## Section8.11Energy Bootcamp

### Subsection8.11.11Miscellaneous

A block leaves a frictionless inclined surface horizontally after dropping off by a height $h$ as shown in Figure 8.11.49. After leaving the inclined surface, the block is in a free fall motion and lands on the floor after falling by height $H\text{.}$ Find the horizontal distance $D$ where it will land on the floor.

Hint

Energy conservation on the incline and then a projectile motion.

$D = 2\sqrt{h H}$

Solution

Strategy. I will treat the motion on the inclined surface by conservation of energy, from which I will get the speed at the bottom of the incline. Then, I will use projectile motion equations to dind $D\text{.}$

From the conservation of energy on the frictionless incline, we get

\begin{equation*} \dfrac{1}{2} m v^2 = mg h. \end{equation*}

Therefore,

$$v = \sqrt{2gh}.\label{eq-prob-block-on-incline}\tag{8.11.1}$$

After the block leaves the incline, the block is in a free fall motion. The vertical motion of the projectile motion starts with zero velocity. Therefore, the time to fall of $H$ under gravity obeys

\begin{equation*} H = \dfrac{1}{2}gt^2. \end{equation*}

This means the flight time is

\begin{equation*} t = \sqrt{ \dfrac{2H}{g} }. \end{equation*}

Multiplying this with the horizontal velocity $v$ in we get $D\text{.}$

\begin{equation*} D = v t = \sqrt{2gh} \times \sqrt{ \dfrac{2H}{g} } = 2\sqrt{h H}. \end{equation*}

A block of mass $m$ after sliding down a frictionless incline strikes another block of mass $M$ that is attached to a spring of spring constant $k$ (see Figure 8.11.51. The blocks stick together upon impact and travel together.

(a) Find the compression of the spring in terms of $m\text{,}$ $M\text{,}$ $h\text{,}$ $g$ and $k$ when the combination comes to rest.

(b) The loss of kinetic energy as a result of the bonding of the two masses upon impact is stored in the so-called binding energy of the two masses. Calculate the binding energy.

Hint

(a) First use conservation of energy, then use conservation of momentum, and then energy of conservation again. (b) Follow the instructions in the problem statement.

(a) $\sqrt{\dfrac{2m^2gh}{k(m+M)}}\text{,}$ (b) $\dfrac{mMgh}{m+M} \text{.}$

Solution 1 (a)

Strategy. Before we solve this problem, we first work out a strategy. It involves three steps as outlined in Figure 8.11.52.

Step 1: Conservation of energy on the incline. We get the speed at the bottom of the incline by

\begin{equation*} \dfrac{1}{2} m v^2 = m g h,\ \ \Longrightarrow\ \ v = \sqrt{2 g h}. \end{equation*}

Step 2: Momentum conservation. We get the speed $V$ of the two blocks together after the collision.

\begin{equation*} m v = (m + M) V, \ \ \Longrightarrow\ \ V = \dfrac{m}{m + M}\, v. \end{equation*}

Step 3: Conservation of energy. Let $\Delta l$ be the compression when the two blocks come to rest. The conservation of energy gives us

\begin{equation*} \dfrac{1}{2} k (\Delta l)^2 = \dfrac{1}{2} (m + M) V^2, \end{equation*}

which we solve for $\Delta l\text{.}$

\begin{align*} \Delta l \amp = \sqrt{\dfrac{m+M}{k}}\, V, \\ \amp = \sqrt{\dfrac{m+M}{k}}\, \dfrac{m v}{m + M}, \\ \amp = \sqrt{\dfrac{2 m^2 g h}{k(m+M)}}. \end{align*}
Solution 2 (b)

In the collision step, step 2 in Figure 8.11.52, kinetic energy is not conserved. The kinetic energy before the collision equals the sum of two energies, the kinetic energy after the collision and the binding energy,$E_B\text{.}$

\begin{equation*} \dfrac{1}{2}mv^2 = \dfrac{1}{2}(M+m) V^2 + E_B. \end{equation*}

Solving for $E_B\text{,}$ substituting for $V\text{,}$ and simplifying we get

\begin{align*} E_B \amp = \dfrac{1}{2}m v^2 - \dfrac{1}{2}(m+M) V^2\\ \amp = \dfrac{1}{2}mv^2\left(1 - \dfrac{m}{m+M} \right) = \dfrac{1}{2}\, \dfrac{mM}{m+M}\, v^2. \end{align*}

We can use $v = \sqrt{2gh}$ to get the following expression.

\begin{equation*} E_B = \dfrac{mM}{m+M}\, g\,h. \end{equation*}

A block of mass $m$ slides in a frictionless track that bends in a circular arc and then straightens out as shown in Figure 8.11.54. If the block is released at rest from a critical height $h\text{,}$ the block will make it through the circular track without falling off; any less height and the block would not complete the circular path. Find the critical value of height $h\text{.}$

Hint

Use energy conservation and equation of motion in a circular motion.

