Laws of Conservation of Energy and Momentum

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

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

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

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

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.

$T = 1000$ J.

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)

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

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 $v_{final} = \sqrt{gL}$.

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

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.

$H \approx 2$ m.

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)} $.

$A \approx -80.2$ J.

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). $

$A \approx 3.5$ J.

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

$F_{res} \approx 1570$ N.

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).

$Q \approx 9.8$ J.

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.

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

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 $x=2x_0$).

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

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.

$k = 0.05$.

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 $k = h/(b + l)$.

a) $p = 0.17$ N·s;

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

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). $

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

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 $h = \frac{v_1^2}{2g} = \frac{(v + 2u)^2}{2g} = \frac{(\sqrt{2gl} + 2u)^2}{2g}.$

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.

$l \approx 2.7$ m.

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

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

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 ($m/M \approx 1$).

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

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

$A \approx 105$ J.

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.

$v_1 \approx 590$ m/s.

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.

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

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}}. $

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

See the solution to problem 24.

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

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.

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

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)

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

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

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. $

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

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}. $

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

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.

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

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.

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

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)}. $

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

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)}. $

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

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 $u_x$.

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

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}.
$

$n = 5/3$.

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 $n = (4+1)/|4-1| = 5/3$).

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

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)

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

b) 1.9%.

See problems 28, 37 and 38.

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

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

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

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

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

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 $a = \dfrac{F_{\text{buoy}} - mg}{m} = g \dfrac{\rho_2 - \rho_1}{\rho_1}$.

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

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 $h_{\text{min}} = F_{\text{max}} \frac{\Delta l}{2mg} = Mg \frac{0.01l}{2mg} = 0.01 Ml / (2m).$

$s \approx 4.9$ m.

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 $s = 8F / (\rho g \pi d^2)$.

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

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 $s = \pi d^2/4$).

$l_1/l_2 = m_2/m_1$, 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.

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 $m_2/m_1$.

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

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}
$

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

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

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

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

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.

At a right angle.

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.

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

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.
$

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

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

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.

**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
$