'$ and $\\vec{v}_2'$ form a right angle (Pythagorean theorem). 2nd method. Direct consideration of the interaction of the balls leads to the same result. Indeed, let's decompose the momentum of the first ball into components along the x-axis (coinciding with the line connecting the centers of the balls) and the y-axis. Along the x-axis, a head-on collision of the balls occurs. It was previously shown that in this case, for equal masses, the balls exchange velocities. Thus, the second ball acquires velocity $v_x$. Along the y-axis, there is no interaction between the balls (the balls are perfectly smooth). Therefore, the velocity of the first ball after the impact is $v_y$. The velocities $v_x$ and $v_y$ are perpendicular.

49. Two balls $A$ and $B$ of unknown masses collide elastically. Before collision, $A$ was stationary and $B$ moved with velocity $v$. After collision, $B$ moves at $0.5v$ perpendicular to its initial motion. Determine $As post-collision direction and velocity $v_A$.

Answer

At an angle $\\alpha = \\arctan(0.5)$ to the direction of motion of ball $B$ before the collision, $v_A = 0.66v$.

Solution

Apply the law of conservation of momentum in projections onto the axes:

Along the $x$-axis (direction of $\\vec{v}$):
$
m_B v = m_A v_{Ax}
$
Along the $y$-axis:
$
m_B \\cdot 0.5v = m_A v_{Ay}
$

Here, $m_A$ and $m_B$ are the masses of the balls; $v_{Ax}$ and $v_{Ay}$ are the mutually perpendicular components of the velocity of ball $A$ after the collision. The ratio $v_{Ay}/v_{Ax} = 0.5$ gives the tangent of the angle of deflection of ball $A$ from the $x$-axis. The total velocity of ball $A$ after the collision is:
$
v_A = \\sqrt{v_{Ax}^2 + v_{Ay}^2} \\approx 1.1v,\\frac{m_B}{m_A}
$

The ratio $m_B / m_A$ can be found from energy conservation:
$
m_B v^2 = m_B \\cdot 0.25v^2 + 1.25v^2 \\cdot \\frac{m_B^2}{m_A^2} \\cdot m_A
$

Solving gives:
$
\\frac{m_B}{m_A} = \\frac{3}{5}
\\quad\\Rightarrow\\quad v_A = 0.66v.
$

50. When helium is bombarded by α-particles with energy $E_0$, an α-particle is deflected by $\\varphi = 60^\\circ$. Assuming elastic collision, find the post-collision energies $W_\\alpha$ (α-particle) and $W_{He}$ (nucleus). Thermal motion energy is much less than $E_0$.

Answer

$W_p = 1/4 E_0$; $W_{He} = 3/4 E_0$. See solution to problem 5.49.

51. A soft lead ball collides with an identical stationary ball. After collision, the second ball moves at angle $\\alpha$ to the initial direction. Find the separation angle $\\beta$ and the fraction of kinetic energy $T$ converted to heat $Q$.

Answer

$\\beta = \\arctan(2 \\tan \\alpha)$; $Q/T = \\dfrac{1}{2} \\cos^2 \\alpha$.

Hint

In the direction passing through the centers of the balls, an inelastic collision occurs such that the component of the momentum of the incident ball $mv \\cos \\alpha$ is distributed equally between the balls and will be equal to $1/2 mv \\cos \\alpha$. The tangential component of the momentum of the incident ball $mv \\sin \\alpha$ does not change during the collision, as the balls are smooth.

52. A ball of mass $m$ moving at velocity $v$ strikes a stationary ball of mass $m/2$. After elastic collision, it continues at $\\alpha = 30^\\circ$ to its initial direction. Find both balls' post-collision velocities.

Answer

**5.52.** $v_1 = \\sqrt{3}v/3;\\quad v_2 = 2\\sqrt{3}v/3;\\quad \\beta = 30^\\circ.$

Solution.
Let the velocities of the balls with masses $m$ and $m/2$ after the collision be $v_1$ and $v_2$, and let the angle between directions of $v_2$ and $v$ be $\\beta$. Write the momentum conservation laws in projections on the $x$-axis (taken along the initial velocity $v$) and the $y$-axis, as well as the conservation of energy:

$
mv = mv_1 \\cos\\alpha + \\frac{m}{2}v_2 \\cos\\beta
\\quad \\Rightarrow \\quad
2v - 2v_1 \\cos\\alpha = v_2 \\cos\\beta \\tag{1}
$

$
0 = mv_1 \\sin\\alpha - \\frac{m}{2}v_2 \\sin\\beta
\\quad \\Rightarrow \\quad
2v_1 \\sin\\alpha = v_2 \\sin\\beta \\tag{2}
$

$
mv^2 = mv_1^2 + \\frac{m}{2}v_2^2
\\quad \\Rightarrow \\quad
2v^2 - 2v_1^2 = v_2^2 \\tag{3}
$

Solving equations (1), (2), and (3) together:
(Squaring equations (1) and (2), summing them, and substituting into (3), using $\\alpha = 30^\\circ$), we get a quadratic:
$
3v_1^2 - 2\\sqrt{3}vv_1 + v^2 = 0
$

Solving this gives:
$
v_1 = \\frac{\\sqrt{3}v}{3}
$

From (3) we find:
$
v_2 = \\frac{2\\sqrt{3}v}{3}
$

From (2) we find:
$
\\beta = 30^\\circ
$

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1. Find the linear speed $v$ of the Earth during its orbital motion. The average radius of the Earth's orbit is $R = 1.5 \\times 10^8$ km.

Answer

$v \\approx 30$ km/s.

2. An airplane propeller with a radius of 1.5 m rotates during landing with a frequency of 2000 min$^{-1}$. The landing speed of the airplane relative to the Earth is 162 km/h. Determine the speed of a point on the tip of the propeller. What is the trajectory of motion of this point?

Answer

$v \\approx 316$ m/s; the point describes a helical line with a pitch $h \\approx 1.35$ m.

3. A disk of radius $R$ rolls without slipping with a constant speed $v$. Find the geometric locus of points on the disk that currently have a speed $v$.

Answer

The velocity $v$ at this moment is possessed by points lying on an arc of radius $R$, whose center is at the point of contact between the disk and the plane (instantaneous center of rotation).

4. A cylindrical roller of radius $R$ is placed between two parallel rails. The rails move in the same direction with speeds $v_1$ and $v_2$ (see figure). Determine the angular velocity of rotation of the roller and the speed of its center, assuming no slipping. Solve the problem for the case where the speeds of the rails are directed in opposite directions. $\"Description\"$

Solution

Let the roller move translationally with velocity $v$ and rotate about its axis with angular velocity $\\omega$. The linear velocity of points on the roller in the coordinate system associated with its axis is $\\omega R$. The points of contact between the roller and the racks also have this velocity. Since there is no slipping, this is also the velocity of the racks. In the coordinate system associated with the Earth, the velocities of the racks are equal to the sum of their velocities relative to the cylinder axis and the velocity of the cylinder axis relative to the Earth. Therefore, $v_1 = v + \\omega R,\\qquad v_2 = v - \\omega R.$ From this, $v = \\frac{1}{2}(v_1 + v_2),\\qquad \\omega = \\frac{1}{2R}(v_1 - v_2).$ For the case where the rack velocities are directed in opposite directions ($v_1 = v + \\omega R$, $v_2 = \\omega R - v$), we get: $v = \\frac{1}{2}(v_1 - v_2);\\qquad \\omega = \\frac{v_1 + v_2}{2R}.$

5. A hoop of radius $R$ rolls without slipping along a horizontal plane with a constant speed $v_c$. What are the speeds and accelerations of various points on the hoop relative to the Earth? Express the speed as a function of the angle between the vertical and the line drawn between the point of contact of the hoop with the plane and the given point on the hoop.

Answer

$v_A = 2v_c \\cos \\alpha$. The accelerations of all points on the rim are centripetal and equal to $a_{cp} = v_c^2/R$.

Solution

The angular velocities of rotation of all points on the rim relative to the axis passing through the point of contact of the hoop with the plane (instantaneous center of rotation) are identical and equal to $\\omega = v_c/R$. The linear velocity of point $A$ is $v_A = \\omega r = \\omega (2R \\cos \\alpha) = \\frac{v_c}{R} 2R \\cos \\alpha = 2v_c \\cos \\alpha$ and is perpendicular to OA (see figure). Since the translational motion of the hoop is uniform, its points only have centripetal acceleration. This becomes obvious when considering the motion of the hoop in an inertial frame of reference moving with velocity $v_c$. $\"Description\"$

6. A car moves with a speed $v = 60$ km/h. With what frequency $n$ do its wheels rotate if they roll on the highway without slipping, and the outer diameter of the tires is $d = 60$ cm? Find the centripetal acceleration $a_{cp}$ of the outer layer of rubber on its tires.

Answer

$n \\approx 8.84 \\text{ s}^{-1}$;

$a_{cp} \\approx 926 \\text{ m/s}^2$ (this acceleration is 94.4 times greater than $g$).

7. A thin-walled cylinder, rotating with speed $\\omega_0$ about its axis, is placed on a horizontal plane. What will be the speed of the cylinder's axis when the slipping of the cylinder relative to the plane stops?

Answer

$v = v_0/2$.

Solution

1st method. When slipping, a constant friction force acts on the cylinder from the plane, which accelerates its translational motion and decelerates its rotational motion. After time $t$, the translational velocity will be $v = at$, and the linear velocity of rotational motion relative to the cylinder axis will be $v_r = v_0 - at$ (since the cylinder is thin-walled, all elementary masses have the same acceleration $a$ relative to the cylinder axis, equal in magnitude to the acceleration of the translational motion).

The velocity of the point on the cylinder touching the plane is equal to the difference in velocities $v_r - v$. Slipping will stop when this velocity becomes zero:
$v_r - v = (v_0 - at) - at = 0,$
whence $at = v_0/2$. Thus, the translational velocity (velocity of the cylinder axis) at this moment will be $v_0/2$ and will not change further.

2nd method. While slipping, the cylinder performs work against the friction force $F$ applied to it: $A = F S_{sl},$
where $S_{sl}$ is the path traveled by the point (more precisely, the line) of application of the friction force on the cylinder surface during its sliding on the plane over time $t$: from the moment of contact, when the sliding velocity $v_{sl} = v_0$, until the moment slipping stops ($v_{\\text{sl}} = 0$). Since the motion is uniformly accelerated, the average sliding velocity is $v_0/2$. Thus,
$A = F S_{\\text{sl}} = F v_0 \\frac{t}{2}.$
According to the work-energy theorem, the work $A$ equals the decrease in the kinetic energy of the cylinder:
$\\frac{mv_0^2}{2} - (\\frac{mv^2}{2} (\\text{trans}) + \\frac{mv^2}{2} (\\text{rot})) = F \\frac{v_0 t}{2}$
(cylinder is thin-walled, therefore $T_{trans} = T_{rot}$, see problem 50). The friction force $F$, which brakes the cylinder's rotation to velocity $v$ over time $t$, simultaneously imparts translational motion with the same velocity $v$. Writing Newton's second law for the translational motion in impulse form $Ft = mv$ and substituting $Ft$ with $mv$ in equation (1), we obtain an equation from which we find $v = v_0/2$.

8. Does the resultant of all forces applied to a body moving uniformly in a circle perform work?

Answer

No. This force, according to the problem statement, ensures the uniform rotation of the body, is a constant magnitude centripetal force, which at each moment in time is directed perpendicular to the displacement of the body and does no work.

9. A load of mass $m$ can slide without friction along a horizontal rod, which rotates around a vertical axis passing through one of its ends. The load is connected to this end of the rod by a spring with spring constant $k$. At what angular velocity $\\omega$ will the spring stretch by 50% of its original length?

Answer

$\\omega = \\sqrt{\\frac{1}{3} \\frac{k}{m}}$.

Solution

The centripetal acceleration $a_{\\text{cp}} = \\omega^2 (l_0 + \\Delta l) = 1.5 l_0 \\omega^2$ ($l_0$ is the initial length of the spring; $\\Delta l$ is its extension) is imparted to the mass by the tension force of the spring, which, according to Hooke's law, is $F = k \\Delta l = 0.5 l_0 k.$ Therefore, $1.5 m \\omega^2 l_0 = 0.5 l_0 k.$ Hence $\\omega = \\sqrt{\\frac{1}{3} \\frac{k}{m}}.$

10. Two point masses $m_1$ and $m_2$ are attached to strings and are located on a perfectly smooth table. The distances from them to the fixed end of the string are $l_1$ and $l_2$ respectively (see figure). The system rotates in a horizontal plane around the axis passing through the fixed end, with an angular velocity $\\omega$. Find the tension forces in the sections of the string $T_1$ and $T_2$. $\"Description\"$

Answer

$T_1 = (m_1 l_1 + m_2 l_2) \\omega^2$;

$T_2 = m_2 \\omega^2 l_2$.

Hint

The force $T_2$ imparts centripetal acceleration to the outermost mass; the force $(T_1 - T_2)$ imparts it to the other mass (see the following figure). $\"Description\"$

11. A person sits on the edge of a circular horizontal platform of radius $R = 4$ m. With what frequency $n$ must the platform rotate around a vertical axis so that the person cannot stay on it with a coefficient of friction $k = 0.27$?

Answer

$n = 6.75 \\text{ min}^{-1}$.

Solution

A person cannot stay on the platform if the maximum possible static friction force with the platform is insufficient to provide the necessary centripetal acceleration, i.e., $F_{fr} = kmg < m\\omega^2 R$ or $n = \\omega/(2\\pi) > \\frac{1}{2\\pi} \\sqrt{kg/R}.$

12. A body of mass $m$ is on a horizontal disk at a distance $r$ from the axis. The disk begins to rotate with a small acceleration. Construct a graph of the dependence of the radial component of the friction force acting on the body on the angular velocity of the disk's rotation. At what value of the angular velocity will the body begin to slide?

Answer

See the following figure. $\"Description\"$

Hint

Before the body starts sliding, the friction force provides it with centripetal acceleration $a = \\omega^2 r.$ Therefore, considering the smallness of the tangential acceleration according to the condition, $F_{fr} \\approx m\\omega^2 r \\le kmg.$ The body starts sliding at $\\omega = \\omega_0$, such that $F_{\\text{cp}} = kmg$, i.e., when $\\omega_0^2 r = kg$.

13. A stone of mass $m = 0.5$ kg, tied to a rope of length $l = 50$ cm, rotates in a vertical plane. The tension force in the rope when the stone passes the lowest point of the circle is $T = 44$ N. To what height $h$ above the lowest point of the circle will the stone rise if the rope is cut at the moment when its velocity is directed vertically upwards?

Answer

$h = l(T - mg)/(2mg) \\approx 2.0$ m.

Hint

The kinetic energy of the stone at the lowest point of the circle completely converts into its potential energy during the ascent: $mv^2/2 = mgh$, where $h$ is the height of ascent above the lowest point of the circle. Further, write the equation of dynamics of rotational motion for the moment of passing the lowest point $mv^2/l = T - mg$ and solve both equations together to find the answer.

14. An athlete throws a hammer (a weight on a cable) a distance $l = 70$ m along a trajectory providing maximum range. What force $T$ acts on the athlete's hands at the moment of release? The mass of the hammer is $m = 5$ kg. Assume the athlete spins the hammer in a vertical circle of radius $R = 1.5$ m. Neglect air resistance.

Answer

$T = mg[\\frac{l}{R} - (1 + \\frac{\\sqrt{2}}{2})] \\approx 2205$ N.

Hint

The flight distance $l$ will be maximum if the hammer's velocity is directed at an angle $\\alpha = 45^\\circ$ to the horizontal. In this case (see problem 63 in Chapter: Kinematics) $l_0 = \\sqrt{v_0^2/g}$, where $l_0$ is the flight distance when the points of takeoff and landing are at the same height above the Earth's surface. In our case (see figure) $l_0 = l - R$. $\"Description\"$

15. A car of mass $M = 3 \\times 10^3$ kg moves with a constant speed $v = 36$ km/h: a) over a horizontal bridge; b) over a convex bridge; c) over a concave bridge. The radius of curvature of the bridge in the last two cases is $R = 60$ m. With what force does the car press on the bridge (in the last two cases) at the moment when the line connecting the center of curvature of the bridge with the car makes an angle $\\alpha = 10^\\circ$ with the vertical?

Answer

a) $F_1 = mg \\approx 29400$ N;

b) $F_2 = mg (\\cos \\alpha - \\frac{v^2}{gR}) \\approx 24000$ N;

c) $F_3 = mg (\\cos \\alpha + \\frac{v^2}{gR}) \\approx 34000$ N.

At the moment the car passes the middle of the bridge, $F_2 = 24400$ N; $F_3 = 34400$ N.

Hint

Two forces act on the car: the force of gravity $F_t = mg$ and the normal reaction force of the bridge $N$, which, according to Newton's third law, is equal in magnitude to the force $F$ of the car's pressure on the bridge. Along the curved bridge, the car moves with centripetal acceleration, imparted to it by the resultant of the force $N$ and the radial component of the gravity force, equal to $mg \\cos \\alpha$ (see figure). In accordance with Newton's second law, one can write $mv^2/R = mg \\cos \\alpha - N_2$ for a convex bridge and $mv^2/R = N_3 - mg \\cos \\alpha$ for a concave bridge. $\"Description\"$

16. A car of mass $m = 2$ t moves over a convex bridge with radius of curvature $R = 90$ m at a speed $v = 54$ km/h. At a point on the bridge, the direction to which from the center of curvature makes an angle $\\alpha$ with the direction to the top of the bridge, the car exerts a force $F = 14400$ N. Determine the angle $\\alpha$.

Answer

$\\alpha = \\arccos (\\frac{F}{mg} + \\frac{v^2}{gR}) \\approx 8.2^\\circ$.

17. A ball of mass $m = 100$ g is suspended on a string of length $l = 1$ m. The ball was spun so that it began to move in a circle in a horizontal plane. In doing so, the angle formed by the string with the vertical is $\\alpha = 60^\\circ$. Determine the total work done in spinning the ball.

Answer

$A = mgl (1 + \\frac{1}{2} \\tan \\alpha \\sin \\alpha - \\cos \\alpha) \\approx 1.23$ J.

Hint

The work goes into increasing the potential and kinetic energy. The speed of the ball is found from the condition $mv^2/(l \\sin \\alpha) = mg \\tan \\alpha$, and $h = l (1 - \\cos \\alpha)$.

18. What is the maximum speed at which a car can move on a turn with a radius of curvature $R = 150$ m, so that it does not skid, if the coefficient of sliding friction of the tires on the road is $k = 0.42$?

Answer

$v = \\sqrt{kgR} \\approx 89$ km/h.

Hint

See problem 11.

19. 1. What must be the maximum coefficient of sliding friction $k$ between the tires of a car and asphalt for the car to be able to negotiate a curve of radius $R = 200$ m at a speed $v = 100$ km/h?

A car with all driving wheels, starting from rest, uniformly gains speed while moving along a horizontal section of road representing an arc of a circle $\\alpha = 30^\\circ$ with radius $R = 100$ m. What is the maximum speed at which the car can exit onto the straight section of the path? The coefficient of friction between the wheels and the ground is $k = 0.3$.