$h = \dfrac{5}{2}\,R\text{.}$

Solution

Strategy. Refer to Figure 8.11.55. We first write Newton's law for circular motion at C. Then apply energy conservation between initial point and point C.

The lowest speed occurs at point C. In order for the block to make through point C in the circle, the normal force would there will barely be zero. That means when the block is at C, we will have the following equation if motion in the vertical direction.

\begin{equation*} m\dfrac{v_C^2}{R} = mg. \end{equation*}

Therefore,

$$v_C^2 = g R.\label{eq-prob-circular-track-1}\tag{8.11.2}$$

Now, between point where the block was released and point C, energy is conserved. That means

\begin{equation*} \dfrac{1}{2}\,m v_C^2 = mg (h - 2R). \end{equation*}

Using $v_c^2$ from Eq. (8.11.2) we get

\begin{equation*} \dfrac{1}{2}R = h - 2 R. \end{equation*}

Therefore,

\begin{equation*} h = \dfrac{5}{2}\,R. \end{equation*}

A small block of mass $m$ is sitting at the top of a sphere of radius $R\text{.}$ When the block is moved infinitesimally to one side, it slides on the spherical surface up to a point when it leaves the spherical surface. Assuming no friction between the block and the surface of the sphere find the point at which the block will fall off the surface.

You can specify the location of the place on the sphere by an angle a line from the center of the sphere to that point will make with the vertical line.

Hint

The instant after the block is no longer in contact, there is no normal force on the block, but the speed of the block is same as it was when it left the surface.

$\cos^{-1}(2/3)$

Solution

Strategy. (1) When the block is sliding on the sphere, it has two forces - the weight and the normal force with centripetal acceleration based on the balance of the components of these forces in the direction towards the center of the sphere. The normal force will become zero at the point the block is not in contact with the sphere. We will set up an equation of motion for this point. (2) Due to the frictionless nature of the surface, we have energy conservation. The height dropped from the top will be $R - R \cos\,\theta\text{.}$

We now work out these steps.

(1) Let $v$ be the speed when the block is about to leave the surface. We will have the following equation of motion at this instant.

\begin{equation*} m g\cos\,\theta = m \,\dfrac{v^2}{R}. \end{equation*}

(2) The energy conservation will give

\begin{equation*} \dfrac{1}{2}\, m v^2 = m g ( R - R \cos\,\theta ). \end{equation*}

We can eliminate $v^2$ from these two equations to get

\begin{equation*} \cos\theta = \dfrac{2}{3}. \end{equation*}

Hence, $\theta = \cos^{-1}(2/3)\text{.}$

A bullet of mass $m$ is fired horizontally with speed $v$ in a block of $M$ that is hanging by a massless string of length $L$ from the ceiling. After the impact the bullet is embedded in the block at its center and the two move together. A pendulum used this way is called a ballistic pendulum.

Find the angle $\theta$ the string will make with the vertical when the bullet/block system comes to rest momentarily. Give your answer in terms of quantities given and $g\text{.}$

Hint

Two steps: momentum conservation in collision and then energy conservation.

$\sin\,\dfrac{\theta}{2} = \dfrac{1}{2\sqrt{gL}}\ \dfrac{mv}{m+M}$

Solution

Strategy. This problem has two steps. (1) An inelastic collision and (2) energy conservation as illustrated in Figure 8.11.58.

Step 1: conservation of momentum gives

\begin{equation*} V = \dfrac{mv}{m+M}. \end{equation*}

Step 2: conservation of energy for the pendulum gives

\begin{equation*} (m+M) g (L - L\cos\,\theta) = \dfrac{1}{2} (m+M) V^2. \end{equation*}

We can simplify this to

\begin{equation*} 1-\cos\,\theta = \dfrac{1}{2gL}\, V^2 \end{equation*}

By writing $1-\cos\,\theta = 2\sin^2(\theta/2)$ and replacing $V$ from the momentum conservation equation, we get

\begin{equation*} \sin\,\dfrac{\theta}{2} = \dfrac{1}{2\sqrt{gL}}\ \dfrac{mv}{m+M}. \end{equation*}

A ball bounces off the floor inelastically so that each bounce takes the ball to a lower height than the previous bounce. Eventually the ball comes to rest.

