Center of Mass and Collision

The ball is dropped from a height of 42 m. The coefficient of restitution, $e$, is 0.4. This number tells us how much energy the ball keeps after bouncing.

At first, the ball is dropped. So, its starting speed ($u$) is 0.

Height after first bounce:

The ball comes back up to a height equal to $e^2$ times the height it fell from. So:

$ h_1 = (0.4)^2 \times 42 $

$ h_1 = 0.16 \times 42 = 6.72~\mathrm{m} $

Height after later bounces:

Each time the ball bounces, it goes up to $e^2$ times the previous bounce's height.

$ h_2 = (0.4)^2 \times 6.72 = 0.16 \times 6.72 $

Total Distance:

When the ball is dropped, it first travels 42 m down.

After that, on every bounce, it goes up and then comes down the same height. So, for all the bounces, we call the sum of all the heights $h_1, h_2, h_3, \ldots$.

Total distance = distance of first fall + 2 × (sum of all bounce heights)

$ = 42 + 2 \times [h_1 + h_2 + h_3 + \ldots] $

Summing the Series:

All the bounce heights make a series called a geometric progression (GP). The first term $a = 42$ and the common ratio $r = 0.16$.

The sum of an infinite GP is given by:

Sum $= \frac{a}{1 - r} = \frac{42}{1 - 0.16}$

$ \text{Sum} = \frac{42}{0.84} = 50 $

Final Calculation:

Now, total distance is:

$ \text{Total distance} = 42 + 2 \times (0.4)^2 \times 50 $

$ = 42 + 2 \times 0.16 \times 50 $

$ = 42 + 16 = 58~\mathrm{m} $

According to conservation of linear momentum

$ \begin{aligned} & \Rightarrow \quad m v+2 m \times 0=m v_1+2 m v_2 \\ & \Rightarrow \quad v=v_1+2 v_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Coefficient of restitution

$ \begin{aligned} e & =\frac{v_2-v_1}{u_1-u_2} \\ \Rightarrow \quad e \quad e & =\frac{v_2-v_1}{v-0} \\ \Rightarrow \quad e v & =v_2-v_1 \\ \Rightarrow \quad v_1 & =v_2-e v \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} & v=v_2-e v+2 v_2 \\ \quad & v=3 v_2-e v \\ \Rightarrow \quad & v(1+e)=3 v_2 \\ \Rightarrow \quad & v_2=\frac{v(1+e)}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} $

Now, from Eq. (ii) and (iii), we get,

$ \begin{aligned} v_1 & =\frac{v(1+e)}{3}-e v=\frac{v(1-2 e)}{3} \\ \Rightarrow \quad v_1 & =\frac{v(1-2 e)}{3} \end{aligned} $

Thus, $\frac{v_1}{v_2}=\frac{1-2 e}{1+e}$

Let the centre of the sphere be at origin ( $x=0$ ).

Then, the centre of the cylinder lies at a distance of $x=0.3 \mathrm{~m}$ from the sphere centre.

CM of the system

$ \begin{aligned} X_{\mathrm{CM}} & =\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\ & =\frac{2 \times 0.3+0.5 \times 0}{2+0.5} \\ X_{\mathrm{CM}} & =\frac{0.6}{2.5}=0.24 \mathrm{~m} \end{aligned} $

or $X_{\mathrm{CM}}=24 \mathrm{~cm}$ (From the centre of sphere)

For block $m$ (moving down)

$ m g-T=m a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For block $n$ (moving up)

$ T-n g=n a \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eqs. (i) and (ii), we get

$ \begin{aligned} m g-n g & =m a+n a \\ (m-n) g & =(m+n) a \\ a & =\frac{(m-n)}{(m+n)} g \end{aligned} $