Answer

1. $k = v^2/(gR) \\approx 0.4$.

$v = \\dfrac{\\sqrt{2kgR\\alpha}}{\\sqrt{1 + 4\\alpha^2}} \\approx 14.5$ m/s.

Hint and Solution

See problem 11.

The acceleration $a$ of the car is composed of the constant tangential acceleration $a_t$, which provides the car's acceleration, and the centripetal acceleration $a_{cp} = v^2/R$ ($v$ is the car's speed). Therefore
$a = \\sqrt{a_t^2 + \\frac{v^4}{R^2}}.$
The external force providing acceleration to the car is the friction force of the wheels on the road. Since the friction force cannot exceed $kmg$ ($m$ is the mass of the car), the maximum acceleration $a$ at the moment the car transitions to the horizontal section of the road cannot be greater than $kg$:
$\\sqrt{a_t^2 + \\frac{v^4}{R^2}} = kg.$
Since
$v = \\sqrt{2a_t l}$
($l = R\\alpha$ is the path traveled by the car; $\\alpha = \\pi/6$), then
$a_t = \\frac{v^2}{2l} = \\frac{v^2}{2R\\alpha},$
and after substitution we get
$\\frac{v^2}{R} \\sqrt{\\frac{1}{4\\alpha^2} + 1} = kg.$
Hence $v = \\frac{\\sqrt{2kgR\\alpha}}{\\sqrt{1 + 4\\alpha^2}}$.

20. A train moves along a curve of radius $R = 800$ m with a speed $v = 72$ km/h. Determine how much higher the outer rail must be than the inner rail so that no lateral force arises on the wheels. The distance between the rails horizontally is taken as $d = 1.5$ m.

Answer

$\\Delta h \\approx 7.65$ cm.

Hint

The resultant force $\\vec{F}$ of the wagon's weight $m\\vec{g}$ and the reaction force of the rails $\\vec{N}$ imparts centripetal acceleration $a_{cp}$ to the wagon; $m a_{\\text{cp}} = m\\vec{g} + \\vec{N};$ the resultant is directed horizontally (since its direction must coincide with the direction of the train's acceleration) and is equal to $mg \\tan \\alpha$ (see figure): $mv^2/R = mg \\tan \\alpha.$ Hence $\\tan \\alpha = \\frac{v^2}{gR}.$ The upper rail exceeds the lower one by the amount $\\Delta h = d \\tan \\alpha = d v^2/(gR).$

In this section, the dimensions of the bodies are considered much smaller than the radius of rotation, i.e., the entire mass of the body is considered concentrated at one point - its center of mass. $\"Description\"$

21. A motorcyclist rides on a horizontal road at a speed of 72 km/h, making a turn with a radius of curvature of 100 m. By how much must he lean in this case so as not to fall on the turn?

Answer

$\\alpha \\approx 22^\\circ$.

Solution

For a motorcyclist to move in a circle, he must lean so that the resultant of the three forces applied to him: gravity $m\\vec{g}$, normal reaction force of the road $\\vec{N}$, and friction force with the road $\\vec{F}_{fr}$ (see figure) - provides him with centripetal acceleration. Since the center of mass of the motorcyclist does not move vertically, $N = mg$. The centripetal acceleration is ultimately provided by the friction force: $mv^2/R = F_{fr}$.

The resultant $\\vec{R} = \\vec{N} + \\vec{F}{fr}$, obviously, must pass through the center of mass of the motorcyclist (the moment of the resultant relative to the center of mass must be zero). Therefore
$\\cot \\alpha = N/F{fr} = mg / (mv^2/R) = gR/v^2.$
$\"Description\"$

22. 1. What is the maximum speed $v$ at which a motorcyclist can ride on a horizontal plane, describing an arc of radius $R = 90$ m, if the coefficient of sliding friction $k = 0.4$?

At what angle $\\phi$ from the vertical direction must he lean in this case?
What will be the maximum speed of the motorcyclist if he rides on a banked track with an angle of inclination $\\alpha = 30^\\circ$ at the same radius of curvature and coefficient of friction?
What must be the angle of inclination $\\alpha_0$ of the track so that the speed of the motorcyclist can be arbitrarily large?

Answer

1. $v \\approx 18.8$ m/s.

$\\phi \\approx 21.8^\\circ$.
$v_{\\text{max}} \\approx 33.5$ m/s.
$\\alpha_0 = \\arctan (1/k)$.

Solution

Centripetal acceleration is provided by the friction force (see previous problem), therefore $mv^2/R = F_{fr}$. But the friction force cannot exceed the value
$F_{\\text{fr.max}} = kN = kmg,$
meaning the motorcyclist's speed cannot be greater than some limiting value, determined by this maximum possible value of the friction force:
$mv_{\\text{max}}^2/R = kmg.$
Hence $v_{\\text{max}} = \\sqrt{kgR}.$
The angle of inclination of the motorcyclist at this speed must be equal to:
$\\phi = \\arctan \\frac{v_{\\text{max}}^2}{gR} = \\arctan k.$
In creating the centripetal acceleration, the horizontal components of the friction force and the track reaction force will participate (see figure):
$mv_{\\text{max}}^2/R = F_{fr.max} \\cos \\alpha + N \\sin \\alpha. \\tag{1}$
The sum of the vertical components of the forces acting on the motorcycle must be zero (there is no acceleration in the vertical direction):
$F_{fr.max} \\sin \\alpha + mg - N \\cos \\alpha = 0.\\tag{2}$
Maximum friction force
$F_{fr.max} = kN.\\tag{3}$
Solving equations (1), (2), and (3) together, we get
$v_{\\text{max}} = \\sqrt{gR \\frac{k + \\tan \\alpha}{1 - k \\tan \\alpha}}.$
Hint. From part 3, it follows that
$v_{\\text{max}} \\to \\infty$ when $1 - k \\tan \\alpha \\to 0$, whence
$\\alpha_0 = \\arctan (1/k).$
$\"Description\"$

23. An airplane performs a turn, moving along an arc of a circle with a constant speed $v = 360$ km/h. Determine the radius $R$ of this circle if the body of the airplane is turned around the direction of flight by an angle $\\alpha = 10^\\circ$.

Answer

$R \\approx 5780$ m.

Solution

The aircraft changes its direction of flight by transitioning to motion along an arc of a circle by banking its fuselage around the direction of flight. In this case, the centripetal acceleration is imparted to the aircraft by the resultant of the gravity force and the lift force, perpendicular to the plane of the wings. This resultant is equal to $mg \\tan \\alpha$ (see figure). According to Newton's second law $mg \\tan \\alpha = mv^2/R,$ whence $R = v^2/(g \\tan \\alpha).$ $\"Description\"$

24. A car moves uniformly on a road turn of radius $R = 100$ m. The center of gravity of the car is at a height $h = 1$ m, the width of the car's track is $a = 1.5$ m. Determine the speed $v$ at which the car might tip over. The car does not slide in the transverse direction.

Answer

$v \\approx 26.1$ m/s.

Solution

When the car turns, the pressure on its wheels, and hence the forces acting on the car wheels from the road, are redistributed. At the maximum permissible speed of the car, the forces acting on it from the road will be applied to the outer wheels and are equal to: a) the reaction force $N = mg$ (numerically equal to the gravity force acting on the car, as there is no acceleration in the vertical direction) and b) the friction force $F_{fr} = mv^2/R$ (as it provides the car with centripetal acceleration). The car will overturn if the direction of the resultant of these forces passes below the center of mass of the car - in this case, there is an overturning moment of force relative to the center of mass. At the critical speed, this resultant passes through the center of mass of the car and therefore (see figure) $N/F_{fr} = \\tan \\alpha = (a/2)/h = a/(2h),$ or $mg / (mv^2/R) = a/(2h),$ whence $v = \\sqrt{gRa/(2h)}.$ $\"Description\"$

25. A driver, driving a car, suddenly noticed a fence ahead, perpendicular to the direction of his movement. What is more advantageous to do to prevent an accident: brake or turn aside?

Answer

Brake.

Solution

During braking, the friction force $F_{fr}$ must perform work $A = F_{fr} x$ ($x$ is the braking distance), equal to the kinetic energy the car had before braking: $F_{fr} x = mv^2/2$. Hence the braking distance $x = mv^2/(2 F_{fr})$. If the driver, without braking, turns aside, the car will move along an arc of a circle, and the centripetal acceleration $a_{cp} = v^2/R$ ($R$ is the radius of curvature) will be provided to the car by the same friction force. From the equation of motion $F_{fr} = mv^2/R$, we find that the turning radius will be $R = mv^2/F_{fr}$, i.e., it is twice the 'braking distance'. This means it is more advantageous to brake than to turn.

26. In a train car, moving uniformly along a curved path with a speed $v = 72$ km/h, a load is weighed on a spring scale. The mass of the load is $m = 5$ kg, and the radius of curvature of the path is $R = 200$ m. Determine the reading of the spring scale (the tension force of the spring $T$).

Answer

$T \\approx 51$ N.

Solution

Centripetal acceleration is imparted to the suspended mass by the resultant of the gravity force and the tension force $T$ of the spring. This resultant is equal to $mg \\tan \\alpha$ (see figure). According to Newton's second law $mg \\tan \\alpha = mv^2/R$, hence $\\tan \\alpha = v^2/(gR)$. The tension force of the spring $T = mg/\\cos \\alpha = mg \\sqrt{1 + \\tan^2 \\alpha} = mg \\sqrt{1 + \\frac{v^4}{g^2 R^2}}.$ $\"Description\"$

27. Find the separating force $F_{sep}$ acting on cream (density $\\rho_c = 0.93$ g/cm$^3$) from skim milk ($\\rho_m = 1.03$ g/cm³) per unit volume, if the separation occurs:

a) in a stationary container;

b) in a centrifugal separator rotating with a frequency of 6000 min$^{-1}$, if the liquid is at a distance $r = 10$ cm from the axis of rotation.

Answer

a) $F_{sep} = (\\rho_m - \\rho_c) g \\approx 980 \\text{ N/m}^3$;

b) $F_{sep} = (\\rho_m - \\rho_c) \\omega^2 r \\approx 3.94 \\times 10^6 \\text{ N/m}^3$.

Hint

a) In a stationary vessel, the separating force is the Archimedean (buoyant) force.

b) If the volume $V$, occupied by a particle of cream, were occupied by milk, it would be in equilibrium. This means that a force $F_1 = \\rho_m \\omega^2 r V,$ directed towards the axis of rotation, acts on the cream particle from the surrounding milk. This force turns out to be greater than the force required for the circular motion of the cream particle, which is $F_2 = \\rho_c \\omega^2 r V.$
The difference between these forces is the separating force.

28. An airplane performs a 'loop-the-loop' with a radius $R = 100$ m and moves along it with a speed $v = 280$ km/h. With what force $F$ will the pilot's body of mass $M = 80$ kg press on the seat of the airplane at the top and bottom points of the loop?

Answer

$F_{\\text{top}} = \\frac{Mv^2}{R} - Mg \\approx 4030$ N, $F_{\\text{bottom}} = \\frac{Mv^2}{R} + Mg \\approx 5630$ N.

Hint

See problem 15. The centripetal acceleration at the top and bottom points of the 'dead loop' is imparted to the pilot by the resultant of the gravity force and the reaction force of the seat.

29. Determine the tension force $T$ of the giant swing cable, if the mass of the person is $M = 70$ kg and the cable forms an angle $\\alpha = 45^\\circ$ with the pole during rotation. With what angular velocity $\\omega$ will the giant swings rotate if the length of the suspension is $l = 5$ m?

Answer

$T = Mg/\\cos \\alpha \\approx 990$ N; $\\omega = \\sqrt{g/(l\\cos\\alpha)} \\approx 1.68$ rad/s.

Hint

Centripetal acceleration is imparted to the person by the resultant of the gravity force and the tension force of the rope; this resultant is horizontal.

30. Find the period $T$ of rotation of a conical pendulum performing circular motions in a horizontal plane. The length of the string is $l$. The angle formed by the string with the vertical is $\\alpha$.

Answer

$T = 2\\pi \\sqrt{l\\cos\\alpha/g}$. For small angles $T \\approx 2\\pi \\sqrt{l/g}$.

Solution

Centripetal acceleration is imparted to the pendulum by the resultant of the gravity force $m\\vec{g}$ and the tension force of the string $\\vec{T}$ (see figure). This resultant lies in the plane of motion of the pendulum and is equal to $mg \\tan \\alpha$. According to Newton's second law, $m\\omega^2 R = mg \\tan \\alpha$ or $\\frac{4\\pi^2}{T^2} l \\sin \\alpha = g \\tan \\alpha$ (since $R = l \\sin \\alpha$ and $\\omega = 2\\pi/T$). Hence $T = 2\\pi \\sqrt{l \\cos \\alpha / g}.$ $\"Description\"$

31. A small weight, suspended on a string, rotates in a horizontal plane such that the distance from the suspension point to the plane in which the rotation occurs is $h$. Find the frequency $n$ of rotation of the weight, considering it constant.

Answer

$n = \\frac{1}{2\\pi} \\sqrt{g/h}$. (The answer does not depend on the length of the suspension.)

32. A chandelier of mass $m = 100$ kg is suspended from the ceiling on a metal chain of length $l = 5$ m. Determine the height $h$ to which the chandelier can be deflected so that the chain does not break during subsequent swings. It is known that the chain breaks when the tension force $T > 1960$ N.

Answer

$h \\le (T - mg) l / (2mg) \\approx 2.5$ m.

Solution

Two forces act on the chandelier at any point of the trajectory (see figure): the force of gravity $mg$ and the tension force of the chain $T$. When the chandelier passes the equilibrium position (point A), the forces are directed along the same line. Let's write the equation of Newton's second law for the chandelier at the moment it passes point A: $ma_{cp} = T - mg,$ where $a_{cp} = v^2/l$ is the centripetal acceleration. Then $mv^2/l = T - mg.\\tag{1}$ From the law of conservation of energy, we find the speed of the chandelier at point A: $mgh = mv^2/2,$ or $v^2 = 2gh.\\tag{2}$ Substituting expression (2) into equation (1), we have $m(2gh)/l = T - mg,$ whence $h = (T - mg) l / (2mg).$ According to the condition, $T < 2mg$, therefore, $h < 2.5 \\text{ m}$. $\"Description\"$

33. A ball of mass $m$ is suspended on an inextensible string. To what minimum angle $\\alpha_{\\text{min}}$ must the ball be deflected so that the string breaks during subsequent motion, if the maximum possible tension force of the string is $1.5 mg$?

Answer

$\\alpha_{\\text{min}} = \\arccos 0.75 \\approx 41.4^\\circ$.

Hint

See the previous problem.

34. A pendulum is deflected to a horizontal position and released. At what angle $\\alpha$ with the vertical will the tension force of the string be equal in magnitude to the force of gravity acting on the pendulum bob? Consider the pendulum as simple.

Answer

$\\alpha = \\arccos (1/3)$.

35. A load of mass $m$, tied to an inextensible string, rotates in a vertical plane. Find the maximum difference in the tension forces of the string.

Answer

$6mg$.

Hint

See the hint for problem 28. The relationship between the speeds of the mass at the top $v_t$ and bottom $v_b$ points is determined by the law of conservation of energy: $\\frac{mv_b^2}{2} - \\frac{mv_t^2}{2} = mg \\cdot 2R.$

36. A gymnast 'swings the giant' on the high bar. The mass of the gymnast is $m$. Assuming that his entire mass is concentrated at the center of gravity, and the speed at the highest point is zero, determine the force acting on the gymnast's hands at the lowest point.

Answer

$5mg$.

Hint

See the previous problem and problem 28.

37. One weight is suspended on an inextensible string of length $l$, and another on a rigid weightless rod of the same length. What minimum speeds must be imparted to these weights so that they rotate in a vertical plane?

Answer

$v_{\\text{min}} = \\sqrt{5gl}$ for the string; $v_{\\text{min}} = 2\\sqrt{gl}$ for the rod.

Hint

In the case of a massless rod, the minimum speed $v_{\\text{min}}$ of the mass at the lowest point is determined by the condition: the change in its potential energy upon rising from the lowest to the highest point equals the initial kinetic energy of the mass: $mg \\cdot 2l = mv^2/2.$ The speed of the mass at the highest point is zero. In the case where the mass rotates on an inextensible string, this condition is insufficient; another condition is needed: the string must remain taut up to the highest point. The speed of the mass at the highest point cannot be zero, but must be such that the centripetal acceleration is provided to the mass only by the force of gravity: $mv_{top}^2/l = mg$. At this moment $T = 0$, which is the condition for the minimum speed.

38. A ball of mass $M$ is suspended on a string. In the taut state, the string was positioned horizontally and the ball was released. Derive the dependence of the string tension force $T$ on the angle $\\alpha$ that the string currently makes with the horizontal direction. Check the derived formula by solving the problem for the case of the ball passing through the equilibrium position, when $\\alpha = 90^\\circ$.

Answer

$T = 3Mg \\sin \\alpha$; $T = 3Mg$.

Hint

See problem 32.

39. A simple pendulum of length $l$ and mass $M$ was deflected by an angle $\\phi_0$ from the equilibrium position and given an initial velocity $v_0$, directed perpendicular to the string upwards. Find the tension force of the pendulum string $T$ as a function of the angle $\\phi$ of the string with the vertical.

Answer

$\\displaystyle T = Mg(3 \\cos \\phi - 2 \\cos \\phi_0 + \\frac{v_0^2}{gl})$ until it comes to rest.

Hint

See problem 32. Consider the initial kinetic energy of the pendulum.

40. A weight, suspended on a string, is pulled aside so that the string assumes a horizontal position, and released. What angle $\\alpha$ with the vertical does the string form at the moment when the vertical component of the weight's velocity is maximum?

Answer

$\\cos \\alpha = \\sqrt{3}/3$; $\\alpha \\approx 54.7^\\circ$.

Solution

The vertical component of the bob's velocity will increase until the vertical component of the resultant of the applied forces, gravity $m\\vec{g}$ and string tension $\\vec{T}$ (see figure), becomes zero: $T \\cos \\alpha - mg = 0.\\tag{1}$ After this, the vertical component of the bob's velocity will decrease. Since the bob moves in a circle, according to Newton's second law $mv^2/R = T - mg \\cos \\alpha.$ Substituting the bob's velocity found from the law of conservation of energy: $mv^2/2 = mgR \\cos \\alpha$ [assuming release from horizontal position], into this equation, and solving the resulting equation together with equation (1), we find: $\\cos \\alpha = 1/\\sqrt{3}.$ $\"Description\"$

41. Identical elastic balls of mass $m$, suspended on strings of equal length to one hook, are deflected to different sides from the vertical by an angle $\\alpha$ and released. The balls collide and bounce off each other. What is the force $F$ acting on the hook:

a) at the extreme positions of the strings;

b) at the initial and final moments of the balls' impact;

c) at the moment of maximum deformation of the balls?

Answer

a) $F = 2mg \\cos^2 \\alpha$;

b) $F = 2mg (3 - 2 \\cos \\alpha)$;

c) $F = 2mg$.