(a) If the coefficient of restitution is $0.95\text{,}$ what is the speed of the ball immediately after the first bounce if it was dropped at rest from a height of $3 \text{ meters}\text{?}$

(b) How long does it take for the ball to come to rest? You might need the following infinite series.

\begin{equation*} \dfrac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \ \ \ (|x| \lt 1) \end{equation*}
Hint

(a) Assume floor remains at rest. (b) Set up the geometric series.

(a) $7.3 \text{ m/s}\text{,}$ (b) $30.5 \text{ sec}\text{.}$

Solution 1 (a)

The ball falls freely by height $3\text{ m}\text{.}$ This gives the speed before striking the floor to be

\begin{equation*} v_0 = \sqrt{2gh} = \sqrt{2\times 9.81 \times 3} = 7.67\text{ m/s}. \end{equation*}

Since resitution $\epsilon=0.95\text{,}$ the speed after the bounce will be

\begin{equation*} v_1 = \epsilon\,v_0 = 0.95\times 7.67 = 7.3\text{ m/s}. \end{equation*}
Solution 2 (b)

Let $t_0$ be the time for initial fall from height $h\text{,}$ and $t_1,\ t_2,\ t_3,\ \cdots$ be the durations between successive bounces. Let $v_0$ be the speed just before the first impact, and $v_1,\ v_2,\ v_3,\ \cdots$ e the speeds after bounces 1, 2, 3, etc.

Notice that

\begin{equation*} v_1 = \epsilon v_0,\ \ v_2 = \epsilon v_1,\ \ \cdots. \end{equation*}

Therefore, we can write

\begin{equation*} v_j = \epsilon^{j-1}\, v_0. \end{equation*}

Using these we get the following for the durations

\begin{align*} \amp t_0 = \dfrac{v_0}{g}\\ \amp t_j = 2\, \dfrac{v_0}{g}\, \epsilon^{j},\ \ \ j=1,\,2,\,3,\,\cdots \end{align*}

where we get the factor 2 for up and down motion between bounces. Adding up the times we get

\begin{align*} T \amp = t_0+t_1+t_2+t_3+\cdots, \\ \amp = \dfrac{v_0}{g} + 2\, \dfrac{v_0}{g}\, \epsilon \left( 1 + \epsilon + \epsilon^2 + \cdots \right), \\ \amp = \dfrac{v_0}{g} + 2\, \dfrac{v_0}{g}\, \dfrac{\epsilon}{1-\epsilon}, \\ \amp = \dfrac{v_0}{g}\, \dfrac{1+\epsilon}{1-\epsilon}, \end{align*}

Now, we use the numerical values given to find the time

\begin{align*} v_0 \amp = \sqrt{2gh} = 7.67\text{ m/s},\\ \amp = \dfrac{7.67}{9.81}\, \dfrac{1+0.95}{1-0.95} = 30.5\text{ sec}. \end{align*}

A ring of mass $M$ and radius $R$ has two metal beads, each of mass $m\text{.}$ The beads can slide frictionlessly on the ring as illustrated in Figure 8.11.61. The ring with the beads is hung from the ceiling with a string and the beads are brought at the top and released from rest.

The beads then slide off on the two sides in a symmetric way. When beads reach particular points on the ring, the ring tends to go up if masses are appropriate. In this problem, you will find the place where this will happen. The following parts will guid to to the solution.

(a) The normal force of the ring on the beads varies with its position on the ring. Suppose the position of a bead on the ring is given by angle $\theta$ with respect to the vertical. Find the magnitude of the normal force on a bead $F_N$ as a function of $\theta$ of the beads.

(b) The tension $F_T$ in the ring also varies depending upon where the beads are located in their motion. Find $F_T$ as a function of $\theta\text{.}$

(c) At some value of $\theta$ of the beads, the tension will become zero if $M/m$ is not too great. Find this angle and the condition on $M/m\text{.}$ This is the place the ring will tend to go up.

(d) Suppose $M = 0.4\text{ kg}$ and $m = 0.6\text{ kg}\text{.}$ What is the angle at which the ring would go up?