$ \therefore \quad a_m=+a \text { and } a_n=-a $

then, $a_{\mathrm{CM}}=\frac{m a-n a}{m+n}=\frac{(m-n)}{(m+n)} \times a$

$ a_{\mathrm{CM}}=\left(\frac{m-n}{m+n}\right)^2 g $

Given,

Mass of ball, $ m_{A}=1.2 \mathrm{~kg} $

Mass of ball, $ m_{B}=3.6 \mathrm{~kg} $

Initial velocity of ball, $ v_{A_{i}}=8.4 \mathrm{~ms}^{-1} $

Initial velocity of ball, $ v_{B_{i}}=0 \mathrm{~ms}^{-1} $

In a one-dimensional elastic collision, the final velocities ( $ v_{A_{f}} $ and $ v_{B_{f}} $ ) of the two masses can be calculated using the following formula

$ v_{A_{f}}=\frac{\left(m_{A}-m_{B}\right) \cdot v_{A_{i}}+2 m_{B} v_{B_{i}}}{m_{A}+m_{B}} $

$ \begin{array}{l} \text { After putting values, we get } \\\\ \qquad \begin{aligned} v_{A_{f}} & =\frac{(1.2-3.6) 8.4}{1.2+3.6}=\frac{-2.4 \times 8.4}{4.8} \\\\ v_{A_{f}} & =-4.2 \mathrm{~m} / \mathrm{s} \\\\ \text { Similarly, } v_{B_{f}} & =\frac{\left(m_{B}-m_{A}\right) v_{B_{i}}+2 m_{A} v_{A_{i}}}{m_{A}+m_{B}} \\\\ v_{B_{f}} & =\frac{0+2 \times 1.2 \times 8.4}{1.2+3.6} \\\\ & =\frac{2.4 \times 8.4}{4.8} \\\\ V_{B_{F}} & =4.2 \mathrm{~m} / \mathrm{s} \end{aligned} \end{array} $

Initial kinetic energy of ball $ A $ is,

$ \begin{aligned} \mathrm{KE}_{A_{i}} & =\frac{1}{2} m_{A}\left(v_{A_{i}}\right)^{2}=\frac{1}{2} \times 1.2 \times 8.4 \times 8.4 \\\\ & =42.33 \mathrm{~J} \end{aligned} $

Final kinetic energy of ball $ B $ is,

$ \begin{aligned} \mathrm{KE}_{B_{f}} & =\frac{1}{2} m_{B}\left(v_{B_{f}}\right)^{2} \\\\ & =\frac{1}{2} \times 3.6 \times 4.2 \times 4.2=31.75 \mathrm{~J} \end{aligned} $

The percentage of kinetic energy

$ \begin{array}{l} \text { transferred from ball } A \text { to ball } B \text { is, } \\\\ \qquad \begin{aligned} \mathrm{KE}_{\text {transferred }} & =\frac{\mathrm{KE}_{A_{i}}}{\mathrm{KE}^{\prime}} \times 100 \% \\\\ & =\frac{31.75}{42.33} \times 100 \%=75 \% \end{aligned} \end{array} $

Let mass of meter scale be $ m $.

Mass of each coin $ =9 \mathrm{~g} $

Total mass of coin $ =18 \mathrm{~g} $

Length of meter scale $ =100 \mathrm{~cm} $

The new balance point is at 35 cm

The torque due to the coins placed at the 10 cm mark

Distance from the new balance point to the coins $ =(35-10) \mathrm{cm} $

$ =25 \mathrm{~cm} $

Torque due to coins

$ =F \cdot r=m g r=18 \times g \times 25 $

The centre of mass of the meter scale is at its mid-point ( 50 cm mark).

Distance from the new balance point to the $ =(50-35) \mathrm{cm} $

Centre of the meter scale $ =15 \mathrm{~cm} $

Torque due to meter scale $ =m \times g \times 15 $

For the scale to be balanced, the net torque will conserver. So, from Eq. (i) and Eq. (ii), we get

$ \begin{array}{l} 18 \times g \times 25=m \times g \times 15 \\\\ m=\frac{18 \times 25}{15} \\\\ m=30 \mathrm{~g} \end{array} $

$ u_1=10 \mathrm{~m} / \mathrm{s} \quad u_2=0 \mathrm{~m} / \mathrm{s} $

Coefficient of restitution,

$ e=0.4 $

Velocity of $m_1$ (ball $P$ ) after collision is given by

$ \begin{aligned} & v_1=\left(\frac{m_1-e m_2}{m_1+m_2}\right) u_1 \\\\ & v_1=\frac{(0.5-0.4 \times 1)}{1.5} \times 10 \\\\ & v_1=\frac{2}{3} \mathrm{~m} / \mathrm{s} \end{aligned} $