Hint

a) The resultant of the string tension force and the ball's weight force is directed perpendicular to the string and imparts an initial acceleration $a = g \\sin \\alpha$ to the ball.

b) The speed of each ball at the initial moment of impact can be found using the law of conservation of mechanical energy. Then, from the laws of conservation of energy and momentum, it follows (see problems 28 and 39 in Chapter: Conservation of Energy) that the speeds of the balls at the final moment of impact will be equal in magnitude to their speeds at the initial moment of impact and directed in opposite directions. Knowing the speeds of the balls, it is easy to find the tension forces of the strings (see, for example, problem 32).

c) At the moment of maximum deformation of the balls, the speed of each ball is zero.

42. A simple pendulum with a flexible inextensible string of length $l$ is given a horizontal velocity $v_0 = \\sqrt{3gl}$ from the equilibrium position. Determine the maximum height $h$ of its rise when moving in a circle, if $v_0^2 = 3gl$. What trajectory will the pendulum bob follow after it reaches the maximum height $h$ on the circle? Determine the maximum height $H$ reached during this motion of the pendulum.

Answer

$h = \\sqrt{3}l$; parabolic trajectory (trajectory of a body thrown at an angle to the horizon); $H = 40l/27$.

Hint

See problems 37 and 32. At the point of maximum ascent of the pendulum, the tension force of the string becomes zero, and the centripetal acceleration is provided to the pendulum by the radial component of the gravity force.

43. A small ball is suspended at point A on a string of length $l$. At point O at a distance $l/2$ below point A, a nail is driven into the wall. The ball is pulled aside so that the string occupies a horizontal position, and released. At what point in the trajectory does the tension force of the string disappear? How will the ball move further? To what maximum height will the ball rise?

Answer

Below the suspension point by $l/6$; parabolic; $2l/27$ below the suspension point.

Hint

See the previous problem.

44. A vessel, having the shape of an expanding truncated cone with a bottom diameter $D = 20$ cm and a wall inclination angle $\\alpha = 60^\\circ$, rotates around the vertical axis $OO_1$. At what angular velocity of rotation $\\omega$ of the vessel will a small ball lying on its bottom be thrown out of the vessel? Neglect friction.

Answer

$\\omega > \\sqrt{2g\\tan\\alpha/D} \\approx 13$ rad/s.

Hint

Two forces act on the ball located on the wall of the rotating vessel: the force of gravity $mg$ and the reaction force $N$ of the vessel wall (see figure). At equilibrium of the ball relative to the vessel, the resultant of these forces provides the ball with centripetal acceleration, i.e., $m\\omega^2 R = mg \\tan \\alpha.$ From this, it follows that the ball can be in equilibrium at a distance $R$ from the axis of rotation of the vessel if the vessel rotates with angular velocity $\\omega = \\sqrt{g\\tan\\alpha/R}.$ At the bottom of the vessel, the ball is still in equilibrium, but no longer presses on the bottom at $\\omega_{cr} = \\sqrt{2g\\tan\\alpha/D};$ if $\\omega$ is increased, then from the equation given above, it follows that for no $R \\ge D/2$ will it be in equilibrium and will be ejected. $\"Description\"$

45. A sphere of radius $R = 2$ m rotates uniformly around its axis of symmetry with a frequency of 30 min⁻¹. Inside the sphere, there is a ball of mass $m = 0.2$ kg. Find the height $h$ corresponding to the equilibrium position of the ball relative to the sphere, and the reaction force $N$ of the sphere.

Answer

$h \\approx 1$ m; $N \\approx 0.4$ N.

Solution

At equilibrium of the ball relative to the sphere, the resultant of the reaction force $N$ of the sphere and the gravity force $mg$ of the ball is directed towards the axis of rotation and provides the ball with centripetal acceleration. The magnitude of this resultant $Q = mg \\tan \\alpha$ (see figure).

According to Newton's second law $m\\omega^2 l = mg \\tan \\alpha,$ where $l = R \\sin \\alpha$ is the distance of the ball from the axis of rotation;
$\\omega^2 R \\sin \\alpha = g \\tan \\alpha.$
This equation has two roots for $\\alpha$:

$\\sin \\alpha = 0$, $\\alpha = 0$;
$\\cos \\alpha = g/(\\omega^2 R)$.

The position of the ball with $\\alpha = \\arccos\\left(\\frac{g}{\\omega^2 R}\\right)$ is possible only when
$\\omega \\ge \\sqrt{g/R}.$
The position $\\alpha = 0$ is possible at any angular velocity of the sphere, but for $\\omega < \\sqrt{g/R}$ the equilibrium is stable, and for $\\omega \\ge \\sqrt{g/R}$ it is unstable. In our case $\\omega > \\sqrt{g/R} \\approx 2.2 \\text{ s}^{-1}$. Height $h = R - R \\cos \\alpha = R\\left(1 - \\frac{g}{\\omega^2 R}\\right)$, and the reaction force of the sphere
$N = mg/\\cos \\alpha = m\\omega^2 R.$
$\"Description\"$

46. Inside a conical surface, moving with acceleration $a$, a ball rotates along a circle of radius $R$. Determine the period $T$ of the ball's motion along the circle. The angle at the apex of the cone is $2\\alpha$.

Answer

$T = 2\\pi \\sqrt{R\\tan\\alpha/(a + g)}$.

Solution

The forces acting on the ball must provide it with both horizontal and vertical acceleration. Let's resolve the force $F$, acting on the ball from the side of the conical surface, into vertical $F' = F \\sin \\alpha$ and horizontal $F'' = F \\cos \\alpha$ components (see figure). The resultant of the vertical component of the force $F'$ and the gravity force provides the ball with acceleration $a$ upwards. The horizontal component of the force $F''$ provides the ball with centripetal acceleration $a_{cp} = \\omega^2 R$ ($\\omega$ is the angular velocity of rotation). Therefore, according to Newton's second law $ma = F \\sin \\alpha - mg$ and $m\\omega^2 R = F \\cos \\alpha$. Eliminating $F$ from these equations, we find: $\\omega = \\sqrt{(a + g)/(R\\tan\\alpha)},$ $T = 2\\pi/\\omega = 2\\pi \\sqrt{R\\tan\\alpha/(a + g)}.$ $\"Description\"$

47. A small body of mass $m$ slides down an inclined chute that transitions into a loop-the-loop of radius $R$ (see figure). Friction is negligible. Determine:

a) what must be the minimum height $h$ of the chute so that the body makes a complete loop without falling;

b) what pressure $F$ does the body exert on the support at the point whose radius vector makes an angle $\\alpha$ with the vertical?

$\"Description\"$

Answer

a) $h = 2.5R$;

b) $F = 3mg (1 - \\cos \\alpha)$.

Solution

a) The body will not detach from the loop if at the highest point it has a speed $v \\ge \\sqrt{gR}$ (see problem 37). Writing the law of conservation of energy $\\frac{1}{2}mv^2+ mg \\cdot 2R = mgh,$ we find that the body must slide down from a height $h = 2.5 R$.

b) At the point whose radius vector makes an angle $\\alpha$ with the vertical (see figure), the centripetal acceleration will be provided to the body by the resultant of the reaction force $N$ of the platform and the radial component of the gravity force $(mg)_r = mg \\cos \\alpha$.

According to Newton's second law
$mv^2/R = N + mg \\cos \\alpha.$
The speed of the body at this point can be easily determined using the law of conservation of energy:
$\\frac{5}{2} mgR = \\frac{mv^2}{2} + mg (R + R \\cos \\alpha),$
$v = \\sqrt{gR (3 - 2 \\cos \\alpha)}.$
Substituting this expression for $v$ into Newton’s second law, we get:
$
N = \\frac{m}{R} \\left[ gR(3 - 2\\cos\\alpha) \\right] - mg \\cos\\alpha = 3mg(1 - \\cos\\alpha)
$
The normal force $F$ exerted by the body on the track numerically equals the normal reaction $N$.
(Determine the value of $N$ for $\\alpha = 0$ and $\\alpha = \\pi$.)
$\"Description\"$

48. A conveyor belt is inclined to the horizontal at an angle $\\alpha$. Determine the minimum speed $v_{\\text{min}}$ of the belt at which a particle of ore lying on it separates from the surface of the belt at the point where it runs onto the drum, if the radius of the drum is $R$.

Answer

$v_{\\text{min}} = \\sqrt{gR \\cos \\alpha}$.

Hint

Separation of the ore from the belt means that the ore has lost contact with the belt and only the force of gravity acts on the ore. Its radial component provides the ore particles with centripetal acceleration at the moment the belt runs onto the drum (see figure) $mv^2/R = mg \\cos \\alpha.$ $\"Description\"$

49. A small body slides from the top of a sphere downwards. At what height $h$ from the top will the body detach from the surface of the sphere of radius $R$? Neglect friction.

Answer

$h = R/3$.

Solution

Before detaching from the surface of the sphere, the body is acted upon by its weight $mg$ and the reaction force of the support $N$. The resultant of the radial component of the weight force and the support reaction force provides the body with centripetal acceleration $mv^2/R = mg \\cos \\alpha - N.$ As the body slides down, the component of the weight force $mg \\cos \\alpha$ (where $\\alpha$ is the angle between the vertical and the direction from the center of the sphere to the body) decreases, and its speed increases, which leads to an even faster decrease in the support reaction force (the support reaction force depends on the nature of the body's motion). At the moment of detachment, $N$ becomes zero and then $mv^2/R = mg \\cos \\alpha.\\tag{1}$ Based on the law of conservation of energy $mv^2/2 = mgh.\\tag{2}$ Also, it is obvious that $\\cos \\alpha = (R - h)/R$. (3) Solving equations (1), (2), (3) together, we get $h = R/3.$

50. Find the kinetic energy of a hoop of mass $m$ rolling with speed $v$. There is no slipping.

Answer

$mv^2$.

Solution

Each point of the hoop participates in two motions: translational motion with velocity $v$ and rotational motion with angular velocity $\\omega$ about the center of the hoop. The linear velocity of rotational motion is $v_{rot} = \\omega R$. In the absence of slipping $\\omega = v/R$ and therefore $v_{rot} = v$. Consider two diametrically opposite identical small elements of the hoop, having mass $\\Delta m$ (see figure). Their velocities are equal to the vector sum of the velocities of translational and rotational motions, and therefore: $v_A^2 = v^2 + v_{rot}^2 - 2vv_{rot} \\cos \\alpha = 2v^2 (1 - \\cos \\alpha);$ $v_B^2 = v^2 + v_{rot}^2 + 2vv_{rot} \\cos \\alpha = 2v^2 (1 + \\cos \\alpha).$ The total kinetic energy of these elements of the hoop is $\\frac{\\Delta m v_A^2}{2} + \\frac{\\Delta m v_B^2}{2} = 2 \\Delta m v^2.$ This relationship does not depend on $\\alpha$, i.e., on the choice of the diameter at the ends of which the chosen elements of the hoop lie. Therefore, the kinetic energy of the entire hoop will be equal to $mv^2$. $\"Description\"$

51. A thin hoop rolls without slipping into a pit having the shape of a hemisphere. At what depth $h$ is the normal pressure force of the hoop on the wall of the pit equal to its weight? The radius of the pit is $R$, the radius of the hoop is $r$.

Answer

$h = (R - r)/2$.

Solution

Sliding down, the hoop moves along a circle. Centripetal acceleration is imparted to it by the sum of the radial components of the forces applied to the hoop (see figure), equal to $N - mg \\cos \\alpha$ ($N$ is the reaction force of the hemisphere, $m$ is the mass of the hoop, and $\\alpha$ is the angle formed by the radius vector of the hoop with the vertical). According to Newton's second law $mv^2/(R - r) = N - mg \\cos \\alpha,\\tag{1}$ where $v$ is the velocity of the center of the hoop. From the law of conservation of energy, it follows that $mg(R-r)(1-\\cos\\alpha) = \\frac{mv^2}{2} + \\frac{I\\omega^2}{2}.$ For a hoop $I=mr^2$. With no slip $v=\\omega r$. So $mg(R-r)(1-\\cos\\alpha) = \\frac{mv^2}{2} + \\frac{m(v/r)^2 r^2}{2} = mv^2.\\tag{2}$ Here $mv^2/2$ corresponds to the kinetic energy of the translational motion of the center of mass of the hoop; $\\frac{m(\\omega r)^2}{2}$ corresponds to the kinetic energy of rotation of the hoop relative to the center of mass (see previous problem); $mg(R-r)(1-\\cos\\alpha)$ is the change in potential energy. From equation (2) we find $v^2 = g (R - r) (1-\\cos \\alpha)$. Substituting this expression for $v^2$ into equation (1) and setting $N = 0$ (condition for detachment), we finally get $\\cos \\alpha = 1/2$ and $h = (R - r) \\cos \\alpha = (R - r)/2.$ $\"Description\"$

52. A small hoop rolls without slipping on the inner surface of a large hemisphere. At the initial moment, the hoop was at rest at its upper edge. Determine:

a) the kinetic energy of the hoop at the lowest point of the hemisphere;

b) what fraction of the kinetic energy is accounted for by the rotational motion of the hoop around its axis;

c) the normal force pressing the hoop to the lowest point of the hemisphere. The mass of the hoop is $m$, the radius of the hemisphere is $R$.

Answer

a) $mgR$;

b) half;

c) $2mg$.

53. Water flows through a pipe located in a horizontal plane and having a bend of radius $R = 2$ m. Find the lateral pressure of the water. The diameter of the pipe is $d = 20$ cm. $M = 300$ tonnes of water flows through the cross-section of the pipe in one hour.

Answer

$p = \\frac{4M^2}{\\pi \\rho a^3 R t^2} \\approx 1.2 \\times 10^5$ Pa.

Solution

The centripetal acceleration of the mass of the selected volume of water (see figure) is provided by the force of lateral pressure $F$ from the pipe walls: $mv^2/R = F.$ The force $F$ is equal to the product of the pressure $p$ and the area of the diametral section ABCD, perpendicular to the radius of the ring, i.e., $F = pdl.$ The mass of the selected volume $m = \\rho \\pi d^2 l / 4.$ The flow velocity of water is determined from the formula $M = \\rho \\pi d^2 v t / 4\\quad (\\tex{time } t = 1 \\text{ h}),$ whence $v = \\frac{4M}{\\rho \\pi d^2 t}.$ Substituting the values of $m$ and $v$ into the formula for $F$, we get the answer. $\"Description\"$

54. A body slides from point A to point B along two inclined surfaces, passing through points A and B once along a convex arc, the second time along a concave one. Both arcs have the same radius of curvature and the coefficient of friction in both cases is the same (see figure). In which case is the speed of the body at point B greater? $\"Description\"$

Answer

When moving along a convex arc.

Solution

Indeed, in this case, the pressure of the body on the surface at each point is less than the pressure on the surface at a point located at the same distance from point A, in the case of the body moving along a concave arc (see problem 15). This means that the friction force acting on the body at each point, and consequently, the work done against friction over the entire path, is less than when moving along the concave arc, and the body at point B has a greater kinetic energy in this case.

55. A rod of negligible mass and length $l$ with two small balls $m_1$ and $m_2$ ($m_1 > m_2$) at the ends can rotate about an axis passing through the center of the rod perpendicular to it. The rod is brought into a horizontal position and released. Determine the angular velocity $\\omega$ and the pressure force $F$ on the axis at the moment the rod passes through the equilibrium position with the balls.

Answer

$\\omega = 2 \\sqrt{\\dfrac{(m_1 - m_2)g}{(m_1 + m_2)l}}$;

$F = g \\dfrac{3(m_1^2 + m_2^2) - 2m_1 m_2}{m_1 + m_2}$.

Solution

The two masses move with the same angular velocity. Let us choose the potential energy reference level at the height where the mass $m_1$ is located at the moment it passes through the equilibrium position. Then, by the law of conservation of energy, we can write: $ (m_1 + m_2)g\\frac{l}{2} = m_2 g l + \\frac{1}{2} m_1 v_1^2 + \\frac{1}{2} m_2 v_2^2 $

Substituting $v_1 = v_2 = \\omega \\frac{l}{2}$, we find:
$
\\omega = 2 \\sqrt{ \\frac{(m_1 - m_2)g}{(m_1 + m_2)l} }
$

The force $F$ exerted on the axis equals the sum of the tension forces $T_1$ and $T_2$ in each half of the rod. We assume both rods are under tension. Then, according to Newton’s second law:
$
T_1 - m_1 g = m_1 \\omega^2 \\frac{l}{2}, \\quad m_2 g + T_2 = m_2 \\omega^2 \\frac{l}{2}
$

Substituting the expression for $\\omega$ at the moment both masses pass through the equilibrium position and solving these equations for $T_1$ and $T_2$, we obtain:
$
T_1 = \\frac{m_1 g}{m_1 + m_2}(3m_1 - m_2), \\quad T_2 = \\frac{m_2 g}{m_1 + m_2}(m_1 - 3m_2)
$
(When $m_1 < 3m_2$, we get $T_2 < 0$. This means that the half of the rod to which the mass $m_2$ is attached is under compression rather than tension, contrary to our assumption.)

The total force on the axis (assuming both rods are in tension):
$
F = T_1 - T_2 = \\frac{3\\left(m_1^2 + m_2^2\\right) - 2m_1 m_2}{m_1 + m_2} g
$

56. A small ring of mass $m$ is placed on a turn of a cylindrical spiral whose axis is vertical. The ring starts sliding down the spiral without friction. With what force $F$ will the ring press on the spiral after it has completed $n$ full turns? The radius of the turn is $R$, the distance between adjacent turns is $h$ (pitch of the spiral). Assume $h \\ll R$.

Answer

$F \\approx mg \\sqrt{1 + \\dfrac{4h^2 n^2}{R^2}}$.

Solution

After making $n$ revolutions, the ring descends by a height $nh$, and from the law of conservation of energy $mghn = \\frac{1}{2}mv^2$, it follows that at this point the ring will have a speed: $ v = \\sqrt{2ghn} $

This speed is directed tangent to the spiral and can be considered as the sum of a vertical component $v_0$ and a horizontal (rotational) component $v_1$.

In particular:
$
v_1 = v \\cos\\alpha,
$
where $\\alpha$ is the inclination angle of the spiral;
$\\alpha \\approx \\sin\\alpha = \\frac{h}{2\\pi R}$.

Since $h \\ll R$, we can take $\\cos\\alpha \\approx 1$ and $v_1 \\approx v$.

The horizontal component of the spiral’s reaction force must provide the ring with centripetal acceleration:
$
a_{\\text{cp}} = \\frac{v_1^2}{R} \\approx \\frac{2ghn}{R}
$
Therefore, the horizontal reaction force is:
$
F_{\\text{hor}} = ma_{\\text{cp}} = \\frac{2mghn}{R}
$

The ring presses on the spiral with the same force in the horizontal direction.
Thus, the total pressure force $F$ of the ring on the spiral consists of the horizontal component $F_{\\text{hor}}$ and the vertical component of gravity, $mg \\cos\\alpha \\approx mg$:
$
F \\approx \\sqrt{(mg)^2 + \\left(\\frac{2mghn}{R}\\right)^2}
= mg \\sqrt{1 + \\frac{4h^2 n^2}{R^2}}
$

57. A closed metal chain lies on a smooth horizontal disk, being freely placed on its centering ring, coaxial with the disk. The disk is set into rotation. Assuming the shape of the chain is a horizontal circle, determine the tension force $T$ along the chain if its mass is $m = 150$ g, length $l = 20$ cm, and the chain rotates with a frequency $n = 20$ s⁻¹.