Hint

(a) Implement equation of motion for centripetal direction and conservation of energy. (b) Use the normal force found in (a) but in the opposite direction when acting on the ring. (c) Set $F_T=0$ and solve for $x=\cos\,\theta$ and notice the discriminant of the quadratic equation solution. (d) Look at the normal forces on the ring,.

(a) $F_N = 3 m g \cos\,\theta - 2 m g,$ (b) $F_T = Mg + 6 mg \cos^2\,\theta- 4 mg \cos\,\theta,$ (c) $\cos\,\theta = \dfrac{1}{12}\left(4 \pm \sqrt{ 4^2 - 24\dfrac{M}{m} } \right),$ (d) $70.5^{\circ}.$

Solution 1 (a)

(a) Let us look at forces on one of the beads at some instant when it has angle $\theta$ with respect to the vertical. The forces are its weight $mg$ and a normal force from the ring, which we have shown pointed radially out in the figure.

Using the centripetal acceleration $v^2/R$ towards the center of the circle we write the radially-inward component of $\vec F = m\vec a$ to get

$$- F_N + m g \cos\,\theta = m\,\dfrac{v^2}{R}. \label{eq-prob-beads-on-ring-centripetal}\tag{8.11.3}$$

Since sliding of the beads is frictionless, we have conservation of energy. The height $h$ a bead falls by is

\begin{equation*} h = R - R\,\cos\,\theta. \end{equation*}

Therefore, we will have

$$\dfrac{1}{2}m v^2 = m g (R - R\,\cos\,\theta).\label{eq-prob-beads-on-ring-energy}\tag{8.11.4}$$

Replacing $mv^2$ from this equation into Eq. (8.11.3) we get

$$F_N = 3 m g \cos\,\theta - 2 m g.\label{eq-prob-beads-on-ring-normal}\tag{8.11.5}$$
Solution 2 (b)

(b) Moving of the ring means its center of mass moves. Vertical component of net force on the ring is zero until the time ring reacts upwards. The forces on the ring are its weight $Mg\text{,}$ the tension $F_T (\theta) \text{,}$ and two normal forces from the beads, whose magnitudes were computed in part (a).

Note that the direction of the normal forces from beads are in exactly opposite direction to the normal force on the beads. They are the third law force pairs. Let $y$ axis be pointed up. We work out $y$ components and sum them to zero, $\sum F_y = 0\text{.}$ This gives us

\begin{equation*} F_T - Mg - 2 F_N\,\cos\,\theta = 0. \end{equation*}

Using the expression for $F_N$ from Eq. (8.11.5) we get

$$F_T = Mg + 6 mg \cos^2\,\theta- 4 mg \cos\,\theta.\label{eq-prob-beads-on-ring-tension}\tag{8.11.6}$$
Solution 3 (c)

(c) Setting $F_T=0$ in Eq. (8.11.6) we get a quadratic equation in $\cos\,\theta\text{.}$

\begin{equation*} Mg + 6 mg \cos^2\,\theta- 4 mg \cos\,\theta = 0. \end{equation*}
Let $x = \cos\,\theta\text{,}$ and dividing by $mg$ we get
\begin{equation*} 6 x^2 - 4 x + \dfrac{M}{m} = 0. \end{equation*}
Solving for $x$ to get
$$x = \dfrac{1}{12}\left(4 \pm \sqrt{ 4^2 - 24\dfrac{M}{m} } \right),\label{eq-prob-beads-on-ring-cos-theta}\tag{8.11.7}$$
which will be real only if
\begin{equation*} 24\dfrac{M}{m} \lt 16. \end{equation*}
That is if
$$3M \lt 2 m.\tag{8.11.8}$$
The angle at which the ring would accelerate up is then given by Eq. (8.11.7)
\begin{equation*} \theta = \cos^{-1} \left[ \dfrac{1}{3}\left(1 \pm \sqrt{ 1 - \dfrac{3M}{2m} } \right) \right]. \end{equation*}
Solution 4 (d)

(d) With $3\times M = 2\times m\text{,}$ we get,

\begin{align*} \theta \amp = \cos^{-1} \left (\dfrac{1}{3} \right) = 70.5^{\circ}. \end{align*}

Suppose you friend claims to have discovered a mysterious force in nature that acts on all particles in the three-dimensional space. He tells you that the force is always pointed towards a definite point in space, which we can call the force center. The magnitude of the force turns out to be proportional to $1/r^3\text{,}$ where $r$ is the distance from the force center to any other point. Let $b$ be the proportionality constant, i.e. the magnitude of the force on a particle can be written as $b/r^3\text{,}$ when the particle is at a distance $r$ from the force center.