Velocity of ball $ Q $ (initially rest) after collision

$ v_{2}=\left[\frac{m_{2}-e m_{1}}{m_{1}+m_{2}}\right] u_{2}+\frac{(1+e) m_{1} u_{1}}{m_{1}+m_{2}} $

$ v_{2}=\frac{(1+e) m_{1} u_{1}}{m_{1}+m_{2}} $

(as $ u_{2}=0 $ )

$ v_{2}=\frac{(1+0.4)(0.5)(10)}{1.5} $

$ v_{2}=\frac{14}{3} \mathrm{~m} / \mathrm{s} $

Ratio of $ v_{1} $ and $ v_{2} $

$ \begin{aligned} \frac{v_{1}}{v_{2}} & =\frac{2}{3} \times \frac{3}{14}=\frac{1}{7} \\\\ v_{1}: v_{2} & =1: 7 \end{aligned} $

When the section is cut out we use area formula of COM.

$ X_{\mathrm{cm}}=\frac{A_{1} x_{1}-A_{2} x_{2}}{A_{1}-A_{2}} $

where $ A_{1} $ area of plate $ P, A_{2} $ area of cut out portion

$ \begin{array}{l} x_{1}=0, x_{2}=2 r, A_{1}=\pi(4 r)^{2}, A_{2}=\pi(r)^{2} \\\\ X_{\mathrm{cm}}=\frac{0-\pi r^{2} \cdot 2 r}{\pi(4 r)^{2}-\pi r^{2}} \\\\ X_{\mathrm{cm}}=\frac{-\pi r \cdot 2}{\pi \mathrm{l} 5} \end{array} $

$ \begin{aligned} & X_{\mathrm{cm}}=-\frac{2 r}{15} \\\\ & \text { Distance }=\left|X_{\mathrm{cm}}\right|=\frac{2 r}{15} \end{aligned} $

Given,

Mass $ m_{1}=1.2 \mathrm{~kg}, u_{1}=12 \mathrm{~m} / \mathrm{s} $. $ m_{2}=1.2 \mathrm{~kg}, u_{2}=0 $

Coefficient of restitution $ =1 / \sqrt{2} $

Now apply conservation of linear momentum equation for the collision.

$ m_{1} u_{1}+m_{2} u_{2}=m_{1} v_{1}+m_{2} v_{2} $

( $ v_{1}, v_{2}= $ velocity after collision)

$ m\left[u_{1}+u_{2}\right]=m\left[v_{1}+v_{2}\right] \quad \text { [same masses] } $

$ \text { as } u_{2}=0, u_{1}=12 \mathrm{~m} / \mathrm{s} $

$ 12=v_{2}+v_{1} $

From coefficient of restitution

$ \begin{aligned} e & =\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \Rightarrow \frac{1}{\sqrt{2}}=\frac{v_{2}-v_{1}}{12} \\\\ \Rightarrow 6 \sqrt{2} & =v_{2}-v_{1} \end{aligned} $

Solving Eqs. (i) and (ii)

$ v_{1}=(6-3 \sqrt{2}) \mathrm{m} / \mathrm{s} $

and

$ v_{2}=(3 \sqrt{2}+6) \mathrm{m} / \mathrm{s} $

Kinetic energy before collision

$ =\frac{1}{2} m_{1} u_{1}^{2}=\frac{1}{2}(m)(144) $

Kinetic energy after collision

$ \begin{array}{l} =\frac{1}{2} m\left[v_{1}^{2}+v_{2}^{2}\right] \\\\ =\frac{1}{2} m[108] \end{array} $

Ratio of final (after collision) kinetic energy to initial

$ =\frac{\frac{1}{2} m[108]}{\frac{1}{2} m[144]}=\frac{3}{4}=3: 4 $

Initial condition,

Centre of mass of two rod, ( $ Y $-axis)