Answer

$T \\approx m n^2 l \\approx 12$ N.

Solution

Consider a small element of the chain of length $\\Delta l = R \\Delta \\alpha$. The mass of this element $\\Delta m = m \\Delta l / l = m R \\Delta \\alpha / l.$ During the rotation of the chain, the centripetal acceleration is imparted to the selected element by the tension forces acting on it from the rest of the chain. The resultant of these forces is $2T \\sin (\\Delta \\alpha / 2)$ (see figure).

According to Newton's second law
$2T \\sin (\\Delta \\alpha / 2) = (\\Delta m) \\omega^2 R.$
Substituting $\\omega = 2\\pi n$ here and taking into account that for small angles $\\sin (\\Delta \\alpha / 2) \\approx \\Delta \\alpha / 2$, we get
$2T \\Delta \\alpha / 2 \\approx m R^2 \\Delta \\alpha (2\\pi n)^2 / l,$
whence $T \\approx m n^2 (2\\pi R)^2 / l = m n^2 l.$
$\"Description\"$

58. A jet airplane $m = 30$ t flies along the equator from west to east with a speed $v = 1800$ km/h. By how much will the lift force acting on the airplane change if it flies with the same speed from east to west?

Answer

$\\Delta F_{lift} \\approx 1.74 \\times 10^3$ N.

Solution

The resultant of the gravity force of the aircraft and the lift force $F_{lift}$ provides it with centripetal acceleration. When moving from west to east, the velocities of the aircraft and the Earth add up. Therefore, we can write $m (\\omega R + v)^2 / R = mg - F_{1lift}\\tag{1}$ where $\\omega$ is the angular velocity of the Earth; $R$ is its radius. If the aircraft flies in the opposite direction, then $m (\\omega R - v)^2 / R = mg - F_{2lift}. \\tag{2}$ From equations (2) and (1) we find the change in lift force: $\\Delta F_{lift} = F_{2lift} - F_{1lift} = 4mv\\omega = 4m (2\\pi v / T),$ where $T$ is the period of rotation of the Earth, i.e., one day.

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1. Find the dimensions of the gravitational constant $G$ in SI units. Using its value and the acceleration due to gravity, determine Earth's mass M. Earth's radius $R \\approx 6400$ km.

Answer

[$G$] = m$^3$/(kg$\\cdot$s$^2$); $M \\approx 6 \\times 10^{24}$ kg.

Hint

Near the Earth's surface, the force of gravity imparts acceleration $g$ to bodies; $mg = G mM/R^2$, whence $M = gR^2/G$, where $R$ is the radius of the Earth.

2. Calculate the gravitational force between two touching lead spheres (diameter 1 m each). Lead density $11.3 \\, \\text{g/cm}^3$.

Answer

$F \\approx 2.34 \\times 10^{-3}$ N.

3. A lead sphere (radius $R$) contains a spherical cavity that touches its surface and passes through its center. Total mass $M$. Using the law of gravitation, find the force on a small mass $m$ placed at distance $d > R$ from the sphere's center along the line connecting centers of the sphere and cavity (on the cavity side).

Answer

$F = \\frac{1}{7} G Mm \\left[\\dfrac{8}{d^2} - \\dfrac{1}{(d-R)^2}\\right]$.

Hint

The sought force is found as the difference between the forces that would act on the small sphere from the entire lead sphere and from the part that was removed from the sphere.

4. At what distance from Earth's surface is gravitational acceleration $1 \\, \\text{m/s}^2$?

Answer

$h \\approx 13600$ km.

5. Determine gravitational acceleration $g$ at height $h = 20$ km above Earth, given surface gravity $g_0 = 981 \\, \\text{cm/s}^2$ and Earth radius $R = 6400$ km.

Answer

$g \\approx 975$ cm/s$^2$.

Solution

The acceleration due to gravity is found from the formulas $g_0 = G \\frac{M}{R^2}$ and $g = G \\frac{M}{(R+h)^2}$. Hence $\\frac{g_0}{g} = (\\frac{R+h}{R})^2 = 1 + \\frac{2h}{R} + \\frac{h^2}{R^2} \\approx 1 + \\frac{2h}{R},$ since $h/R \\ll 1$, and therefore $h^2/R^2 \\ll 2h/R$. Thus, $g = g_0 \\times (\\frac{R}{R+h})^2 \\approx \\frac{g_0 R}{R+2h}.$

6. Prove that gravitational force on a mass $m$ inside Earth is $F = mgr/R_0$, where $r$ is distance from center and $R_0$ is Earth's radius. Assume constant Earth density.

Answer

Hint. Let's show that the gravitational force acting from an arbitrary spherical shell on a material point inside it is zero. For this, let's divide the spherical shell into thin spherical layers. A cone with a small solid angle with its vertex at this point cuts out elements from the layer, and it can be shown that $\\Delta m_1 / (\\Delta m_2) = r_1^2 / r_2^2.$ Therefore, the resultant gravitational forces towards these elements are equal to zero. Dividing the entire layer into such elements, we obtain the initial statement. Consequently, only the mass of that part of the Earth enclosed within the sphere of radius $r$ acts on the point.

7. A body falls through a straight shaft drilled along Earth's rotation axis. Find its maximum velocity (ignore air resistance).

Answer

$v = \\sqrt{gR}$.

Solution

If the mass of the body falling through the shaft is $m$, then the acceleration $a$ of the body depends on the distance $r$ to the center of the Earth according to the law $a =\\frac{F}{m} = \\frac{mgr/R}{m} = g \\frac{r}{R}$ where $R$ is the radius of the Earth (see problem 6).

The acceleration is maximum at $r = R$ and becomes zero at $r = 0$. This means that the speed of the body is maximum at $r = 0$. After the body passes the center of the Earth, its acceleration is directed opposite to the velocity, and the speed of the body will decrease. To find the speed of the body at $r = 0$, let's use the law of conservation of energy. Assuming that the potential energy of the body is zero at the center of the Earth, we get $mv^2/2 = \\Pi$, where $\\Pi$ is the potential energy of the body on the surface of the Earth. It is equal to the work that must be done to very slowly move the body from the center to the surface of the Earth. Let's plot the graph of the dependence of the force that needs to be applied to the body on the distance $r$ (see figure). The work done in moving the body is numerically equal to the area of the shaded triangle (see problems 5 and 17 in Chapter: Work, Power, Energy ), therefore
$\\Pi = mgR/2\\ \\text{ and }\\ mv^2/2 = mgR/2.$
Hence
$v = \\sqrt{gR}.$
$\"Description\"$

8. What direction and horizontal speed must an airplane fly along the equator to compensate for reduced weight due to Earth's rotation?

Answer

In the direction from east to west with a speed $v = 464$ m/s.

Hint

The aircraft must be stationary in the coordinate system associated with the fixed stars. For this, the angular velocity of the aircraft must be equal to the angular velocity of the Earth and directed in the opposite direction. When flying over the equator $v = \\omega R_E = 2\\pi R_E / T,$ where $R_E$ is the radius of the Earth; $T$ is its period of rotation.

9. Why are space rockets typically launched west-to-east? Why are equatorial launches most advantageous?

Answer

The rotational speed of the Earth is used.

10. At a planet's equator, bodies weigh half as much as at poles. Planet density $ρ = 3 \\times 10^3 \\, \\text{kg/m}^3$. Determine its rotation period.

Answer

$T = \\sqrt{6\\pi/(\\rho G)} \\approx 2$ h 41.6 min.

Solution

The weight $P$ of a body is the force with which the body acts on a suspension or support that is stationary relative to the body. According to Newton's third law, the support acts on the body with a force $P'$, equal and opposite to $P$. Thus, two forces are applied to bodies at rest on the planet: the force of gravity $F_g$ and the reaction force of the support $P'$. These forces impart the necessary centripetal acceleration to the body during the rotation of the planet around its axis. At the pole and the equator, these forces are directed along the same line. According to Newton's second law $\\frac{mv^2}{R} = F_g - P',$ where $v$ is the speed of the rotational motion of the body together with the Earth around its axis. At the pole $v = 0$ and the centripetal acceleration is absent. The reaction force of the support and, consequently, the weight of the body at the pole are equal to the force of gravity: $P_0 = F_g.$ At the equator $v = \\dfrac{2\\pi R}{T}$, where $T$ is the period of revolution. The weight of the body at the equator $P_e = F_g - m \\frac{4\\pi^2 R}{T^2}.$ According to the condition, $P_e = 0.5 P_0 = 0.5 F_g,$ then $\\frac{4\\pi^2 m R}{T^2} = 0.5 F_g.\\tag{1}$ According to the law of universal gravitation $F_g = G \\frac{mM}{R^2}.$ Here $m$ is the mass of the body; $M$ is the mass of the planet; $R$ is the radius of the planet, but $M = \\frac{4}{3} \\pi R^3 \\rho,$ where $\\rho$ is the density of the planet. Substituting the expressions for $F_g$ and $M$ into formula (1), we get $0.5 G \\frac{m \\cdot 4/3 \\pi R^3 \\rho}{R^2} = \\frac{4\\pi^2 m R}{T^2}.$ After simplification, we find $T = \\sqrt{6\\pi/(\\rho G)}.$

11. Find the average density of a planet where spring scales show 10% less weight at the equator than at poles. Day length $T = 24$ hours.

Answer

$\\rho = 30\\pi/(T^2G) \\approx 180$ kg/m$^3$.

12. How long should an Earth day be for equatorial weightlessness?

Answer

$t \\approx 1$ h 25 min.

Hint

See problem 10.

13. Derive the dependence of body weight on geographic latitude.

Solution

Two forces act on the weighed body: the force of gravity $F_g$, directed towards the center of the Earth, and the reaction force of the support $P'$ (see figure); $P'$ is numerically equal to the weight of the body. The resultant $Q$ of the forces $F_g$ and $P'$ imparts centripetal acceleration $a = \\omega^2 r$ to the body. Since $r = R \\cos \\phi$ ($\\phi$ is the geographical latitude), then $a = \\omega^2 R \\cos \\phi\\quad\\text{and}\\quad Q = m\\omega^2 R \\cos \\phi$ where $m$ is the mass of the body. In turn, for the force of gravity $F_g$, we can write $F_g = G mM/R^2 = mg_0,$ where $g_0 = G M/R^2$ ($M$ is the mass of the Earth) is a coefficient with the dimension of acceleration, numerically close to the acceleration of free fall at the pole. According to the cosine theorem from the triangle of forces, we get $P' = \\sqrt{F_g^2 + Q^2 - 2F_g Q \\cos \\phi} = F_g \\sqrt{1 + (\\frac{Q}{F_g})^2 - \\frac{2Q}{F_g} \\cos \\phi}.$ Considering that $Q/F_g \\ll 1$, we can neglect the square of this quantity in the radicand. Using the known approximation $\\sqrt{1 - \\alpha} \\approx 1 - \\alpha/2$ for $\\alpha \\ll 1$, we obtain the final expression: $P' \\approx F_g (1 - \\frac{Q}{F_g} \\cos \\phi) = m(g_0 - \\omega^2 R \\cos^2 \\phi)$. $\"Description\"$

14. Calculate the Sun-Earth mass ratio given: Moon completes 13 orbits/year; average Sun-Earth distance is 390× greater than Moon-Earth distance.

Answer

336,000 times.

Solution

The equations of motion of the Earth around the Sun and the Moon around the Earth will be: $G \\frac{m_E M_S}{R_{ES}^2} = m_E \\omega_E^2 R_{ES};$ $G \\frac{m_M m_E}{R_{ME}^2} = m_M \\omega_M^2 R_{ME},$ or $G \\frac{M_S}{R_{ES}^3} = \\omega_E^2;$ $G \\frac{m_E}{R_{ME}^3} = \\omega_M^2.$ Dividing the first equation by the second and considering that $\\omega = 2\\pi/T$, we find $\\frac{M_S}{m_E} = \\frac{R_{ES}^3}{R_{ME}^3} \\frac{T_M^2}{T_E^2}.$

15. Determine the Sun's mass using gravitational constant γ, Earth's orbital period $T$, and Earth-Sun distance $L = 1.5 \\times 10^{11}$ m.

Answer

$m_S \\approx 2 \\times 10^{30}$ kg.

Hint

See problem 14.

16. Can a satellite orbit in a plane not passing through Earth's center?

Answer

No. Because in this case, the force of gravity would have a component not lying in the plane of the orbit.

17. A satellite orbits Earth at height $h$ (Earth radius $R$). For circular orbit, express orbital velocity and period in terms of $h$, $R$, and surface gravity $g$.

Answer

$v = R\\sqrt{\\dfrac{g}{R+h}$; $T = 2\\pi \\dfrac{R+h}{R} \\sqrt{\\dfrac{R+h}{g}}$.

Solution

Assume that the satellite moves in a circular orbit, the center of which is at the center of the Earth. In this case, the satellite moves with centripetal acceleration $a_{\\text{cp}} = \\frac{v^2}{R+h},$ where $v$ is the speed of the satellite in orbit; $R$ is the average radius of the Earth; $h$ is the average altitude of the satellite. This centripetal acceleration is provided to the satellite by the force of gravity $F = G \\frac{mM}{(R+h)^2},$ where $M$ is the mass of the Earth; $m$ is the mass of the satellite; $G$ is the gravitational constant. According to Newton's second law, we have $\\frac{m\\,v^2}{R+h} = G \\frac{mM}{(R+h)^2},$ whence $v^2 = G \\frac{M}{R+h}.$ The value $G M$ can be found from the following relation: the weight of the body on the surface of the Earth $mg = G \\frac{mM}{R^2},$ whence $G M = gR^2.$ Substituting the value $G M$ into the formula for the speed of the satellite, we get $v = R\\sqrt{\\frac{g}{R+h}}.$ The period of revolution of the satellite $T = 2\\pi/\\omega,$ where $\\omega = v/(R+h)$ is the angular velocity of the satellite in orbit. Consequently, $T = \\frac{2\\pi(R+h)}{v} = 2\\pi \\frac{R+h}{R} \\sqrt{\\frac{R+h}{g}}.$

18. Find average angular $\\omega$ and linear $v$ velocities of a satellite with 105-minute orbital period.

Answer

$\\omega = 2\\pi/T \\approx 0.001$ rad/s; $v = \\sqrt[3]{2\\pi g R_E^2 / T} \\approx 7.4 \\times 10^3$ m/s.

Solution

The radius of the orbit $R$ can be found from Newton's second law equation $mv^2/R = G \\frac{Mm}{R^2}.$ Expressing the linear velocity $v$ in terms of the angular velocity $\\omega$, we get $v = \\omega R$. Using the relation $G M = gR_E^2$ (see problem 17), we obtain $R = \\sqrt[3]{\\frac{gR_E^2}{\\omega^2}},$ and further, $v = \\omega R = \\sqrt[3]{gR_E^2 \\omega} = \\sqrt[3]{2\\pi g R_E^2 / T}.$

19. What orbital radius and velocity would give a satellite Earth's rotation period? What trajectory would it appear to have from Earth? What orbital plane would make it appear stationary?

Answer

$R = \\sqrt[3]{gR_E^2/\\omega^2} \\approx 42000$ km; $v = \\sqrt[3]{gR_E^2 \\omega} \\approx 3.1$ km/h.

Hint

See the previous problem. The angular velocity of rotation of the satellite must be equal to the angular velocity $\\omega$ of the Earth's rotation. From the Earth, it will seem that the satellite moves forward and backward along the meridian. It will appear stationary if its orbit lies in the plane of the equator.

20. What is the first cosmic velocity for a planet with twice Earth's mass and radius?

Answer

$v_1 \\approx 8$ km/s, the same as Earth's.

Hint

The first cosmic velocity is the speed of a satellite moving in a circular orbit around a planet near its surface. Therefore $mv_1^2/R = G mM/R^2,$ whence $v_1 = \\sqrt{\\frac{G M}{R}}.$ Since $M/R = M_E/R_E$, and $G M_E = gR_E^2$, we get $v_1 = \\sqrt{gR_E}.$

21. What is the first cosmic velocity for a planet with Earth's density but half its radius?

Answer

$v_1 \\approx 4$ km/s, i.e., half that of Earth.

22. A satellite required $A = 3.2 \\times 10^{10}$ J work to reach low circular orbit. Find its mass (Earth radius $R_3 = 6400$ km).

Answer

1000 kg.

Solution

Neglecting the change in the potential energy of the satellite, we can write $A = mv^2/2$. To move in an orbit with radius $R \\approx R_E$, the satellite must have a speed $v = \\sqrt{gR_E}$ (see problem 17), then $A = mgR_E/2.$ Hence $m = 2A/(gR_E).$

23. Compare energy costs for: a) lifting a satellite to $h_1 = 3200$ km vs. orbital insertion at that height; b) same for $h_2 = 6400$ km.

Answer

In the first case 1:1, in the second 2:1.

Solution

If the potential energy of a body of mass $m$ in the gravitational field is considered zero infinitely far from the Earth (mass $M$), then at a distance $R$ from the center of the Earth it should be equal to $-G Mm/R$ (this formula can be easily obtained based on the analogy of gravitational and electrostatic fields or by considering the work done when moving the body over a small segment $\\Delta R$). To lift the satellite to a height $h$ above the Earth's surface, work must be done equal to the change in the potential energy of the satellite: $A = \\Pi_{R_E+h} - \\Pi_{R_E} = -G \\frac{Mm}{R_E+h} - (-G \\frac{Mm}{R_E}) = mg \\frac{hR_E}{R_E+h}$ (since $G M/R_E^2 = g$). To launch the satellite at this altitude, it needs to be given kinetic energy $mv^2/2$, i.e., perform work $A_1 = mv^2/2.$ For the satellite to rotate in orbit at height $h$ from the Earth, it needs to be given a speed $v = R_E \\sqrt{g/(R_E+h)}$ (see problem 17), then $A_1 = \\frac{1}{2} \\frac{mgR_E^2}{R_E+h}.$ The ratio of the works is $A/A_1 = 2h/R_E.$ For $h = 3200$ km $A/A_1 = 1$, for $h = 6400$ km $A/A_1 = 2$.

24. When would an astronaut experience weightlessness during interplanetary flight?

Answer

The astronaut will be in a state of weightlessness from the moment the spacecraft engines are turned off in outer space.

Solution

Let us consider the forces acting on the spacecraft and on the bodies located within it. The spacecraft and the bodies within it always move with the same acceleration. This acceleration for the spacecraft is determined by the thrust force $\\vec{F}_T$, the drag force $\\vec{F}_d$, and the resultant gravitational force $\\vec{F}_g$, which is proportional to its mass $M$. For any body inside the spacecraft, the acceleration is determined by the normal force $\\vec{N}$ (equal in magnitude to the force exerted on the support) and the resultant gravitational force $\\vec{F}'_g$, which is proportional to the body's mass $m$: $ \\vec{a} = \\frac{\\vec{F}_T + \\vec{F}_d + \\vec{F}_g}{M} = \\frac{\\vec{N} + \\vec{F}'_g}{m} $ However, the gravitational force per unit mass (the gravitational field) is the same for both the spacecraft and the body: $ \\frac{\\vec{F}_g}{M} = \\frac{\\vec{F}'_g}{m} $ From this, it follows that: $ \\vec{N} = \\frac{m}{M}(\\vec{F}_T + \\vec{F}_d) $ Consequently, the force exerted by the body on the support (the \"apparent weight\" of the body) is: $ \\vec{P} = -\\vec{N} = -\\frac{m}{M}(\\vec{F}_T + \\vec{F}_d) $ This relationship is valid for any stage of the spacecraft's motion. It can also be used to determine the g-forces experienced by the cosmonaut during takeoff and during entry into the dense layers of the atmosphere. When the spacecraft enters outer space ($\\vec{F}_d = 0$) and its engines are shut down ($\\vec{F}_T = 0$), then $\\vec{P} = 0$. That is, all bodies on the spacecraft are in a state of weightlessness and do not experience the interactions and internal stresses associated with weight.