Find the potential energy of a particle when it is at a distance D from the force center, assuming the potential energy to be zero when the particle is at infinity.

Hint

Do $\Delta U = -W_{if}$ in the radially inward direction from reference to final position.

$-\dfrac{b}{2}\,\dfrac{1}{D^2}\text{.}$

Solution

Potential energy will be negative of the work done by the force in bringing a particle from the reference to the point of interest. Suppose we bring the particle on a radially inward path. Then, on an infinitesimal segment of the path we will have

\begin{equation*} d\vec r = \hat r\, dr,\ \ \vec F = -\dfrac{b}{r^3}\, \hat r, \end{equation*}

where $\hat r$ is a unit vector from origin to infinity radially out. The dot product force with the displacement is

\begin{equation*} \vec F\cdot d\vec r = -\dfrac{b}{r^3}\, dr. \end{equation*}
\begin{align*} U_D \amp = - W_{\infty,D} = -\int_{\infty}^{D} \left( -\dfrac{b}{r^3} \right) \, dr,\\ \amp = \int_{\infty}^{D} \dfrac{b}{r^3} dr = -\dfrac{b}{2}\,\dfrac{1}{D^2}. \end{align*}

A cyclist in a race must climb a $5^{\circ}$ hill at a speed of $8 \text{ m/s}$ (see Figure 8.11.64). The mass of the bike and the biker together is $80 \text{ kg}\text{.}$ In addition of providing the increase in his potential energy, he must also overcome air drag. Suppose air drag takes away energy at $200\text{ J/s}$ on that speed. What must be the power output of the biker to achieve the goal?

Hint

Find energy needed in an interval $\Delta t\text{.}$

$747 \text{ W} \text{.}$

Solution

We examine the work and energy picture during an interval $t$ to $t+\Delta t\text{.}$ During this time, the biker will go a distance $d = v \Delta t$ and his height will increase by $h = d\,\sin\,\theta$ as illustrated in Figure 8.11.65.

The biker must provide the energy equal to the change in his and bike's potential energy and the dissipation of energy through air drag.

\begin{equation*} E_\text{provided} = mgh + 200\,\Delta t. \end{equation*}

Here

\begin{equation*} h = v\Delta t\,\sin\,\theta = 0.697\, \Delta t. \end{equation*}

Therefore,

\begin{equation*} E_\text{provided}= 80\times 9.81\times 0.697\,\Delta t + 200\,\Delta t = 747\,\Delta t. \end{equation*}

Dividing by the duration will give us the power output of the biker.

\begin{equation*} P = 747\text{ W}. \end{equation*}

A spring-block system is resting on a frictionless floor as shown in the Figure 8.11.67. The spring constant is $2.0\text{ N m}^{−1}$ and the mass of the block is $2.0\text{ kg}\text{.}$ Ignore the mass of the spring.

Initially the spring is in an unstretched condition. Another block of mass 1.0 kg moving with a speed of $2.0\text{ m s}^{−1}$ collides elastically with the first block. The collision is such that the $2.0\text{ kg}$ block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is $D\text{.}$ Find $D\text{.}$ (Indian JEE Advanced, 2018)

Hint

After the collision, the block attached to spring will return in half its period.

$2.09\text{ m}\text{.}$

Solution

First we need to find the speeds of the two blocks after the elastic collision. Let positive $x$ axis is pointed to the right. Let $v_1$ and $v_2$ be speeds of the striking block and spring-attached block after the collision. Then, we have the following momentum and kinetic energy conservation equations applicable for the elastic collision. (Ignoring the units; note the signs in the momentum conservation equation.)

\begin{align} \amp 1\times 2 + 2\times 0 = - 1\times v_1 + 2\times v_2 \label{eq-pr-moving-block-striking-a-block-attached-to-spring-mom}\tag{8.11.9}\\ \amp 1\times 2^2 + 2\times 0^2 = 1\times v_1^2 + 2\times v_2^2 \text{ (canceled out 1/2.)} \label{eq-pr-moving-block-striking-a-block-attached-to-spring-ke}\tag{8.11.10} \end{align}