$Y_{\mathrm{cm}}=\frac{m_1 y_1+m_2 y_2}{m_1+m_2}$

As rod are identical $ m_{1}=m_{2}=m $

$ \begin{array}{l} y_{1}=\frac{L}{2}, y_{2}=L+\frac{a}{2} \\\\ Y_{\mathrm{cm}}=\frac{m\left[\frac{L}{2}+L+\frac{a}{2}\right]}{2 m}=\frac{3 L+a}{4} \end{array} $

Final case,

Centre of mass of two rod,

$ \begin{aligned} Y_{\mathrm{cm}}^{\prime} & =\frac{m_{1} y_{1}^{\prime}+m_{2} y_{2}^{\prime}}{m_{1}+m_{2}} \\\\ y_{1}^{\prime} & =\frac{a}{2}, y_{2}^{\prime}=a+\frac{L}{2} \end{aligned} $

$ Y_{\mathrm{cm}}^{\prime}=\frac{1}{2}\left[\frac{3 a+L}{2}\right]=\frac{3 a+L}{4} $

Difference in shift

$ Y_{\mathrm{cm}}-Y_{\mathrm{cm}}^{\prime}=\frac{1}{2}[L-a] $

From the figure, velocity of centre of mass,

$ \begin{aligned} v_{\mathrm{COM}} & =\frac{m \times(6 \hat{\mathbf{j}})+2 m(-12 \hat{\mathbf{j}})+3 m(8 \hat{\mathbf{i}})}{m+2 m+3 m} \\\\ & =\frac{6 m \hat{\mathbf{j}}-24 m \hat{\mathbf{j}}+24 m \hat{\mathbf{i}}}{6 m} \\\\ & =-3 \hat{\mathbf{j}}+4 \hat{\mathbf{i}} \end{aligned} $

Magnitude of velocity of centre of mass

$ \begin{aligned} & =\sqrt{3^2+4^2} \\\\ & =5 \mathrm{~m} / \mathrm{s} \end{aligned} $

Here,

$ \begin{aligned} m \cdot v_{A_1} & =m \cdot v_{B_2}+2 m \cdot v_{B_2} \\\\ v_{B_2}-v_{A_2} & =0.4 \times v_{A_1} \\\\ m \cdot v_{A_1} & =m \cdot v_{A_2}+2 m \times\left(0.4 \times v_{A_1}+v_{A_2}\right) \end{aligned} $

For $A$

$ \begin{aligned} & v_{A_1}=3 \cdot v_{A_2}+0.8 \times v_{A_1} \\\\ & v_{A_2}=\frac{0.2}{3} \times v_{A_1} \end{aligned} $

For $B$

$ \begin{aligned} v_{B_2} & =\frac{0.4 \times 3+0.2}{3} \times v_{A_1} \\\\ & =\frac{1.4}{3} v_{A_1} \end{aligned} $

The ratio of both masses

$ \begin{aligned} & \frac{v_{A_2}}{v_{B_2}}=\frac{\left(\frac{0.2}{3}\right) \times v_{A_1}}{\left(\frac{1.4}{3}\right) \times v_{A_1}} \\\\ & \frac{v_{A_2}}{v_{B_2}}=\frac{1}{7} \end{aligned} $

Original system (with 4 particles)

Let coordinates of the particle as

$ \begin{aligned} & A=(0, a) \\\\ & B=(a, a) \\\\ & C=(a, 0) \\\\ & D=(0,0) \end{aligned} $

The centre of mass of system

$ \begin{gathered} X_{\text {com }}=\frac{m x_A+m x_B+m x_C+m x_D}{4 m} \\\\ Y_{\text {com }}=\frac{m y_A+m y_B+m y_C+m y_D}{4 m} \end{gathered} $

Solving,

$ \begin{aligned} & X_{\mathrm{com}}=\frac{0+a+a+0}{4}=\frac{a}{2} \\\\ & Y_{\mathrm{com}}=\frac{a+a+0+0}{4}=\frac{a}{2} \end{aligned} $

