Momentum. Law of Conservation of Momentum

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

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

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 $2mv \sin \alpha$.

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.

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.

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.

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).

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

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).

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.

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.

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

$s \approx 1695$ m.

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.

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

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.

$s_2 = 5000$ m.

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 $s_2 = s_1 + x_2$.

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

a) 4 times; b) 3 times.

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.

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

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 $v_p = v_k - v_{rel}$.

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

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

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 $h = \frac{(m_1 v_1 \sin \alpha + m_2 v_2 \sin \beta)^2}{2g(m_1 + m_2)^2}.$

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.

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 $x_1 = \frac{M}{M+m} l.$

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

See the solution to problem 3.15, 1st method.

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

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.

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

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

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

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: $u = \frac{m}{M_0} (\frac{S_2}{S_1} - 1) v_1.$

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

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.

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 $v = \sqrt{gh}$.

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 $v = v_x = \sqrt{2gh} \cos \alpha$.

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

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 $k = \frac{\rho S v^2}{mg \cos \alpha}.$

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

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

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.)

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

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.

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.

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

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