From the first equation $v_1 = 2 v_2 - 2\text{.}$ Use this in the second equation to get

\begin{equation*} 2v_2(3 v_2 -4) = 0. \end{equation*}

From this, we choose the solution $v_2=4/3$ since this block is not at rest after the strike. The striking block returns with speed $v_1$ give by

\begin{equation*} v_1 = 2v_2 - 2 = \frac{2}{3}. \end{equation*}

To find the time for block attached to the spring, we note that it will be equal to half its period of the periodic motion it will execute. The period of a mass $m$ attached to a spring of spring constant $k$ is $T = 2\pi \sqrt{ m/k }\text{.}$ Therefore the time we are interested in is

\begin{equation*} \Delta t = \frac{T}{2} = \pi \sqrt{ m/k } = \pi. \end{equation*}

In this time striking block will move by amount

\begin{equation*} v_1 \Delta t = \frac{2}{3} \times \pi = 2.09 \text{ m}. \end{equation*}

A particle of mass $m$ is initially at rest at the origin. It is subjected to a force and starts moving along the $x$-axis. Its kinetic energy $K$ changes with time as $dK/dt = \gamma t\text{,}$ where $\gamma$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true ?

1. The force applied on the particle is constant.
2. The speed of the particle is proportional to time.
3. The distance of the particle from the origin increases linerarly with time.
4. The force is conservative.

Hint

Use $F_x = dp_x/dt\text{.}$

(A) and (D).

Solution

To decide about force, we use Newton's second law, $\vec F = d\vec p/dt\text{,}$ with only the $x$-components. This becomes $F_x = d p_x/dt\text{.}$ We drop the subscript $x$ from symbols in our calculations below.

Now, to commect $dK/dt$ to $dp/dt\text{,}$ we write $K = p^2/2m\text{.}$ This gives

\begin{equation*} \frac{p}{m}\,\frac{dp}{dt} = \gamma t. \end{equation*}

Therefore, we get the following integral using the initial condition of rest.

\begin{equation*} \int_0^p p\, dp = \gamma\, m \int_0^t t dt. \end{equation*}

This gives $p^2 = \gamma m t^2\text{,}$ where we will keep only positive root since motion is towards positive $x$-axis and $p$ is $x$-component. Writing this as $x$-component as a reminder.

\begin{equation*} p_x = \sqrt{\gamma m} t. \end{equation*}

Take derivative with $t$ to get $F_x\text{,}$

\begin{equation*} F_x = \frac{dp_x}{dt} = \sqrt{\gamma m},\ \ \text{(constant)}. \end{equation*}

We also know that $F_y=F_z=0\text{.}$ Therefore, we have a constant force. Also, all constant forces are also conservative. Answer choices are (A) and (D).

A small ball of mass $m$ rests on a larger ball of mass $M$ as shown in Figure 8.11.70. The two are let go such that large ball falls a height $h$ before striking the floor vertically and elastically. Within a very short period of time, the velocity of the large ball points up while small ball is still in the falling mode. Thus, there is a secondary collision between the large ball and the small ball. If this collision is also elastic, find the maximum height the small ball will fly before coming back down. Use $R$ for the radius of the large ball.

Hint

After the bigger ball bounces back it collides with the smaller ball which is still in the falling motion.

$2R + \frac{v^2}{2g} = 2R + \left( \frac{3M-m}{M+m} \right)^2\, h\text{.}$

Solution

Figure 8.11.71 shows two collisions that we treat as elastic collisions. In the first elastic collision, since floor can be taken to be infintely more massive than the large ball, the speed of the large ball after collision will be same as before. Since the ball has fallen a height $h\text{,}$ the speed before and after collision will be

\begin{equation*} V_\text{after} = V_\text{before} = \sqrt{2gh}. \end{equation*}
Now, we look at the second collision. This is an elastic collision between small ball falling at speed $v_\text{before}=\sqrt{2gh}$ and the large ball going up at speed $V_\text{before} = \sqrt{2gh}\text{.}$ In addition to momentum conservation, since the collision is elastic, coefficient of restitution will be $1\text{.}$ Let $v$ and $V$ denote the speeds of the small and large balls after the seond collision. Let both balls be going up after the collision and use up as positive direction.
\begin{align*} \amp M \sqrt{2gh} - m \sqrt{2gh} = M V + m v. \\ \amp \frac{ |V - v|}{2\sqrt{2gh} } = 1. \end{align*}
With $v\gt V\text{,}$ the second equation becomes
\begin{equation*} v - V = 2\sqrt{2gh}. \end{equation*}
We can solve this equation with the first equation to obtain
\begin{align*} \amp V = \frac{M-3m}{M+m}\,\sqrt{2gh}.\\ \amp v = \frac{3M-m}{M+m}\,\sqrt{2gh}. \end{align*}
Therefore, small ball will fly a height $\frac{v^2}{2g}$ above the collision point. Taking into account the diameter of the big ball, the height from the ground will be
\begin{equation*} H = 2R + \frac{v^2}{2g} = 2R + \left( \frac{3M-m}{M+m} \right)^2\, h. \end{equation*}