The centre of mass of original system at $\left(\frac{a}{2}, \frac{a}{2}\right)$

New centre of mass, when one particle (B) is removed,

$ \begin{gathered} X_{\text {com }}^{\prime}=\frac{m \cdot x_A+m \cdot x_C+m \cdot x_D}{3 m} \\\\ Y_{\text {com }}^{\prime}=\frac{m \cdot y_A+m \cdot y_C+m \cdot y_D}{3 m} \\\\ X_{\text {con }}^{\prime}=\frac{0+a+a}{3}=\frac{2 a}{3} \\\\ Y_{\text {com }}^{\prime}=\frac{a+0+0}{3}=\frac{a}{3} \end{gathered} $

Now, shift in position

$ \sqrt{\left(\frac{2 a}{3}-\frac{a}{2}\right)^2+\left(\frac{a}{3}-\frac{a}{2}\right)^2}=\frac{a}{3 \sqrt{2}} $

Thus, centre of mass of 3 system shift to $\frac{a}{3 \sqrt{2}}$

Given,

Initial mass, $m=16 \mathrm{~kg}$

Mass of $1^{\text {st }}$ piece, $m_1=4 \mathrm{~kg}$

Mass of $2^{\text {nd }}$ piece, $m_2=12 \mathrm{~kg}$

Velocity of $2^{\text {nd }}$ piece, $v_2=4 \mathrm{~m} / \mathrm{s}$

Using law of conservation of linear momentum.

Initial momentum = Final momentum

$ \begin{aligned} & m \times v=m_1 v_1+m_2 v_2 \\ & \quad(v=0 \text { Initially at rest }) \\ & m_1 v_1=-m_2 v_2 \\ & 4 \times v_1=12 \times 4 \\ & v_1=-12 \mathrm{~m} / \mathrm{s} \end{aligned} $

Kinetic energy is given by for $2^{\text {nd }}$ particle

$ \begin{aligned} (\mathrm{KE}) & =\frac{1}{2} m_2 v_2^2 \\ & =\frac{1}{2} \times 4 \times(12)^2=\frac{1}{2} \times 4 \times(12)^2 \\ & =288 \mathrm{~J} \end{aligned} $

Hence, the correct option is (d).

Given,

Height $=h$

Coefficient of restitution $=e$

The ball reaches the $u=\sqrt{2 g h}$, when dropped from height $h$.

If collision on the floor is elastic, the ball will rebound with same velocity. But here coefficient of restitution is $e$; the velocity $v$ (less than $u$ ) is given by

$ v=e u \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

As $v$ is velocity of separation and $u$ is the velocity of approach.

In above case ball will rebound to a height $h_1$.

$ \begin{aligned} h_1 & =v^2 / 2 g \\ v & =e u \Rightarrow h_1=\frac{e^2 u^2}{2 g} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Initial height

$ h=\frac{u^2}{2 g} $

Put in $2^{\text {nd }}$ equation

$ h_1=e^2 h $

So, total distance travel by ball before $2^{\text {nd }}$ drop is $h+e^2 h+e^2 h=h\left(1+2 e^2\right)$

Acceleration of the given system will be

$ a=\frac{m_2-m_1}{m_2+m_1} \mathrm{~g} $

Putting values of $m_1$ and $m_2$, we get

$ a=\frac{g}{3} \mathrm{~m} / \mathrm{s}^2 $

Acceleration of centre of mass will be

$ a_{\mathrm{CM}}=\left(\frac{m_2-m_1}{m_2+m_1}\right) a=\frac{g}{9} $

Putting values of $a_{\mathrm{CM}}$ into second equation, we get

$ \begin{aligned} S_{\mathrm{CM}} & =\frac{1}{2} a_{\mathrm{CM}} t^2 \\ & =\frac{1}{2}\left(\frac{g}{9}\right)(2)^2 \\ & =2.22 \mathrm{~m} \end{aligned} $

Total mechanical energy is conserved in this collision

$ \begin{aligned} & \text { } \begin{aligned}Therefore, (\mathrm{KE}+\mathrm{PE})_i= & (\mathrm{KE}+\mathrm{PE})_f \\ & =\Delta K+\Delta U=0 \\ \Rightarrow \quad \Delta K & =-\Delta U \end{aligned} \end{aligned} $