25. How would the operation of pendulum clocks ('walkers') and spring (wrist) watches change inside an interplanetary spacecraft?

Answer

Pendulum clocks will stop after the rocket engine is turned off. The movement of the spacecraft will not affect the running of wristwatches.

26. How can body mass be measured under weightlessness conditions?

Answer

Many methods are possible, based on the application of Newton's second and third laws. For example:

measure the ratio of the velocities acquired by bodies with known and unknown masses after burning the thread compressing a spring between the bodies;
act on the body with a known force $F$ and measure the acceleration $a$ obtained by the body in this case, then $m = F/a$;
measure the tension force of a pre-calibrated spring when rotating the body attached to this spring with a known constant angular velocity, etc.

27. Is it possible to create artificial gravity inside a spaceship?

Answer

Yes, by rotating the spacecraft or turning on the engines.

28. Does the potential energy of objects relative to Earth change when they move inside an orbiting satellite?

Answer

Yes. The force of gravity acts on bodies in a state of weightlessness (see problem 24). The change in potential energy is still given by the formula $\\Delta \\Pi = mg_h \\Delta h,$ where $g_h$ is the acceleration due to gravity at the given altitude.

29. Do Pascal's and Archimedes' laws remain valid under weightlessness conditions?

Answer

Formally, both laws are applicable, but Archimedes' law loses its meaning, since both the body and the displaced liquid are weightless.

30. How would the linear and angular velocities of a satellite change under conditions of weak atmospheric drag? Assume a circular orbit.

Answer

The linear and angular velocities of the satellite will increase.

Hint

The total energy of the satellite is $E = \\Pi + K.$ The potential energy of the satellite $\\Pi = -G mM/r$ (see problem 23). We find the kinetic energy $K$ from the relation $\\frac{mv^2}{r} = G \\frac{mM}{r^2},\\quad T = G \\frac{mM}{2r}.$ Thus, the total energy $E = -G \\frac{mM}{2r}.$ Under conditions of motion with weak friction, the work done against friction forces is produced at the expense of the total energy $E$ of the satellite; in this case, as seen from the last relation, $|E|$ must increase, which is possible if $r$ decreases, and consequently $v$ and $\\omega$ (see problem 17) increase.

31. In which case does a spacecraft experience greater heating due to atmospheric friction: during launch or during re-entry? Explain why.

Answer

When falling to Earth. When launched, its speed in the denser layers of the atmosphere near the Earth's surface is low, while during descent it is high.

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1. A lantern of mass $M = 10$ kg is suspended above the center of a street of width $l = 10$ m by a cable. The allowable tension force in the cable is $T = 500\\ \\mathrm{N}$. Determine the height $H$ at which the ends of the cable should be attached if the lantern must be at a height of $h = 5$ m.

Answer

$H = h + \\frac{lMg/2}{\\sqrt{4T^2 - (Mg)^2}} \\approx 5.5$ m.

Hint

The resultant of the cable tension forces must be equal to the weight of the lantern.

2. Is it possible to stretch a cable horizontally so that it does not sag?

Answer

No. In this case, the tension force of the rope would be infinite.

3. What should the force $F$ be in order to move a box of mass $M = 60$ kg uniformly along a horizontal surface, if the coefficient of friction between the box and the surface is $k = 0.27$, and the force is applied at an angle of $\\alpha = 30^{\\circ}$ to the horizontal?

Answer

$F = kMg/(\\cos \\alpha + k \\sin \\alpha) \\approx 162$ N.

4. What angle $\\alpha$ should the force make with the horizontal so that the force $F$ needed to move a load uniformly along a horizontal plane is minimized? The force is applied at the center of mass of the load, and the coefficient of friction is $k$.

Answer

$\\alpha = \\arctan k$ (see figure).

Solution

$F = \\dfrac{kMg}{\\cos \\alpha + k \\sin \\alpha}$ (see problem 3).

The expression on the right-hand side is minimized when the value of the denominator $\\cos \\alpha + k \\sin \\alpha$ is maximized.

Denoting $k$ through $\\tan \\beta$, we can transform the expression as follows:

$
\\cos \\alpha + \\tan \\beta \\sin \\alpha = \\dfrac{1}{\\cos \\beta} (\\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta) = \\dfrac{1}{\\cos \\beta} \\cos (\\alpha - \\beta).
$

The last expression is maximized when $\\cos (\\alpha - \\beta) = 1$, that is, when $\\alpha - \\beta = 0$, from which it follows that $\\alpha = \\beta = \\arctan k$.
$\"Description\"$

5. A spool lies on a table (see figure). In which direction will it move if a thread is pulled with force $F_1$, $F_2$, or $F_3$ (the extension of the force line of $F_2$ passes through the contact point of the spool with the table)? $\"Description\"$

Answer

The action of the moment of force $F_1$ relative to the instantaneous axis will lead to the rotation of the spool around point O clockwise, and it will roll to the right.

The moment of force $F_2$ relative to the instantaneous axis is zero, so in this case, the thread will unwind, leaving the spool in place.

The moment of force $F_3$ will lead to the rotation of the spool counterclockwise, and it will roll to the left.

6. A rod $AB$ is hinged at point $A$ and rests with its end $B$ on a platform (see figure). What is the minimum force $F$ needed to move the platform? The mass of the rod is $m$, the coefficient of friction between the rod and the platform is $k$, and the angle the rod makes with the vertical is $\\alpha$. Neglect rolling friction of the platform’s wheels and axle friction. $\"Description\"$

Answer

$F = \\dfrac{kmg \\tan \\alpha}{2(k + \\tan \\alpha)}$ to move the cart left;

$F' = \\dfrac{kmg \\tan \\alpha}{2(\\tan \\alpha - k)}$ to move it right if $\\tan \\alpha > k$. If $\\tan \\alpha < k$, it's impossible to move the cart right.

Solution

The force needed to pull the cart equals the magnitude of the friction force acting on the cart from the rod.

Consider first the case where force $F$ moves the cart left (see figure above). In this case, forces $N$ and $F_{\\text{fr}}$ act on the rod, as shown in the figure. Since the rod is in equilibrium, for the moments of forces acting on it, we can write
$\\frac{1}{2} mg \\sin \\alpha - Nl \\sin \\alpha - F_{\\text{fr}} l \\cos \\alpha = 0$
($l$ is the length of the rod). Since $F_{\\text{fr}} = kN$, i.e., $N = F_{\\text{fr}}/k$, from this equation we find that
$F = F_{\\text{fr}} = \\frac{mg \\sin \\alpha}{2(\\cos \\alpha + \\frac{1}{k} \\sin \\alpha)} = \\frac{kmg \\tan \\alpha}{2(k + \\tan \\alpha)}.$

If now force $F'$ must move the cart right, the friction force acting on the rod is directed right, and from the equilibrium condition of the rod, we get
$\\frac{1}{2} mg \\sin \\alpha - Nl \\sin \\alpha + F'{\\text{fr}} l \\cos \\alpha = 0.$
Assuming the cart moves, i.e.,
$F'{\\text{fr}} = kN$ or
$N = F'{\\text{fr}}/k,$
we find
$F' = F'{\\text{fr}} = \\frac{mg \\sin \\alpha}{2(\\sin \\alpha - k \\cos \\alpha)} = \\frac{kmg \\tan \\alpha}{2(\\tan \\alpha - k)}.$
But if $\\tan \\alpha < k$, $F'{\\text{fr}} \\to \\infty$. The same happens if $\\tan \\alpha < k$. From the equilibrium equation of the rod, it follows that $F'{\\text{fr}} = (N - mg/2) \\tan \\alpha$. If $\\tan \\alpha < k$, the friction force $F'{\\text{fr}} < kN - (mg/2)k$, i.e., $F'{\\text{fr}} < kN$. This means that if $\\tan \\alpha < k$, the friction force does not reach the value $kN$ for any value of force $F'$ applied to the cart, and it's impossible to move the cart. 'Jamming' occurs.

7. A ball of mass $m$ is suspended from a vertical smooth wall at point $A$ by a rope of length $l$ (see figure). What is the tension $T$ in the rope and the force $F$ exerted by the ball on the wall, if its radius is $R$? Neglect friction with the wall. $\"Description\"$

Answer

$T = mg \\dfrac{l+R}{\\sqrt{l^2 + 2lR}}$; $F = mg \\dfrac{R}{\\sqrt{l^2 + 2lR}}$.

Solution

For equilibrium of the body, two conditions must be met:

the sum of all forces acting on the body must be zero:
$\\sum \\vec{F}_i = 0 \\tag{1}$
(this equality must hold for the projections of all forces onto any line);
the sum of the moments of all forces acting on the body relative to any fixed axis must also be zero:
$\\sum M_i = 0.$
As the fixed axis (see figure below), let's choose a horizontal line parallel to the wall and passing through the center of the sphere. Forces $\\vec{F}$ and $m\\vec{g}$ are directed along the radii of the sphere, so their moments relative to the chosen axis are zero. From the last condition, it follows that the moment of force $\\vec{T}$ must also be zero, which means this force must be directed along the radius of the sphere. Let's now write equation (1) for the projections of forces onto the vertical and horizontal directions:
$mg - T \\cos \\alpha = 0,$
$F - T \\sin \\alpha = 0,$
where $\\sin \\alpha = R/(l+R)$, whence we find:
$T = mg/\\cos \\alpha = mg(l+R)/\\sqrt{l^2+2lR};$
$F = T \\sin \\alpha = mg \\frac{R}{\\sqrt{l^2+2lR}}.$
According to Newton's third law, the pressure force of the sphere on the wall is equal in magnitude to force F and directed in the opposite direction.

$\"Description\"$

8. A cylinder of radius $r$ stands on an inclined plane with angle $\\alpha$ to the horizontal. What is the maximum height of the cylinder such that it does not topple, assuming it is made of a homogeneous material?

Answer

$h = 2r \\cot \\alpha$.

Hint

The cylinder does not tip over if the vertical line passing through the center of gravity of the cylinder passes through its base.

9. A metal bar was weighed using several dynamometers, each with a maximum load of 50 N. The total mass of the bar turned out to be 17.5 kg. How was the weighing performed and what is the minimum number of dynamometers needed?

Answer

Four dynamometers are sufficient for weighing. The sum of their readings equals the weight of the bar. The metal bar can be suspended from dynamometers located, for example, at the same distance from each other.

10. What should the coefficient of friction $k$ be so that a wedge driven into a log does not pop out? The wedge angle at the tip is $30^{\\circ}$.

Answer

$k \\ge \\tan 15^\\circ \\approx 0.268$.

Hint

The force pushing the wedge out is the resultant of the reaction forces $\\vec{F}_1$ and $\\vec{F}_2$ of the log (see figure) (due to symmetry $|\\vec{F}_1| = |\\vec{F}_2|$), and the balancing force is the resultant of the friction forces $F_{fr1} = kF_1$ and $F_{fr2} = kF_2$, directed along the wedge (the force of gravity compared to these forces can be neglected). $\"Description\"$

11. A pipe of mass $M = 1.2 \\cdot 10^{3}$ kg lies on the ground. What force $F$ must be applied to lift the pipe from one end using a crane?

Answer

$F = Mg/2 \\approx 6 \\times 10^3$ N.

12. A car with a mass of 1.35 tons has a wheelbase length of $3.05\\ \\mathrm{m}$. Its center of gravity is located 1.78 m behind the front axle. Determine the force acting on each of the front wheels and each of the rear wheels from the horizontal surface.

Answer

2754 N on the front wheels and 3861 N on the rear wheels.

13. A mass $m$ is suspended from two identical springs, once connected in series and once in parallel (see figure). Find the extension $\\Delta x$ of the springs in both cases if the stiffness of each spring is $k$. Will the distance $\\Delta l$ by which the load descends be the same in both cases? $\"Description\"$

Answer

a) $\\Delta x = mg/k$; $\\Delta l = 2\\Delta x$;

b) $\\Delta x = mg/(2k)$; $\\Delta l = \\Delta x$.

14. Two springs with stiffness coefficients $k_1$ and $k_2$ are connected once in series and once in parallel (see figure). What should be the equivalent stiffness $k_{\\text{eq}}$ of a single spring that could replace the system of two springs? $\"Description\"$

Answer

$k_{eq} = k_1 k_2 / (k_1 + k_2)$ for series connection of springs and $k_{eq} = k_1 + k_2$ for parallel connection.

Solution

In series connection of springs, the forces stretching them are equal and equal to the force F stretching the system of springs, and the extension of the system is equal to the sum of the extensions of the springs: $\\Delta x = \\Delta x_1 + \\Delta x_2.$ Since $F = \\Delta x k_{eq}$, then $\\Delta x = F/k_{eq}.$ Similarly, $\\Delta x_1 = F/k_1$ and $\\Delta x_2 = F/k_2$. Therefore $\\frac{1}{k_{eq}} = \\frac{1}{k_1} + \\frac{1}{k_2}$ and $k_{eq} = \\frac{k_1 k_2}{k_1 + k_2}.$ In parallel connection of springs $\\Delta x = \\Delta x_1 = \\Delta x_2$, and $F = F_1 + F_2$. Substituting $F = \\Delta x k_{eq}$, $F_1 = \\Delta x k_1$ and $F_2 = \\Delta x k_2$ here, we get $k_{eq} = k_1 + k_2.$

15. A mass $m$ is suspended from the end of a spring of initial length $l$, causing the spring to stretch by $0.1\\,l$. At what point along the unstretched spring should a mass of $2m$ be suspended so that its attachment point is equidistant from the spring’s ends? The mass $m$ remains attached to the lower end of the spring. Neglect the mass of the spring.

Answer

$l_1/l_2 = 11/13$.

Solution

Let $l_1$ and $l_2$ be the lengths of the parts into which the inextensible spring is divided by point A, where a mass $2m$ needs to be suspended, and $k_1$ and $k_2$ be the stiffnesses of these parts of the spring. Since when suspending a weight, the extension of the whole spring is $n$ times greater than the extension of the $1/n$ part of this spring, the stiffness of the whole spring is $n$ times less than the stiffness of the $1/n$ part of the spring (see problem 14), i.e., $k_1 = lk/l_1\\quad\\text{and}\\quad k_2 = lk/l_2,$ where $k = mg/(0.1l)$. When suspending a mass $2m$, a weight $3m$ will act on the upper part of the spring and it will extend by $\\Delta l_1 = \\frac{3mg}{k_1} = 3mg \\frac{l_1 } {lk}.$ The lower part of the spring will extend by $\\Delta l_2 = mg/k_2 = mg \\frac{l_2}{lk},$ while $l_1 + \\Delta l_1 = l_2 + \\Delta l_2,$ i.e., $l_1 (\\frac{3mg}{lk} + 1) = l_2 (\\frac{mg}{lk} + 1).$ Hence $\\frac{l_1}{l_2} = \\frac{mg+lk}{3mg+lk} = \\frac{mg + mg/0.1}{3mg + mg/0.1} = \\frac{11}{13}.$

16. What is the minimum coefficient of friction $k_{\\text{min}}$ between the walls of a cube and a horizontal surface so that the cube can be tipped over an edge using a horizontal force applied to the top face? What should the applied force $F$ be? The mass of the cube is $M$.

Answer

$F > Mg/2$, $k_{min} = 1/2$.

Hint

Apply the second condition (see problem 7), choosing the edge about which the cube rotates as the fixed axis. The maximum friction force must be no less than the force F applied to the body.

17. What is the minimum force $F_{\\text{min}}$ required to tip a cube over an edge on a horizontal surface? What must be the minimum coefficient of friction $k$ between the cube and the surface? The mass of the cube is $M$.

Answer

$F_{min} = Mg\\sqrt{2}/4$; $k \\ge 1/3$.

Hint

The force required to tip the cube is minimal if it is perpendicular to the diagonal plane passing through the edge about which the cube tips, and is applied to the second edge of the cube lying in this diagonal plane.

18. A tall rectangular block with a square base stands on a horizontal surface. How can the coefficient of friction between the block and the surface be approximately determined using only a ruler?

Answer

Solution: If a horizontal force $F=kmg$ is applied to the block near its base, it will start moving. If this force is applied at height $h$ from the floor, then $Fh = mga/2$ and it will start tipping over around the edge of the base. Therefore $kmgh = mga/2$ and $k = a/(2h)$; $a$ is the side of the base.

19. An iron rod of mass $M$ is bent in half so that its two parts form a right angle (see figure). The rod is suspended by one end on a hinge. Find the angle $\\alpha$ that the upper part of the rod makes with the vertical in the equilibrium position. $\"Description\"$

Answer

$\\tan \\alpha = 1/3$.

Hint

The rod will hang such that the vertical line passing through the suspension point A of the rod passes through its center of gravity O (see figure). Let the length of the rod be $2l$, find the lengths of the segments: $AB = l/2 + DB = l/2 + l/4 = \\frac{3l}{4}$ and $OB = l/4.$ $\"Description\"$

20. A uniform beam of mass $M$ and length $l$ is suspended by its ends on two springs (see figure). Both springs have the same length when unloaded, but the stiffness of the left spring is $n$ times greater than that of the right spring (under equal loads, the extension of the right spring is $n$ times that of the left). At what distance $x$ from the left end of the beam should a mass $m$ be suspended to keep the beam horizontal? Assume $n = 2$. $\"Description\"$

Answer

$x = l \\frac{M(n-1)+2mn}{2m(n+1)}$.

Hint

The extensions of the springs must be the same. The force $F_1 = \\frac{Mg}{2} + \\frac{l-x}{l} mg$ acts on the left spring at equilibrium, and the force $F_2 = \\frac{Mg}{2} + \\frac{x}{l} mg$ acts on the right spring (see figure above).

21. A ball of mass $m = 4.9$ kg rests between two smooth inclined planes. The left plane forms an angle $\\alpha = 35^{\\circ}$ with the horizontal, and the right plane an angle $\\beta = 20^{\\circ}$. Find the forces $F_1$ and $F_2$ that the ball exerts on the planes. Solve the problem in two ways: a) by resolving forces and b) using the moment method.

Answer

$F_1 = mg \\sin \\beta / \\sin(\\alpha + \\beta) = 20$ H; $F_2 = mg \\sin \\alpha / \\sin(\\alpha + \\beta) = 34$ H.

Hint

As fixed axes, it is convenient to choose horizontal lines passing through the points of contact of the sphere with the planes.

