Question

A particle of mass $m$ moving with a speed $v$ collide elastically with the end of a uniform rod of mass $M$ and length $L$perpendicularly as shown in the figure. If the particle comes to rest after collision, find the value of $\frac{m}{M}$.$3$

• A. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• B. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• C. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
• D. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

Answer 1

The correct option is C

$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Step $1$:Given data and drawing of the situation

Given data:

mass of particle $=m$

mass of the uniform rod $=M$

Before collision:

velocity of particle $=v$

velocity of rod $=0$

After collision:

velocity of particle (rest)

velocity of rod $=v_1$

Step $2$: Applying the law of conservation of angular momentum about the center of mass of the rod

$mv\left(\frac{L}{2}\right)=M\left(\frac{L^2}{12}\right)\omega$ …….…….$\left(1\right)$

Step$3$ : Applying the law of conservation of linear momentum

Initial momentum = Final momentum

$mv=M{v}_1$ …………..$\left(2\right)$

Step$4$: Using coefficient of restitution for elastic collision

Coefficient of restitution, $e=\frac{relativespeedaftercollision}{relativespeedbeforecollision}$

For elastic collision,

$$\begin{gathered} relativespeedaftercollision=\left[v_1+\omega \left(\frac{L}{2}\right)\right] \\ relativespeedbeforecollision=v \end{gathered}$$

$\Rightarrow e=1=\frac{\left[v+\omega \left(\frac{L}{2}\right)\right]}{v}$ ……………$\left(3\right)$

Step $5$: Putting $v_1$ from $\left(2\right)$ and $\omega L$ from $\left(1\right)$ in $\left(3\right)$

$$\begin{gathered} v_{1}from\left(2\right), \\ \left[v_1=\frac{mv}{M}\right] \end{gathered}$$

$$\begin{gathered} \omega Lfrom\left(1\right), \\ \omega L=\frac{mvL}{2}\times \frac{12}{ML} \\ \omega L=\frac{6mv}{M} \end{gathered}$$

Putting these values in $\left(3\right)$

$$\begin{gathered} 1=\left[\frac{mv}{M}+\left(\frac{6mv}{M}\right)\frac{1}{2}\right]\times \frac{1}{V} \\ v=\left[\frac{mv}{M}+\frac{3mv}{M}\right] \\ V=\left[\frac{4mv}{M}\right] \\ \frac{V}{V}=\frac{4m}{M} \\ 1=\frac{4m}{M} \\ \frac{1}{4}=\frac{m}{M} \end{gathered}$$

Hence, 'Option$\left(C\right)$ is the correct.