Let initial velocity be $v$, then

$ \mathrm{KE}_1=\frac{1}{2} m v^2 $

Final potential energy due to spring,

$ (\mathrm{PE})_f=\frac{1}{2} k x^2 $

From conservation of energy,

$ \begin{array}{rlrl} & & \frac{1}{2} m v^2 & =\frac{1}{2} K x^2 \\ \Rightarrow & v & =\sqrt{\frac{K}{m}} x \end{array} $

Where, $x=$ Compression $=L$

Therefore, $v=L \sqrt{\frac{K}{m}}$

Hence, maximum momentum,

$ p=m v=L \sqrt{M K} \quad[\because m=M] $

If a ball is dropped from height $h$, then it will strike the floor with velocity $(\eta)=\sqrt{2 g h}$

Now, after striking the floor, it will bounce back with velocity $(u)=e v$ Now, it will travel total $2 h$ distance before again striking the ground where

$ h_1=\frac{e^2 v^2}{2 g} $

Again the ball will bounce back with $e^2 v$ velocity and now it will travel $2 h_2$ distance where, $h_2=\frac{e^4 v^2}{2 g}$

This process continuous for long till ball comes to rest.

So, total distance travelled, $d$

$ \begin{aligned} &=h+1 h_1+2 h_2+\ldots \\ & d=\frac{v^2}{2 g}+\frac{2 e^2 v^2}{2 g}+\frac{2 e^4 v^2}{2 g}+\frac{2 e^6 v^2}{2 g} \ldots \\ &=\frac{v^2}{2 g}\left(1+e^2+e^4+e^6+\ldots\right) \\ &=\frac{v^2}{2 g}+\frac{2 e^2 v^2}{g}\left(\frac{1}{1-e^2}\right) \\ &=\left(\frac{1-e^2+2 e^2}{1-e^2}\right) \frac{v^2}{2 g}=\left(\frac{1+e^2}{1-e^2}\right) \frac{v^2}{2 g} \\ &=h\left(\frac{1+e^2}{1-e^2}\right) \end{aligned} $

$ \text { According to FBD given below, } $

$ \begin{aligned} &T-\frac{m g}{2}=m a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ &\text { (for left side block) } \end{aligned} $

$ \begin{aligned} &\frac{\sqrt{3}}{2} m g-T=m a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ &\text { (for right side block) } \end{aligned} $

Addign Eqs. (i) and (ii), we get

$ a=\frac{1}{2 m}\left(\frac{\sqrt{3} m g}{2}-\frac{m g}{2}\right)=\left(\frac{\sqrt{3}-1}{4}\right) g $

Acceleration COM was $k_e$ as follows.

So net acceleration on COM is

$ \mathbf{a}_{\mathrm{CM}}=\frac{\mathbf{a}_1+\mathbf{a}_2}{2}=\frac{\sqrt{2} a}{2}=\frac{a}{\sqrt{2}} $

$\Rightarrow \quad \mathbf{a}_{\mathrm{CM}}=\left(\frac{\sqrt{3}-1}{4 \sqrt{2}}\right) g$ in $x$ direction

Given,

Two masses $m_1$ and $m_2$ moves toward CM by $d$ distance.

We know position of CM

$ X_{\mathrm{CM}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2} $

Shift in position of centre of mass,

$ \Delta x_{\mathrm{CM}}=\frac{m_1 \Delta x_1+m_2 \Delta x_2}{m_1+m_2} $

But $\Delta x_{C M}=0$ (Given)

$ \begin{aligned} m_1 \Delta x_1+m_2 \Delta x_2 & =0 \\ \left(\frac{m_1}{m_2}\right) \Delta x_1 & =-\Delta x_2 \\ \Delta x_2 & =-\frac{m_1}{m_2} d \end{aligned} $

Since, collision is head on, then particle $A$ will continue moving along the same line as before the collision, but there will be a change in the magnitude of its velocity vector. Let particle $A$ starts moving with velocity $v_1$ and particle $B$ with velocity $v_2$ after collision, then from conservation of linear momentum,

$ \begin{aligned} m u & =m v_1+m v_2 \\or\,\,\,\,\,\,\,\, u & =v_1+v_2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

From the given condition,

$ \begin{aligned} & \text { i.e } \quad \eta=\frac{\frac{1}{2} m u^2-\left(\frac{1}{2} m v_1^2+\frac{1}{2} m v_2^2\right)}{\frac{1}{2} m u^2} \\ & \Rightarrow \quad \eta=1-\frac{v_1^2+v_2^2}{u^2} \\ & \Rightarrow \quad v_1^2+v_2^2=(1-\eta) u^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Now solving Eqs. (i) and (ii), we get $v_1=5 \mathrm{~m} / \mathrm{s}$

In an elastic collision of two billiard balls, both kinetic energy and linear momentum remain conserved.