22. A wheel of radius $R$ and mass $m$ stands before a step of height $h$. What is the minimum horizontal force $F$ that must be applied to the axle $O$ of the wheel for it to climb the step? Neglect friction.

Answer

$F \\ge mg \\frac{\\sqrt{h(2R-h)}}{R-h}$.

Hint

For the wheel to roll over the step, the moment of the force $\\vec{F}$ applied to the wheel relative to the axis passing through point A (see figure) must be greater than or equal to the moment of the gravity force $m\\vec{g}$ of the wheel relative to the same axis. $\"Description\"$

23. Which is easier: to move a stationary railway car by pushing on its body or by pushing at the top of a wheel's rim?

Answer

To the rim of the wheel. In this case, the moment of the applied force is twice as large.

24. Why does the front of a car dip down during sudden braking?

Answer

During braking, the friction force $F_{\\text{fr}}$ acts on the car wheels. Let's apply forces $\\vec{F}_1$ and $\\vec{F}_2$ to the center of gravity of the car, equal to $F_{\\text{fr}}$ and directed in different directions (see figure). This can be done, since the sum of these forces and their moments relative to any axis are zero.

Now $\\vec{F}1$ can be considered as the braking force, and $F{\\text{fr}}$ and $\\vec{F}_2$ as a couple of forces causing the car to rotate around the axis passing through the point of contact of the wheels with the road, as a result of which the front end of the car lowers.
$\"Description\"$

25. A wooden plate floats on water. A couple of forces (equal and opposite but not along the same line) is applied horizontally. Around which point does the plate rotate?

Answer

The couple of forces will cause the plate to rotate around its center of mass.

Hint

Since the sum of the forces acting on the plate is zero, its center of mass will remain at rest.

26. A heavy uniform board of mass $M$ and length $l$ leans against a corner between the wall and floor, with a rope attached to the other end. Determine the tension force $F$ in the rope if the angle between the board and rope is $\\beta = 90^{\\circ}$. How does this force change as the angle $\\alpha$ between the board and floor increases, keeping $\\beta$ constant?

Answer

$F = 0.5 Mg \\cos \\alpha$; decreases.

27. A ladder of mass $m$ leans against a smooth vertical wall, forming an angle $\\alpha$ with the horizontal floor. Its center of gravity is at the midpoint. What are the directions and magnitudes of the forces acting on the ladder from the wall and floor? Use construction to find the direction of the reaction from the floor.

Answer

$N_A = \\dfrac{mg}{2} \\cot \\alpha$; $Q = \\dfrac{mg}{2} \\sqrt{4 + \\cot^2 \\alpha}$.

Solution

Let's write the equilibrium conditions for the ladder (see problem 7; see figure).

1st condition: $x$-axis: $N_A - F_{\\text{fr}} = 0$; y-axis: $N_B - mg = 0$;

2nd condition: for moments of forces relative to the horizontal axis passing through the points of contact of the ladder with the support,
$N_A l \\sin \\alpha = mgl \\cos \\alpha / 2.$

The resultant of forces $N_B$ and $F_{\\text{fr}}$ is
$Q = mg/2 \\sqrt{4 + \\cot^2 \\alpha}.$

Force $Q$ must pass through point O of intersection of the lines passing through forces $N_A$ and $mg$, since the sum of the moments of all forces relative to this point must be zero (see figure).
$\"Description\"$

28. A rod $AB$ of mass $m = 5$ kg is hinged at point $A$ and can rotate in a vertical plane (see figure). A thread is attached to point $B$ and runs over pulley $C$ with a mass $m_1 = 2.5$ kg hanging from it. The axes of pulley $C$ and hinge $A$ lie on the same vertical line, and $AC = AB$. Find the angle $\\alpha$ between the rod and vertical at equilibrium. What is the force $F_{AB}$ along the rod at point $A$? Is the equilibrium stable? $\"Description\"$

Answer

$\\sin(\\alpha/2) = m_1/m$; $\\alpha = 60^\\circ$; $F_{AB} = mg \\cos^2(\\alpha/2) \\approx 36.75$ N; unstable equilibrium.

Hint

To determine the angle $\\alpha$, consider the moments of the forces acting on the rod $Q=m_1 g$ and $mg$ relative to point A (see figure above). Having considered the moments of the forces acting on the rod relative to point C, find the force $F_A$ acting in the hinge A (see problem 27; the line of action of force $F_A$ passes through the intersection point of the lines of action of forces $Q$ and $mg$, lying on the midpoint of side BC of the equilateral triangle ABC): $F_A l \\sin(\\alpha/2) = mgl \\sin \\alpha / 2 \\implies F_A = mg \\cos(\\alpha/2).$ Its projection onto the direction of the rod $F_{AB} = F_A \\cos(\\alpha/2)$. The equilibrium is unstable: for any small deviation from the found position of equilibrium, the system cannot return to it on its own, moreover, the changed ratio of the moments of forces acting on the rod will contribute to the further displacement of the system in the same direction.

29. A ladder stands against a wall. The friction coefficient with the wall is $k_1 = 0.4$ and with the ground is $k_2 = 0.5$. The ladder's center of gravity is at its midpoint. Find the minimum angle $\\alpha$ the ladder can make with the horizontal without slipping.

Answer

$\\alpha_{min} = \\arctan \\frac{(1-k_1 k_2)}{2k_2} \\approx 38^\\circ 40'$.

Hint

See the solution to problem 27. The forces acting on the ladder are shown in the following figure. The limiting angle $\\alpha_{min}$, which the ladder can form with the horizon without slipping, will correspond to the maximum possible static friction forces $F_A$ and $F_{\\text{fr}}$, i.e., they must simultaneously have the largest possible values, equal respectively to $k_1 N_A$ and $k_2 N_B$. $\"Description\"$

30. A ladder of length $l = 4$ m leans against a smooth wall at an angle $\\alpha = 60^{\\circ}$ with the floor. The maximum friction force between the ladder and the floor is $F_{\\text{fr}} = 200\\ \\mathrm{N}$. How high $h$ can a person of mass $m = 60$ kg climb before the ladder starts to slip? Neglect the ladder’s mass.

Answer

$h = l F_{\\text{fr}} \\sin \\alpha \\tan \\alpha / (mg) \\approx 2$ m.

Hint

See problem 27. The forces acting on the ladder are shown below. $\"Description\"$

31. A cube is placed near a wall so that one of its faces makes an angle $\\alpha$ with the floor. What coefficient of friction with the floor is required for this to be possible, assuming negligible friction with the wall?

Answer

For $k \\ge 1/2 (\\cot \\alpha - 1)$.

Hint

Write down the condition for the sum of moments of the reaction forces of the wall and the force of gravity relative to the axis passing through the line of contact of the cube with the floor to be zero, and take into account that the reaction force of the wall must be equal in magnitude to the friction force $kN_1$, and the force $N_1$ is the reaction force of the floor, equal to the force of gravity.

32. 1. A rod hangs horizontally on a rope loop. Will equilibrium be disturbed if the rod is bent to the right of the loop?

Suppose the rod is thicker on one side. Are the masses to the left and right of the loop equal?

Answer

1. Will be disturbed.

The mass of the thick end of the rod is greater.

33. Prove that the center of gravity of a flat triangle lies at the intersection point of its medians.

Answer

Hint: By lines parallel to any of the sides, the triangle is divided into thin strips. Obviously, the center of gravity of the triangle lies on the line passing through the midpoints of these strips, i.e., on the median.

34. Prove that the center of gravity of a triangle made of uniform thin rods lies at the center of the incircle of a triangle whose vertices are midpoints of the original triangle’s sides.

Hint

The center of mass of each rod lies at its midpoint. Connecting the midpoints of the rods, we get a triangle whose side lengths are half the side lengths of the given triangle and, consequently, proportional to the masses of the corresponding rods. The angle bisectors of the resulting triangle divide its sides into parts proportional to the other two sides.

35. Ten spheres with masses $1, 2, 3, \\ldots, 10$ g are fixed along a weightless rod of length 90 cm, spaced 10 cm apart. Find the center of mass of the system.

Answer

At a distance of 30 cm from the 10 g ball.

Hint

The system will be balanced if a force equal to the sum of the weights of the balls is applied to the center of mass of the system. Write down the condition for the sum of moments of all forces acting on the system relative to one of the endpoints to be zero, find the position of the center of mass.

36. A uniform thin plate is shaped like a circle of radius $R$, with a circular hole of half the radius cut from it, touching its edge. Where is the center of gravity?

Answer

$R/6$ from the center.

Solution

Let the distance from the center of mass of the plate to the center of the circle be $x$. From symmetry considerations, it is obvious that the center of mass of the plate lies on its diameter passing through the center of the hole (see figure). If we fill the hole with the material from which the plate is made, the center of the resulting solid plate will be at the center of the circle. Therefore, the equality $m_1 R/2 = x m_2$ must hold. Substituting $m_1 = \\pi (R/2)^2 \\rho = \\pi R^2 \\rho / 4$ and $m_2 = \\pi R^2 \\rho - m_1 = 3/4 \\pi R^2 \\rho$ here, where $\\rho$ is the surface density of the plate material, we get $x = R/6$. $\"Description\"$

37. Where is the center of gravity of a cube from which a small cube with edge $a/2$ has been removed (see figure)? $\"Description\"$

Answer

$a\\sqrt{3}/28$ from the geometric center of the cube.

Hint

See the solution to the previous problem.

38. A rod of length $l = 15$ cm and mass $m = 0.025$ kg is placed in a tall, smooth cylindrical glass of radius $R = 6$ cm. What are the forces the rod exerts on the bottom and the wall of the glass? Neglect friction.

Answer

Horizontal forces are equal to $mgR/\\sqrt{l^2 - 4R^2} \\approx 0.16$ N. Vertical force is $mg \\approx 0.25$ N.

39. Two identical spheres of radius $r$ and mass $m$ are placed in a hollow vertical cylinder open on both sides, with radius $R$ where $r > R/2$. The system is on a horizontal surface. What is the minimum mass $M$ of the hollow cylinder to prevent the spheres from toppling it?

Answer

$M \\ge 2m(1 - r/R)$.

Solution

The cylinder will not tip over if the moment of its weight $Q$ relative to point $O$ (see figure) is not less than the tipping moment of forces $\\vec{F}_1$ and $\\vec{F}_2$ of the balls' pressure on the cylinder: $F_1 AO - F_2 B_1 O \\le MgR\\quad (Mg = Q).\\tag{1}$ These forces, according to Newton's third law, are equal in magnitude to the forces $\\vec{N}_1$ and $\\vec{N}_2$ acting on the balls. On the upper ball, besides force $\\vec{F}_1$, act the weight $m\\vec{g}$ and the reaction force of the lower ball $\\vec{N}_1$, directed along the line of centers. The forces acting on the lower ball are shown in the figure. From the equilibrium condition of the balls, we find that $F_1 = F_2 = mg \\tan \\alpha.$ Substituting these values into inequality (1) and considering that $AO - B_1 O = 2(R-r) \\cot \\alpha$, we find that $Mg \\ge 2mg(1 - r/R).$ $\"Description\"$

40. Two identical cylindrical logs lie side by side on the ground. A third identical log is placed on top. What is the minimum coefficient of friction $k$ between them so that they don’t roll apart (assuming no sliding on the ground)?

Answer

$k \\ge 1/(2+\\sqrt{3})$.

Hint

The logs will be in equilibrium if the moment of the pressure force from the upper log on each of the lower logs does not exceed the moment of the friction force of the upper log with the corresponding lower log relative to the line of contact of the log with the ground.

41. A round hole of radius $R/2$ is drilled parallel to the axis of a cylinder at a distance $R/2$ from its center. The cylinder lies on a board that is slowly lifted from one end (see figure). Find the limiting angle $\\alpha$ at which the cylinder remains in equilibrium. The friction coefficient is $k = 0.2$. $\"Description\"$

Answer

The cylinder will roll down at an angle of inclination of the plank $\\alpha = \\arcsin(1/6)$.

Solution

The center of mass of the cylinder with the hole (point O) is located at a distance $r = R/6$ from the axis of the cylinder (see problem 36). When the plank is tilted, the cylinder rotates so that its center of mass is on the vertical line passing through the point of contact of the cylinder with the plank. However, if this vertical line passes from the axis of the cylinder at a distance greater than $r=R/6$ (see the following figure), equilibrium is impossible. This means that the cylinder will start rolling when the angle $\\alpha$ is such that $\\sin \\alpha_1 = r/R = (1/6)R/R = 1/6;\\quad \\alpha_1 = \\arcsin(1/6).$ At the same time, the cylinder will start sliding at such an angle of inclination of the plank at which $\\tan \\alpha_2 = k = 1/5,$ hence $\\alpha_2 = \\arctan(1/5)$; $\\alpha_1 < \\alpha_2$, therefore rolling will occur before sliding. $\"Description\"$

42. A hemisphere and a cylinder of the same radius made of the same material are joined as shown in the following figure. The system rests on a horizontal surface. For what height $x$ of the cylinder will the system be in neutral equilibrium? The center of gravity of the hemisphere lies on the axis, offset by $3/8$ of the radius from its center. $\"Description\"$

Answer

$x = R/\\sqrt{2}$.

Hint

The system is in indifferent equilibrium if its center of gravity coincides with the geometric center of the hemisphere.

43. Water is poured into a cylindrical glass. At what water level does the center of gravity of the glass with water reach its lowest point?

Answer

The center of gravity of the glass with water is in the lowest position when it coincides with the water level in the glass. (Explain!)

44. A heavy block is held by friction between two horizontal rods $A$ and $B$ (see figure). What must be the distance from the block’s center of gravity to its contact point with rod $A$ so it doesn’t slip out? The values of distance $a$, angle $\\alpha$, and friction coefficient $k$ are given. $\"Description\"$

Answer

$x \\ge \\frac{a}{2} (\\frac{\\tan \\alpha}{k} - 1)$.

Solution

The block will not slide out if the component of the gravity force along the block does not exceed the maximum friction force (see figure): $mg \\sin \\alpha \\le F_{fr.max} = N_A k + N_B k$. (1) The pressure forces $N_A$ and $N_B$ of the rods on the block are found from the condition that the sum of the moments of forces relative to points A and B is zero, respectively: $mg x \\cos \\alpha - N_B a = 0$; $mg(a-x)\\cos \\alpha - N_A a = 0$. Substituting the values of $N_A$ and $N_B$ found from here into inequality (1), we find that the block will not slide out if $x \\ge \\frac{a}{2} (\\frac{\\tan \\alpha}{k} - 1)$. $\"Description\"$

45. How much work is required to rotate onto a different face: a) a cube of mass 200 kg; b) a hollow cube half-filled with water? The cube’s mass is negligible compared to the water. Each edge is 1 m. Ignore gravitational work after the cube passes the point of unstable equilibrium.

Answer

a) 410 J; b) 1100 J.

Hint

The minimum work required to be done during infinitely slow tipping of the cube is equal to the difference in potential energies of the cube in the initial position and in the position of unstable equilibrium.

46. A cube-shaped box is moved a certain distance—once by dragging and once by tipping (i.e., rolling over its edge). For what value of the sliding friction coefficient $k$ is the work required equal in both cases?

Answer

$A_1 = mgka$; $A_2 = 0.207 mga$ for $k=0.207$ (see hint for problem 45).

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1. Write the equation of harmonic oscillation if the amplitude of the oscillation is 4 cm and the period is 0.01 s, given \$ x_0 = 0 \$.

Answer

$x = 4 \\sin 200\\pi t$.

2. During what fraction of the period \$ T \$ does a body performing harmonic oscillations travel the entire path from the mean position to the extreme position? The first half of the path? The second half?

Answer

$T/4$; $T/12$; $T/6$.

Solution

1st method. The following figure shows the graph of the displacement of the oscillating body versus time. At the initial moment, the displacement is zero. The phases in radians are marked on the abscissa axis. From the graph, it is clear that the entire path from the mean position to the extreme maximum displacement takes $T/4$ for the oscillating body. The oscillating body covers the first half of the path in $T/12$, which corresponds to the phase $\\pi/6$ ($\\sin(\\pi/6) = 0.5$). The second half of the path is covered in time $T/4 - T/12 = T/6$.

2nd method. The displacement $x$ of the body is determined by the formula $x = a \\sin(2\\pi t / T)$, where $a$ is the amplitude, equal to the displacement to the extreme position; $t$ is the time counted from the moment of passing the equilibrium position. The time $t_1$ required for the body to travel from the mean position to the extreme position is determined from the condition $x = a$: $a = a \\sin(2\\pi t_1 / T)$, whence $2\\pi t_1 / T = \\pi/2$ and $t_1 = T/4$. The time $t_2$ required to travel the first half of this path is determined from the condition $x = a/2$, i.e., $a/2 = a \\sin(2\\pi t_2 / T)$, whence $2\\pi t_2 / T = \\pi/6$ and $t_2 = T/12$. The time $t_3$ to travel the second half of the path is $t_3 = t_1 - t_2 = T/6$.
$\"Description\"$

3. For what fraction of the period is the pendulum bob within 1 cm of the equilibrium position if the amplitude of its oscillations is 2 cm?

Answer

$T/3$.

Hint

See the solution to the previous problem.

4. Show that the period of motion of a conical pendulum (a mathematical pendulum moving in a circle) is equal to the period of its oscillations in a plane for small angles of deviation.

Solution

Circular motions of a 'conical' pendulum can be obtained if the deflected pendulum is given a push (velocity) in a direction perpendicular to the plane of possible oscillations after the initial deflection. Thus, 'conical' motions can be considered as the result of the superposition of two independent mutually perpendicular oscillations. The periods of these oscillations are the same (any plane of swing is no different from any other). Consequently, the period of the complex motion - the revolution of the pendulum along the cone - will be the same as the period of oscillation of a simple pendulum. The period of revolution of a 'conical' pendulum performing circular motion at small angles at the apex of the cone is $T = 2\\pi \\sqrt{l/g}$ (see problem 30 in Chapter: Circular Motion).

5. By what fraction of its length must a mathematical pendulum be shortened so that its period of oscillation at an altitude of 10 km equals its period on the Earth's surface? (Ignore the Earth's rotation.)

Answer

Approximately 0.3% longer.

Hint

The condition for the equality of the periods of oscillation of pendulums of different lengths at different heights is expressed as follows: $l_0/g_0 = l/g_h$. Forming the derived proportion, we get $\\frac{l_0 - l}{l_0} =\\frac{g_0 - g_h}{g_0}.$ Considering that $g_h = \\frac{\\gamma M}{(R+h)^2} \\approx g_0 (\\frac{R}{R+2h})$ (see problem 5 in Chapter: Conservation of Energy), it is easy to find $\\frac{l_0 - l}{l_0} \\approx 2h/(R+2h) \\approx 2h/R \\approx 3 \\times 10^{-3}.$

6. Determine how much a pendulum clock will lose in a day if it is raised to an altitude of 5 km above the Earth's surface. (Ignore the Earth's rotation.)

Answer

$\\Delta t \\approx 67.5$ s.