A block of mass $M$ has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table as shown in Figure 8.11.73.

Initially the right edge of the block is at $x = 0\text{,}$ in a co-ordinate system fixed to the table. A point mass $m$ is released from rest at the topmost point of the path as shown and it slides down.

When the mass loses contact with the block, its position is $x$ and the velocity is $v\text{.}$ At that instant, which of the following options is/are correct?

1. The position of the point mass $m$ is: $x = -\sqrt{2}\,\frac{mR}{M+m}\text{.}$
2. The velocity of the point mass $m$ is: $v = \sqrt{\frac{2gR}{1+\frac{m}{M}}}\text{.}$
3. The $x$ component of displacement of the center of mass of the block $M$ is: $-\frac{mR}{M+m}\text{.}$
4. The velocity of the block $M$ is: $-\frac{m}{M}\sqrt{2gR}\text{.}$

Hint

Use conservation of momentum along horizontal direction and cnservation of energy for the combined system. Also, note horizontal CM of the combined system will not move.

(B) and (C).

Solution

First we notice that there will be no external force in the horizontal direction on the conbined system of mass and block together since the table is frictionless. That means $x$-momentum of the combined system will be conserved. In the problem, we are using symbol $v$ for $v_x\text{,}$ let us use $V$ for $V_x$ of the block with respect to the table. Therefore,

$$m v + MV = 0.\label{eq-momentum-conservation}\tag{8.11.11}$$

Also, note that energy of the combined system will be conserved since there are no non-conservative forces. Thus, equating loss of potential energy of mass $m$ to the sum of kinetic energy gained, we get

$$\frac{1}{2}m v^2 + \frac{1}{2} MV^2 = m g R.\label{eq-energy-conservation}\tag{8.11.12}$$

Using Eq. (8.11.11), we get

\begin{equation*} m v^2 + M \left( m/M\right)^2 v^2 = 2 g R. \end{equation*}

Solving this, we get

\begin{equation*} v = \pm \sqrt{ \frac{2gR}{ 1+ \frac{m}{M} } }. \end{equation*}

This is $dx/dt\text{.}$ We expect block to move towards positve $x$-axis and block to move towards negative $x$-axis. Therefore, this will be

\begin{equation*} v = + \sqrt{ \frac{2gR}{ 1+ \frac{m}{M} } }. \end{equation*}

This is choice (B). With this choice for $v\text{,}$ we can rule out choice (D) for the block since, using Eq. (8.11.11), the block should have velocity

\begin{equation*} V = - \frac{m}{M}\,v = - \frac{m}{M}\,\sqrt{ \frac{2gR}{ 1+ \frac{m}{M} } }. \end{equation*}

Now, to work on position, we note that the $X_\text{cm}$ of the combined system will not move during the motion of the mass $m\text{.}$ $X_\text{cm}$ at initial instant was at

\begin{equation*} X_\text{cm} = \frac{MX_0 + m x_0}{ M + m}, \end{equation*}

where $x_0 = - R$ and $X_0$ was located at a distance $X_0$ from the right edge of the block. These are shown in Figure 8.11.74.

When the mass has moved down to the exit point, the $x$ coordinate of mass $m$ is $x$ and the $x$ coordinate of the block is $x+X_0\text{.}$

\begin{equation*} X_\text{cm} = \frac{ M(x+X_0) + m x }{ M + m}, \end{equation*}

Equating these two expressions of $X_\text{cm}$ gives

\begin{equation*} MX_0 + m x_0 = M(x+X_0) + m x. \end{equation*}

We solve this for $x$ to get

\begin{equation*} x = \frac{mx_0}{M+m} = \frac{-m R }{M+m}. \end{equation*}

This is how much block's $X_\text{cm}$ has also moved. Therefore, choice (C) is correct. This also rules out (A).