In elastic collision, momentum is conserved but total energy of any one of ball is not conserved i.e. exchange of energy takes place.

Hence, A is true but R is false.

Let the mass of the moving particle $=m$

∴ The mass of the stationary particle

$ =\frac{1}{n} \times m=\frac{m}{n} $

Also, let the speed of moving particle $=v$

Now, by the law of conservation of linear momentum

$ \begin{aligned} & \therefore \quad m v+\left(\frac{m}{n} \times 0\right)=m v^{\prime}+\frac{m}{n} v^{\prime} \\ & \Rightarrow \quad m v=m\left(v^{\prime}+\frac{v^{\prime}}{n}\right) \\ & \Rightarrow \quad v=v^{\prime}+\frac{v^{\prime}}{n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Also, coefficient of restitution,

$ e=\frac{-\left(v^{\prime \prime}-v^{\prime}\right)}{(0-v)} $

Since, it is elastic collision,

$ \begin{array}{ll} \therefore & e=1 \\ \Rightarrow & 1=\frac{-\left(v^{\prime \prime}-v^{\prime}\right)}{-v} \Rightarrow v^{\prime \prime}=v+v^{\prime} \\ \Rightarrow & v^{\prime}=v^{\prime \prime}-v \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array} $

From Eqs. (i) and (ii), we get

$ \begin{gathered} v=v^*+\frac{v^*}{n} \\ v^*=\left[\frac{2 v}{1+\frac{1}{n}}\right] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered} $

Now, kinetic energy transferred to the stationary particle

$=(\text { kinetic energy })_{\text {final }} -(\text { kinetic energy })_{\text {initial }}$

$ \Rightarrow \quad K_f^{\prime}-K_i^{\prime}=\frac{1}{2} \times \frac{m}{n} \times v^{\prime 2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

Also, initial kinetic energy of mass $m$

$ =\frac{1}{2} m v^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)$

∴ The fraction of KE that gets transferred to the stationary particle from the moving particle

$ \begin{aligned} & =\frac{\frac{1}{2} \times \frac{m}{n} \times v^{n^2}}{\frac{1}{2} m v^2}=\frac{v^{v^2}}{n v^2}=\frac{1}{n}\left(\frac{v^0}{v}\right)^2 \\ & =\frac{1}{n}\left(\frac{2 v}{\left(1+\frac{1}{n}\right) v}\right)^2 \quad \text { [From Eq. (iii)] } \\ & =\frac{1}{n} \frac{4 v^2}{\left(1+\frac{1}{n}\right)^2 v^2}=\frac{4}{n\left(\frac{n+1}{n}\right)^2}=\frac{4 n}{(1+n)^2} \end{aligned} $

$ \text { The given situation is shown as } $

We know,

$ \begin{aligned} & X_{C M}=\frac{m_1 x_1+m_2 x_2+m_3 x_3+m_4 x_4}{m_1+m_2+m_3+m_4} \\ & =\frac{(3 M \times-1)+(2 M \times 0)+(M \times 1)+(4 M \times 0)}{3 M+2 M+M+4 M} \\ & =\frac{-3 M+0+M+0}{10 M}=\frac{-2 M}{10 M}=\frac{-1}{5} \\ & \text { Also, } Y_{C M}=\frac{m_1 y_1+m_2 y_2+m_3 y_3+m_4 y_4}{m_1+m_2+m_3+m_4} \\ & =\frac{(3 M \times 0)+(2 M \times 1)+(M \times 0)+(4 M \times-1)}{10 M} \\ & =\frac{0+2 M+0-4 M}{10 M}=\frac{-2 M}{10 M}=\frac{-1}{5} \end{aligned} $

∴ Position of centre of mass from the centre of the circle

$ r_{C M}=X_{C M} \hat{\mathbf{i}}+Y_{C M} \hat{\mathbf{j}}=\left(\frac{-1}{5} \hat{\mathbf{i}}-\frac{1}{5} \hat{\mathbf{j}}\right) $

The given situation is shown in the following figure.