Solution

The period of oscillation of a pendulum at height $h$ is $T = 2\\pi \\sqrt{\\frac{l}{g_h}}.$ The acceleration of gravity at this height is $g_h = g_0 (\\frac{R}{R+h})^2,$ where $g_0$ is the acceleration at the Earth's surface and $R$ is its radius (see problem 5 in Chapter: Conservation of Energy). Then $T = 2\\pi \\sqrt{\\frac{l}{g_h}} = 2\\pi \\sqrt{\\frac{l}{g_0}} \\left(\\frac{R+h}{R}\\right) = T_0 \\frac{R+h}{R} = T_0 \\left(1 + \\frac{h}{R}\\right),$ where $T_0$ is the period of oscillation at the Earth's surface. Thus, over time $T_0$, the clock lags by $\\Delta T = T - T_0 = h\\frac{T_0}{R}.$ Over time $t = 24$ h, the clock will lag by some time $\\Delta t$, which relates to $t$ as $\\Delta T$ relates to $T_0$, and obviously, $\\Delta t / t = \\Delta T / T_0$, whence $\\Delta t = t \\frac{\\Delta T}{T_0} = t\\frac{h}{R}.$

7. How can the change in the period of a pendulum placed above a spherical ore deposit with density \$ \\rho \$ be used to estimate the volume of the deposit? The Earth's density is \$ \\rho_0 \$ (\$ \\rho_0 > \\rho \$).

Hint

For estimation, assume the deposit has the shape of a sphere of radius $r$. Then the force of gravity acting on a body of mass $m$ near the deposit should be less than far from it (on the Earth's surface) by the amount $\\frac{\\gamma m (\\rho_0 - \\rho) \\cdot 4/3 \\pi r^3}{r^2} = \\frac{4}{3} \\pi \\gamma m (\\rho_0 - \\rho) r$. The acceleration $g'$ of free fall near the deposit will therefore be equal to $g - \\frac{4}{3} \\pi \\gamma (\\rho_0 - \\rho) r$, and the period of oscillation of the pendulum $T' = 2\\pi \\sqrt{l/g'} = 2\\pi \\sqrt{l/g} \\sqrt{g/g'} = T \\sqrt{g/g'}$, where $T$ is the period of oscillation of the pendulum far from the deposit. Knowing $g'$, it is easy to find $r$.

8. Will the period of oscillation of a pendulum change if it is placed in water? Assume the pendulum has an ideally streamlined shape and that water friction is negligible.

Answer

The period of oscillation of the pendulum will increase.

Hint

The action of the buoyant force on the pendulum (Archimedes' force) reduces the tension force of the string, which is equivalent to a decrease in $g$.

9. A pendulum with a period \$ T = 1 \\, \\text{s} \$ hangs in a stationary elevator. With what acceleration is the elevator moving if the pendulum's period becomes \$ T_1 = 1.1 \\, \\text{s} \$? In what direction is the elevator moving?

Answer

$a = g \\left[1 - \\left(\\dfrac{T}{T_1}\\right)^2\\right] \\approx 0.17g$. The acceleration is directed downwards. The direction of motion of the lift does not matter.

Soltuion

When the lift moves with constant acceleration $a$, the tension force of the pendulum string $F_N$ in its equilibrium position relative to the cabin is determined from Newton's second law: $ma = mg - F_N$, whence $F_N = m(g - a)$. In the last formula, $a$ is an algebraic quantity: positive when the lift's acceleration is directed downwards, and negative when the acceleration is directed upwards. It follows that when the pendulum is deflected, the force restoring it to the equilibrium position will be proportional not to $g$, but to $(g - a)$. This means that in the lift moving with acceleration $a$, a pendulum of length $l$ has a period $T_1 = 2\\pi \\sqrt{l/(g-a)}$. Taking the ratio of the periods of oscillation of the pendulum in the lift moving with acceleration and in a stationary lift and squaring it, we get $\\left(\\frac{T_1}{T}\\right)^2 = \\frac{g}{g-a},$ whence we find the answer: $a = g \\times \\left[1 - \\left(\\frac{T}{T_1}\\right)^2\\right] \\approx 0.17g;$ the answer is positive, meaning the lift moves with acceleration directed downwards; the direction of velocity does not matter.

10. Find the period of oscillation \$ T \$ of a mathematical pendulum of length \$ l \$ suspended in a train car moving horizontally with acceleration \$ a \$.

Answer

$T = 2\\pi \\sqrt{l/\\sqrt{g^2 + a^2}}$.

Hint

In the equilibrium position relative to the wagon, the tension force of the pendulum string $F_N$ will be equal to: $F_N = m\\sqrt{g^2 + a^2}$.

11. Determine the wavelength \$\\lambda \$ of a sound wave in water produced by a source with a frequency of 200 Hz, given the speed of sound in water is 1450 m/s.

Answer

$\\lambda = 7.25$ m.

12. Which tuning fork sounds longer: one clamped in a vise or one placed on a resonance box?

Answer

The tuning fork clamped in the vise sounds longer. The radiation of sound wave energy per unit time from the tuning fork standing on the resonator box is greater.

13. The speed of sound in water is 1450 m/s. What is the distance between the nearest points oscillating in opposite phases if the oscillation frequency is 725 Hz?

Answer

1 m.

14. At what speed \$ v \$ of a train will a pendulum of length \$ l = 44 \\, \\text{cm} \$ suspended in the train car oscillate most strongly, given the length of the rails is 25 m?

Answer

$v \\approx 67.5$ km/h.

Hint

When the wagon wheels hit the rail joints, the wagon receives an impulse that has both vertical and horizontal components. If the period between impacts is equal to the period of oscillation of the pendulum, the latter will swing particularly strongly.

15. Waves propagate at a speed of 360 m/s with a frequency of 450 Hz. What is the phase difference between two points 20 cm apart?

Answer

$\\pi/2$.

Hint

If two points are separated by a distance equal to $\\lambda$, the phase difference is $2\\pi$, and if by $l$, then $2\\pi l / \\lambda$.

16. A body is located at point \$ A \$ on the inner surface of a sphere. In which case will it reach the lowest point \$ B \$ faster: by sliding along the sphere's surface or along an inclined plane \$ AB \$? Assume negligible friction, zero initial velocity, and that the distance \$ AB \$ is much smaller than the sphere's radius.

Answer

When moving along the surface of the sphere.

Solution

When moving along the surface of the sphere $t_1 = T/4 = \\frac{1}{4} \\cdot 2\\pi \\sqrt{\\frac{R}{g}}$ ($T$ is the period of oscillation of a pendulum of length $R$. Explain!). When moving along the inclined plane $t_2 = \\sqrt{\\frac{2s}{a}} = \\sqrt{2\\frac{2R \\sin \\alpha} {g \\sin \\alpha}} = 2 \\sqrt{\\frac{R}{g}} > t_1$ ($\\alpha$ is the angle of inclination of the plane).

17. A weight suspended from a spring causes it to elongate by \$ \\Delta l = 4 \\, \\text{cm} \$. Find the period \$ T \$ of the spring's natural oscillations with the weight.

Answer

$T = 2\\pi \\sqrt{\\Delta l / g} \\approx 0.4$ s.

18. Find the period \$ T \$ of natural oscillations in the systems described in problem 14 in Chapter: Statics (see figure). $\"Description\"$

Answer

For series connection of springs $T = 2\\pi \\sqrt{m\\dfrac{k_1+k_2}{k_1 k_2}$. For parallel connection $T = 2\\pi \\times \\sqrt{\\dfrac{m}{k_1+k_2}}$.

Hint

See problem 14 in Chapter: Statics.

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1. A hollow cube with edge length \$ a \$ is filled to the top with a liquid of density \$ \\rho \$. Determine the forces acting on the faces of the cube.

Answer

On the bottom $\\rho g a^3$, on the side face $(1/2) \\rho g a^3$.

Hint

The pressure $p$ in a liquid at depth $h$ is $\\rho g h$, i.e., it varies linearly with height (see figure). Therefore, the total pressure force on the entire side face of the cube is equal in magnitude to the average pressure (i.e., the pressure at depth $a/2$) multiplied by the area of the face. $\"Description\"$

2. A vessel shaped like a truncated cone with a detachable bottom is submerged in water. If 200 g of water is poured into the vessel, the bottom detaches. Will the bottom detach if a 200 g weight is placed on it? If 200 g of oil is poured in? If 200 g of mercury is poured in?

Answer

If the container narrows upwards, the weight and mercury will not detach the bottom, but the oil will. If the container narrows downwards, the opposite is true.

Hint

The pressure force of the liquid acts on the bottom from the inside: $\\rho g h S$ (where $\\rho$ is density, $h$ is the liquid level height, $S$ is the bottom area). If the container narrows upwards, this force is greater than the weight of the liquid poured into the cone by the amount of the weight of the liquid occupying the shaded volume (figure, a). Therefore, mercury and the weight will not detach the bottom. Oil will detach the bottom, because in the case of a cone narrowing upwards, the product $ph$ for oil is greater than for water (oil will occupy a larger volume than the same amount of water. This relative increase in level height $h$ is greater than the relative decrease in density $\\rho$). If the container narrows downwards, the pressure force of the liquid poured into the container on the bottom is less than its weight by the amount of the weight of the liquid in the shaded volume (figure, b). Therefore, in this case, oil will not detach the bottom, but the weight and mercury will. $\"Description\"$

3. A tube with cross-sectional area \$ S = 2 \\, \\text{cm}^2 \$ is inserted into a vessel with water. Then, 72 g of oil (\$ \\rho_M = 900 \\, \\text{kg/m}^3 \$) is poured into the tube. Find the difference in the levels of the oil and water.

Answer

$\\Delta h = 4$ cm.

Hint

Apply the equilibrium condition for non-uniform liquids in communicating vessels.

4. To lift a load of mass \$ m = 2 \\, \\text{t} \$ using a hydraulic press, work \$ A = 40 \\, \\text{J} \$ is expended. The small piston makes \$ n = 10 \$ strokes, moving \$ h = 10 \\, \\text{cm} \$ per stroke. By what factor is the area of the large piston greater than that of the small piston, assuming the press's efficiency is 1?

Answer

$S_2/S_1 = n h m g / A \\approx 500$ times.

5. Water is poured into communicating vessels with diameters \$ D_1 \$ and \$ D_2 \$. By how much will the water level change in the vessels if a piece of wood with mass \$ m \$ is placed in the first vessel? In the second vessel? The density of water is \$ \\rho_0 \$.

Answer

The water levels in both cases rise by $\\Delta h = \\frac{m}{\\frac{\\pi}{4} (D_1^2 + D_2^2) \\rho_0}$.

6. Water and alcohol, separated by mercury, are poured into the arms of a U-shaped tube. The mercury levels in both arms are equal. At a height of 24 cm from the mercury level, the arms are connected by a horizontal tube with a valve (see figure). Initially, the valve is closed. Determine the height of the alcohol column \$ h_2 \$ (\$ \\rho_c = 800 \\, \\text{kg/m}^3 \$) if the height of the water column \$ h_1 = 32 \\, \\text{cm} \$. What happens if the valve is opened? At what position of the tube will equilibrium be maintained when the valve is opened? $\"Description\"$

Answer

$h_2 = 40$ cm. Alcohol will flow over. The tube should be positioned inclined, descending towards the arm with water.

Hint

See problem 3. By plotting the graph of pressure versus height (initial level - mercury level) for both arms (see the following figure), it is easy to see that at the height where the tube is located, the pressure in the arm with alcohol is greater than in the arm with water. Therefore, alcohol will flow over. For equilibrium not to be disturbed when the tap is opened, the tube must be positioned inclined. The difference in heights of the tube connection points can be found using the graph. $\"Description\"$

7. An ice floe with cross-sectional area \$ S = 1 \\, \\text{m}^2 \$ and height \$ H = 0.4 \\, \\text{m} \$ floats in water. What work must be done to fully submerge the ice floe in water?

Answer

$A = gSH^2 (\\rho_w - \\rho_i)^2 / (2\\rho_w) \\approx 7.84$ J.

Hint

During uniform immersion of the ice floe, the buoyant force $Q$ acting on it increases uniformly, and therefore the force $F = Q - mg = (\\rho_w - \\rho_i) gSh$ ($m$ is the mass of the ice floe, $\\rho_w$ is the density of water, $\\rho_i$ is the density of ice, $h$ is the depth of immersion) changes linearly with the immersion height from zero (when the ice floats) to the value $F_H = (\\rho_w - \\rho_i) gSH.$ Therefore, the work required to immerse the ice floe is equal in magnitude to the work of the average force $(1/2)F_H$ over the path equal to the thickness of the part of the ice floe protruding above the water.

8. A piece of ice floats in a glass of water. Will the water level change when the ice melts? Also consider the cases: 1) when the ice contains an air bubble; 2) when the ice contains a lead plate.

Answer

Ice displaces water whose weight is equal to the weight of the ice. When the ice melts, the same amount of water is formed, so the level will not change.

The presence of an air bubble in the piece of ice does not change the previous reasoning (the mass of air is negligible), therefore, in this case too, the water level in the vessel will remain the same.
If the ice contains a piece of lead, then the volume of water formed when the ice melts, together with the volume of the piece of lead, will be less than in the case of a piece of pure ice of the same weight. Consequently, the level will decrease.

9. One bottle is filled with water, the other with mercury. Will the water-filled bottle sink if placed in water? Will the mercury-filled bottle sink if placed in mercury?

Answer

The bottle with water will sink, but the one with mercury will not.

Hint

The weight of the bottle with water is greater than the weight of water whose volume is equal to the external volume of the bottle, while the bottle with mercury weighs less than the same volume of mercury.

10. A rectangular tin box with mass \$ m = 76 \\, \\text{g} \$, base area \$ S = 38 \\, \\text{cm}^2 \$, and height \$ H = 6 \\, \\text{cm} \$ floats in water. Determine the height \$ h \$ of the box's part above the water.

Answer

$h = H - m/(\\rho_0 S) = 4$ cm, where $\\rho_0$ is the density of water.

11. A 2-liter pot is completely filled with water. A body with volume \$ 0.5 \\, \\text{L} \$ and mass \$ 0.6 \\, \\text{kg} \$ is placed into the pot. How much water will spill out?

Answer

0.5 kg.

12. A tin can with a load floats on the surface of water in a vessel, with the water level at \$ H_1 \$. Will the new water level \$ H_2 \$ be higher or lower than \$ H_1 \$ if the load is moved from the can to the bottom of the vessel? The density of the load is greater than that of water.

Answer

Less.

Solution

When the weight is removed from the beaker, the volume of water displaced by the beaker will decrease by $V_1 = m/\\rho_1$ ($m$ is the mass of the weight, $\\rho_1$ is the density of water). When the weight is immersed in water, the volume of displaced water will increase by the volume of the weight $V_2 = m/\\rho_2$ ($\\rho_2$ is the density of the weight). Since $\\rho_2 > \\rho_1$, then $V_2 < V_1$, which means $H_2 < H_1$.

13. A liquid with density \$ \\rho \$ is poured into a vessel with vertical walls and base area \$ S \$. By how much will the liquid level change if a body of arbitrary shape with mass \$ m \$ (which does not sink) is placed into the vessel?

Answer

$\\Delta h = m/(\\rho S)$.

Solution

1st method. When the body is placed in the vessel, the pressure force on the bottom increases by $\\Delta F = mg.$ But the body itself does not touch the bottom of the vessel, so the pressure force changes due to the change in water pressure. If the water level in the vessel rises by $\\Delta h$, then $\\Delta p = \\rho g \\Delta h$ and $\\Delta F = \\rho g \\Delta h S.$ Therefore $\\rho g \\Delta h S = mg$. Hence $\\Delta h = m/(\\rho S).$

2nd method. A body of mass $m$, floating on the surface of a liquid, according to Archimedes' law, will displace (occupy) a volume $\\Delta V$ in it, which would be occupied by the same liquid of mass $m$. Thus, this immersion is equivalent (in the sense of changing the liquid level in the vessel) to adding liquid of mass $m$ to the vessel. Then $\\Delta h = \\Delta V / S = m / (\\rho S)$.

14. A liquid with density \$ \\rho \$ is poured into a U-shaped tube with cross-sectional area \$ S \$. By how much will the liquid level rise in the right arm relative to the initial level if a body with mass \$ m \$ and density \$ \\rho_1 < \\rho \$ is placed into the left arm?

Answer

$\\Delta h = m/(2\\rho S)$.

Hint

See problem 13. The changes in liquid levels in the arms are the same.

15. A mushroom-shaped concrete structure is installed at the bottom of a river (dimensions are given in the following figure). The river's depth is \$ H \$. With what force does the structure press on the riverbed? The density of concrete is \$ \\rho \$, and that of water is \$ \\rho_0 \$. $\"Description\"$

Answer

$g \\frac{\\pi}{4} \\{ \\rho_0 d^2 (H-h) + (\\rho - \\rho_0)(D^2 - d^2)h + \\rho d^2 h_1 \\}$.

Hint

The pressure force of the structure on the bottom consists of the pressure force of water on the upper surface of the part, the weight of the structure, and the pressure force of water on the lower surface, directed upwards and equal to $\\frac{\\pi}{4} (D^2 - d^2)[H - h_1 + h] \\rho_0 g.$

16. A wooden cube lies at the bottom of a vessel. Will it float if water is poured into the vessel (assuming water does not seep under the cube)?

Answer

No, because the buoyant (Archimedean) force does not arise - hydrostatic pressure is not transmitted to the lower face of the cube due to the absence of liquid under it.

17. A round hole with area \$ S_1 \$ in the bottom of a vessel is loosely covered by a conical plug with base area \$ S_2 \$. What is the maximum density \$ \\rho \$ of the plug material for which the plug can be made to float by adding water to the vessel? The density of water is \$ \\rho_0 \$.

Answer

$p = p_0 \\{ 1 + 2(S_1/S_2)^{1/2} - 3S_1/S_2 \\}$.

Hint

The buoyant (Archimedean) force is maximum when the water reaches the top of the plug. At this point, it is equal to the weight of the water in the shaded volume of the plug (see the following figure). $\"Description\"$

18. An empty open bottle is submerged neck-down into water to a depth \$ h \$ and released. The bottle neither rises nor sinks but remains in equilibrium. Why? Is this equilibrium stable? Determine the depth of submersion if the bottle's capacity is \$ V_0 = 0.5 \\, \\text{L} \$, its mass is \$ m = 0.4 \\, \\text{kg} \$, atmospheric pressure is \$ p_0 = 101 \\, \\text{kPa} \$, and the temperature is constant. Neglect the volume of the bottle's walls.

Answer

$h = p_0 (V_0 - m/\\rho) / (mg) \\approx 2.6$ m.

Hint

At the sought depth $h$, the water pressure is $p_1 = p_0 + \\rho g h$, where $p_0$ is the air pressure above the water surface; $\\rho$ is its density. For the air in the bottle, which compresses upon immersion, Boyle's law can be written: $p_0 V_0 = p_1 V_1,$ where $V_0$ and $V_1$ are the volumes of air in the bottle before and after immersion, respectively. Considering, finally, that the bottle at depth $h$ is in equilibrium, supported by the Archimedean force $mg = \\rho V_1 g,$ we can obtain $p_1 = p_0 V_0 \\rho / m.$

19. A hollow sphere (outer radius \$ R_1 \$, inner radius \$ R_2 \$) made of material with density \$ \\rho_1 \$ floats on the surface of a liquid with density \$ \\rho_2 \$. What should be the density \$ \\rho \$ of the substance filling the hollow part of the sphere for it to be in neutral equilibrium inside the liquid?