By the law of conservation of linear momentum, we have Total momentum before collision = Total momentum after collision

$ \begin{aligned} & m_1(u)+m_2(0)=m_1\left(\frac{2 u}{3}\right)+m_2(v) \\ & \Rightarrow \quad m_1 u+0=\frac{2}{3} m_1 u+m_2 v \end{aligned} $

$ \begin{align*} & \Rightarrow \quad m_1 u-\frac{2}{3} m_1 u=m_2 v \Rightarrow \frac{1}{3} m_1 u=m_2 v \\ & \Rightarrow \quad \frac{m_1}{m_2}=\frac{3 v}{u} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Since, the collision is perfectly elastic, then coefficient of restitution $=1$

$ \begin{aligned} e & =1 \\ \left|\frac{\mathbf{v}_2-\mathbf{v}_1}{\mathbf{u}_1-\mathbf{u}_2}\right| & =1 \Rightarrow \frac{v-\frac{2 u}{3}}{u-0}=1 \\ \Rightarrow \quad \frac{v-\frac{2 u}{3}}{u} & =1 \Rightarrow v=\frac{5 u}{3} \end{aligned} $

∴ From Eq. (i), we get

$ \frac{m_1}{m_2}=\frac{3 \times 5 u / 3}{u}=5 $

The given situation is shown in the following figure

Let $v_1$ and $v_2$ are final velocities of bullet and block, respectively, as block rises upto height $h$.

Hence, according to conservation of energy,

$ \begin{array}{rlrl} & & \frac{1}{2} M v_2^2 & =M g h \\ \Rightarrow & v_2 & =\sqrt{2 g h} \end{array} $

Given, $h=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$

$ \begin{aligned} \Rightarrow \quad v_2 & =\sqrt{2 \times 10 \times 20 \times 10^{-2}} \\ & =2 \mathrm{~m} / \mathrm{s} \end{aligned} $

Now, conservation of momentum gives

$ \begin{aligned} m u & =m v_1+M v_2 \\ \frac{25}{1000} \times 250 & =\frac{25}{1000} \times v_1+1 \times 2 \\ \frac{25}{4} & =\frac{25}{1000} \times v_1+2 \\ \Rightarrow \quad v_1 & =170 \mathrm{~ms}^{-1} \end{aligned} $

Let mass per unit area of the $\operatorname{disc}=m$

Total mass of disc, $M=\pi R^2 m$

Mass of the scooped out portion of the disc,

$ M^{\prime}=\pi r^2 m $

For small objects, centre of gravity and centre of mass are at same position.

We take the centre $O$ of the disc as the origin. The masses $M$ and $M^{\prime}$ can be supposed to be concentrated at the centres of the disc and scooped out portion respectively. The mass $M^{\prime}$ of the removed portion is taken negative.

The $x$-coordinate of the centre of mass of the remaining portion of the disc will be

$ \begin{aligned} & x_{\mathrm{CM}}=\frac{M x_1-M^{\prime} x_2}{M-M^{\prime}}=\frac{M \times 0-M^{\prime} \times 3}{\pi R^2 m-\pi r^2 m} \\ &=\frac{-3 M^{\prime}}{\pi m\left(R^2-r^2\right)} \\ &=\frac{-3 \pi r^2 m}{\pi m\left(R^2-r^2\right)} \\ &=-\frac{-3 r^2}{R^2-r^2}=\frac{-3 \times 3^2}{6^2-3^2}=-1 \mathrm{~cm} \\ & \Rightarrow \quad x_{\mathrm{CM}}=-1 \mathrm{~cm} \end{aligned} $

Hence, distance of centre of gravity of the resulting flat body from the centre of the original disc will be 1 cm left side.