Answer

$\\rho = R_1^3 (\\rho_2 - \\rho_1) + \\dfrac{R_2^3 \\rho_1}{R_1^3}$.

Hint

The sphere will be in suspension inside the liquid if its mass (weight) is equal to the mass (weight) of the liquid whose volume is equal to the volume of the sphere.

20. A hollow cast iron sphere floats in water, submerged exactly halfway. Find the volume \$ V \$ of the hollow part if the sphere's mass is \$ m = 5000 \\, \\text{g} \$ and the density of cast iron is \$ \\rho = 7.8 \\, \\text{g/cm}^3 \$.

Answer

$V = m (\\frac{1}{\\rho} - \\frac{1}{\\rho_w}) \\approx 9350 \\text{ cm}^3$.

21. A vessel with water is balanced on a scale. How will the balance change if a block with dimensions \$ 5 \\times 3 \\times 3 \\, \\text{cm}^3 \$, suspended by a thread, is fully submerged into the water without touching the bottom? What weight and on which side must be added to restore balance?

Answer

45 g; on the other pan.

Hint

In accordance with Newton's third law, a force equal in magnitude to the Archimedean (buoyant) force acting on the block, and directed downwards, will act on the vessel with water from the side of the block. This force arises due to the increase in hydrostatic pressure on the bottom (and walls) of the vessel as a result of the rise in the water level in the vessel after immersing the block (see also problem 13).

22. Solid aluminum and iron balls are balanced on a lever. Will the balance be disturbed if the balls are submerged in water? Consider two cases: a) balls of equal mass; b) balls of equal volume.

Answer

In both cases, it will be disturbed; the iron ball will outweigh.

Hint

See the previous problem. In the second case, the arms are not equal and the moments of the buoyant forces are also not equal.

23. The weight of a piece of iron in water is \$ P = 1.67 \\, \\text{N} \$. Find its volume \$ V_{\\text{Fe}} \$. The density of iron is \$ \\rho_{\\text{Fe}} = 7.8 \\, \\text{g/cm}^3 \$.

Answer

$V_x = 25.1 \\text{ cm}^3$.

Solution

Since the weighed piece of iron is in equilibrium, we can write: $m_i g = P + m_w g$, where $m_i$ and $m_w$ are the masses of iron and water in volumes equal to $V_x$. The right side of the equality is the sum of the forces applied to the body: Archimedean force and the reaction force of the support (suspension), equal to the weight of the body in water $P$. Expressing the masses of iron and water through the volume $V_x$ and the corresponding densities, we get $V_x \\rho_i g = P + V_x \\rho_w g,$ whence $V_x = \\frac{P/g}{\\rho_i - \\rho_w}.$

24. The weight of a body in water is three times less than its weight in air. What is the density of the body's material?

Answer

1.5 g/cm$^3$.

Solution

1st method. See the previous problem.

2nd method. Assume that the weight of the body in air is three units, then in water, according to the condition, the weight is one unit, i.e., the loss of weight in water is two units, which is equal to the weight of water in the volume of the body. Hence the density is $3/2$ g/cm$^3 = 1.5$ g/cm$^3$.

25. A wooden block floats in water. How will the depth of the block's submersion change if oil is poured on top of the water?

Answer

It will decrease, as the liquid pressure on the lower face of the block increases.

26. A certain body floats on the surface of water in a closed vessel. How will the depth of submersion change if air is pumped into the vessel?

Answer

It will not change if the compressibility of the body is the same as that of water; it will increase if the compressibility is greater than that of water, and decrease otherwise.

27. One end of a thread is fixed to the bottom, and the other is attached to a cork float. Here, 0.75 of the float's total volume is submerged in water. Determine the tension force \$ F \$ in the thread if the float's mass is 2 kg and the density of cork is \$ 0.25 \\, \\text{g/cm}^3 \$. Neglect the mass of the thread.

Answer

$F \\approx 40$ N.

28. A bucket hangs from a dynamometer's hook. Will the dynamometer's reading change if the bucket is filled with water and submerged in water?

Answer

It will decrease by the amount of the weight of water displaced by the walls and bottom of the bucket.

29. A vessel, completely filled with water, hangs from a dynamometer. Will the dynamometer's reading change if a weight suspended by a thread is submerged into the water without touching the bottom?

Answer

The dynamometer reading will not change.

Hint

The water displaced by the weight will spill out of the vessel. Thus, the weight of the water in the vessel will decrease, which exactly compensates for the force arising from the weight on the vessel with water, which is equal to the buoyant (Archimedean) force acting on the weight. (See also problem 10.21).

30. A vessel with water and a stand with a copper weight of mass \$ m = 100 \\, \\text{g} \$ (see figure) are balanced on a lever scale. Then, the weight, suspended by a thread, is lowered into the water. How can the balance be restored? The density of copper is \$ \\rho_M = 8.9 \\, \\text{g/cm}^3 \$. $\"Description\"$

Answer

Place a weight of mass $M = 2m \\rho_w / \\rho_m \\approx 24$ g on the right pan of the balance.

Hint

When the weight is immersed, the force acting on the left arm of the balance beam increases, and on the right - decreases by an amount equal to the buoyant force $F = V \\rho_w g = m \\rho_w g / \\rho_m$ (see problem 10.21).

31. A thin uniform rod is hinged at its upper end. The lower part of the rod is submerged in water, and equilibrium is achieved when the rod is inclined to the water's surface, with half of it underwater. What is the density of the rod's material?

Answer

0.75 g/cm$^3$.

Solution

Since the stick is thin, we can assume that the buoyant force $Q$ is applied at the midpoint of the part of the stick submerged in water (see figure). By virtue of the equilibrium of the stick, the moments of forces $mg$ and $Q$ relative to point A must be equal, i.e., $mgx = Q \\cdot (3/2)x$, whence $Q = (2/3)mg$. Substituting $m = \\rho V$ and $Q = (1/2) \\rho_w V g$ here ($V$ is the volume of the stick, $\\rho$ is the density of the stick material and $\\rho_w$ is the density of water), we find $\\rho = (3/4) \\rho_w$. $\"Description\"$

32. Two balls with radii \$ r_1 \$ and \$ r_2 \$, made of materials with densities \$ \\rho_1 \$ and \$ \\rho_2 \$, are connected by a weightless rod of length \$ l \$. The system is placed in a liquid with density \$ \\rho \$ (\$ \\rho < \\rho_1 \$ and \$ \\rho < \\rho_2 \$). At what point on the rod should it be suspended for the system to remain in equilibrium with the rod horizontal?

Answer

At a distance $x = l \\frac{\\rho_2 - \\rho}{(\\rho_2 - \\rho) + (\\rho_1 - \\rho)(r_1/r_2)^3}$ from the sphere with density $\\rho_1$.

Hint

The differences between the forces of gravity and Archimedean forces acting on the weights are respectively: $F_1 = \\frac{4}{3} \\pi r_1^3 (\\rho_1 - \\rho)g$ and $F_2 = \\frac{4}{3} \\pi r_2^3 (\\rho_2 - \\rho)g$. Since the rod with the weights is in equilibrium, $F_1 x = F_2 (l - x)$ ($x$ is the distance from the weight with density $\\rho_1$ to the suspension point of the rod).

33. A tube with radius \$ r \$ and height \$ h \$ emerges from a vessel filled with water (see figure). The tube is closed by a circular plate of radius \$ R \$ and mass \$ M \$, pressed against the tube by water pressure. With what force \$ F \$ must the plate be pushed at point \$ A \$ to make it pivot and open the tube? The vessel is filled with water to height \$ H \$. The plate's thickness is negligible. $\"Description\"$

Answer

$F > \\dfrac{\\rho(H-h)\\pi r^3 - M(R-2r)}{2(R-r)} g$.

Solution

The plate will rotate around point B (see the following figure). The water pressures from both sides of the plate balance each other everywhere except for the area above the tube. Here the water pressure force is $F = \\rho g (H-h) \\pi r^2$. Its moment relative to point B is $\\rho g (H-h) \\pi r^3$. The plate will turn over if $\\rho g (H-h) \\pi r^3 < Mg(R-2r) + F(2R-2r);$ $F > \\frac{\\rho(H-h)\\pi r^3 - M(R-2r)}{2(R-r)} g.$ $\"Description\"$

34. A body submerged in a liquid is balanced on a scale. Will the scale's reading change if the liquid and the submerged body are heated?

Answer

The answer depends on the ratio of the thermal expansion coefficients of the body and the liquid. If the thermal coefficient of the body is less than that of the liquid (as is usually the case), the body will outweigh.

35. A solid homogeneous body with volume \$ V \$ and material density \$ \\rho \$ floats at the boundary between a heavy liquid (density \$ \\rho_1 \$) and a lighter liquid (density \$ \\rho_2 \$). What fraction of the body's volume \$ V_1 \$ will be submerged in the heavy liquid?

Answer

$V_1/V = (\\rho - \\rho_2)/(\\rho_1 - \\rho_2)$.

Solution

The mass of the body must be equal to the mass of the displaced liquids: $\\rho V = \\rho_1 V_1 + \\rho_2 V_2$, where $V_1 + V_2 = V$. Solving these equations together, we find $V_1 = V \\frac{\\rho - \\rho_2}{\\rho_1 - \\rho_2}$.

36. A wooden cube with side length 10 cm floats between oil and water, with 2.5 cm of its height below the oil's surface. The cube's bottom surface is 2.5 cm below the interface. What is the cube's mass \$ m \$, given the oil's density is \$ 0.8 \\, \\text{g/cm}^3 \$? Determine the pressure forces \$ F_1 \$ and \$ F_2 \$ on the cube's top and bottom faces. Will the cube's submersion depth in water change if more oil is added?

Answer

$m \\approx 0.85$ kg; $F_1 \\approx 1.96$ N; $F_2 \\approx 10.29$ N. Does not change.

Hint

See the solution to the previous problem.

37. A steel cube (density \$ 7.8 \\, \\text{g/cm}^3 \$) floats in mercury (density \$ 13.6 \\, \\text{g/cm}^3 \$). Water is poured over the mercury until it covers the cube with a thin layer. What is the height \$ H \$ of the water layer? The cube's edge length is \$ a = 10 \\, \\text{cm} \$. Determine the pressure \$ p \$ on the cube's bottom face.

Answer

$H \\approx 4.6$ cm; $p \\approx 7650$ Pa.

38. A piece of cork weighs \$ 0.147 \\, \\text{N} \$ in air, and a piece of lead weighs \$ 1.1074 \\, \\text{N} \$. If these pieces are tied together, suspended from a scale, and submerged in kerosene, the scale reads \$ 0.588 \\, \\text{N} \$. Determine the density of the cork, given the density of kerosene is \$ 0.8 \\, \\text{g/cm}^3 \$ and that of lead is \$ 11.3 \\, \\text{g/cm}^3 \$.

Answer

0.2 g/cm$^3$.

Solution

Let: mass of the cork $m_1$, its density $\\rho_1$, mass of lead $m_2$, its density $\\rho_2$. Then the volume of the cork $V_1 = m_1/\\rho_1$, volume of lead $V_2 = m_2/\\rho_2$, total volume $V = V_1 + V_2 = m_1/\\rho_1 + m_2/\\rho_2$. According to Archimedes' law, the buoyant force $F$ acts on the body immersed in the liquid [in our case $F = (m_1 + m_2)g - Q$, where $Q$ is the reading of the scales], equal to the weight of the displaced liquid in the volume of the immersed body $F = V \\rho_3 g$, where $\\rho_3$ is the density of kerosene. Hence $(m_1 + m_2)g - Q = (m_1/\\rho_1 + m_2/\\rho_2) \\rho_3 g$. Solving the obtained equation for the unknown quantity $\\rho_1$, we get $\\rho_1 = \\frac{m_1 g \\rho_2 \\rho_3}{[(m_1 + m_2)g - Q]\\rho_2 - m_2 g \\rho_3}$.

39. An open cylindrical glass is submerged into water in two ways: once with the base up and once with the base down, both to the same depth. In which case is the work required to submerge the glass greater? No water spills from the vessel, and no water enters the glass when submerged base-down.

Answer

When the beaker is immersed in the inverted position, water, compressing the air inside it, partially fills its internal volume. With the same immersion of the beaker in water, in the first case, the displaced volume of water is less than in the second case, and therefore the work is less.

40. Two balloon envelopes of equal mass—one made of elastic rubber, the other of rubberized fabric—are filled with the same amount of hydrogen and occupy equal volumes near Earth. Which balloon will rise higher, assuming no hydrogen escapes?

Answer

The thin rubber balloon will rise to a great height. At the same height above the Earth, the volume of the thin rubber balloon, and hence the buoyant force acting on it, will be greater than for the fabric balloon.

41. By what factor will the lifting force of the gas filling an aerostat (airship) change if helium is used instead of hydrogen?

Answer

Decreases by a factor of 1.08.

Solution

The lift force $F_L$ of a certain volume of gas is equal to the difference between the weight $m_{air}g$ of the air displaced by the gas and the weight $m_g g$ of the gas itself: $F_L = (m_{air} - m_g)g$. The ratio of the lift forces $(m_{air} - m_{water}) / (m_{air} - m_{helium})$ is equal to the ratio of the differences in molar masses, since the pressure $p$, volume $V$, and temperature $T$ are the same ($pV = mRT/\\mu$). Consequently, $F_1/F_2 = (\\mu_{air} - \\mu_{H2}) / (\\mu_{air} - \\mu_{He}) = (29 - 2) / (29 - 4) = 27/25 = 1.08$.

42. A thin-walled tube of a mercury barometer is suspended from a dynamometer. What does the dynamometer show? Will its reading change with atmospheric pressure variations?

Answer

The dynamometer shows the sum of the weight of the tube and the force of atmospheric pressure on the tube, equal to the weight of mercury in the tube.

43. Estimate the mass of the gaseous envelope surrounding the Earth.

Answer

$m = p \\cdot 4\\pi R^2 / g \\approx 5 \\times 10^{18}$ kg, where $p$ is atmospheric pressure; $R$ is the radius of the Earth.

44. An L-shaped tube, with its long arm open, is filled with hydrogen. In which direction will the rubber film covering the short arm bulge?

Answer

The pressures of air and water at level I-I (see figure) are respectively equal: $p_1 = p_0 - \\rho_{air} g h$, $p_2 = p_0 - \\rho_{water} g h$, where $p_0$ is the pressure at level 0-0; $\\rho_{air}$ and $\\rho_{water}$ are the densities of air and water. Since $\\rho_{water} > \\rho_{air}$, then $p_1 > p_2$, and the film will be bent outwards. $\"Description\"$

45. Water flows through a pipe with a constriction. An elastic rubber ball is released into the pipe. How will its diameter change as it passes through the narrow section?

Answer

It will increase, because the speed of water in the narrow part of the pipe is greater, and the pressure is correspondingly less than in the wide part of the pipe.

46. A body with mass \$ m = 2 \\, \\text{kg} \$ and volume \$ V = 1000 \\, \\text{cm}^3 \$ is located in a lake at depth \$ h = 5 \\, \\text{m} \$. What work must be done to lift it to a height \$ H = 5 \\, \\text{m} \$ above the water's surface? Is this work equal to the change in the body's potential energy? Explain the result.

Answer

$A = (m - \\rho V)gh + mgH \\approx 150$ J.

Solution

To move the body in water, a force $F = mg - F_{buoy} = mg - \\rho V g$ ($\\rho$ is the density of water) must be applied to it. Therefore, to lift the body from depth $h$ to height $H$ above the water surface, work must be done $A = (m - \\rho V)gh + mgH \\approx 150$ J. This work is less than the change in potential energy of the body $\\Delta W_p = mg(h+H)$ by the amount $\\rho V gh \\approx 50$ J. The found difference is precisely the amount by which the potential energy of the water decreased when the body was lifted - volume $V$ of water descends from the surface to depth $h$.

47. A vertical tube with a piston is fixed in a reservoir such that its lower end is submerged in water. The piston, initially lying on the water's surface, is slowly raised to height \$ H = 15 \\, \\text{m} \$. What work is required to do this? The piston's area is \$ S = 1 \\, \\text{dm}^2 \$, and atmospheric pressure is \$ p = 101 \\, \\text{kPa} \$. Neglect the piston's weight.

Answer

$A = \\frac{pS}{2} (2H - \\frac{p}{\\rho_0 g}) \\approx 9800$ J, where $\\rho_0$ is the density of water.

Solution

The water pressure directly under the pump piston is equal to $p - \\rho_0 g h$ ($h$ is the piston lift height). The force lifting the piston is equal to the difference between the pressure forces above and below the piston, i.e., the value $F = [p_0 - (p_0 - \\rho_0 g h)]S = \\rho_0 g h S$. When the piston is lifted to a height $h = p_0 / (\\rho_0 g) = 10$ m, this force increases linearly to the value $p_0 S$. Further, this force remains constant. Therefore, the work done in lifting the piston will be equal to: $A = \\frac{p_0 S}{2} \\frac{p_0}{\\rho_0 g} + p_0 S (H - \\frac{p_0}{\\rho_0 g}) = \\frac{p_0 S}{2} (2H - \\frac{p_0}{\\rho_0 g})$.

48. A submarine is at depth \$ h = 100 \\, \\text{m} \$. At what speed will water rush in through a hole in the submarine's hull? How much water will enter in one hour if the hole's diameter is \$ d = 2 \\, \\text{cm} \$? The air pressure inside the submarine equals atmospheric pressure. Ignore changes in internal pressure.

Answer

$v \\approx 44.3$ m/s; $V = 50$ m$^3$.

Solution

Applying Torricelli's formula for the outflow velocity from the opening $v^2 = 2gh$, we find $v = \\sqrt{2gh}$. In time $t$, a volume of water $V = \\pi d^2 v t / 4$ will enter the boat.

49. A water jet shoots from a hose. The flow rate is \$ Q = 60 \\, \\text{L/min} \$. What is the cross-sectional area \$ S_1 \$ of the jet at height \$ h = 2 \\, \\text{m} \$ above the hose's end if near the end it is \$ S_0 = 1.5 \\, \\text{cm}^2 \$?

Answer

$S_1 \\approx 4.37$ cm$^2$.

Solution

The speed of water at the end of the fire hose is $v_0 = Q/S_0$. Flying out of the fire hose, the water moves with uniform deceleration with acceleration $g$, therefore at height $h$ the speed of water $v_h = \\sqrt{v_0^2 - 2gh}$. From the continuity equation of the stream $v_0 S_0 = v_h S_h$, which expresses that the mass of liquid flowing through any cross-section of the stream per unit time is the same, we get $S_h = \\frac{v_0 S_0}{v_h} = \\frac{Q}{\\sqrt{(Q/S_0)^2 - 2gh}}$.

50. Why does a fast-moving bullet only make two small holes in an empty plastic cup, while a water-filled cup shatters into pieces when hit by a bullet?

Answer

During the time the bullet is in the glass, the water level in it does not have time to change. The resulting high-pressure region upon compression of the water spreads in all directions with the speed of sound, reaches the walls of the glass, and it shatters. This experiment illustrates the low compressibility of liquids.

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Physics Problems and Questions Mechanics

By Unknown Author

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