Question

Two balls are fastened to two light strings of same length of $l$ and of different masses $m$ and $2m$ as shown in the figure. The other end of the strings are fixed at $O$. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is perfectly elastic. Find the ratio of the speeds of the balls, just after the collision.

• A. $5:1$
• B. $2:1$
• C. $\sqrt{2}:1$
• D. $\sqrt{5}:1$

Answer 1

The correct option is A $5:1$

Just before collision:


From WPE - Initial TE = Final TE
$\Rightarrow mgl=\frac{1}{2}m{v}^2$
$v=\sqrt{2gl}$, so just before collision speed of both balls will be same.
No horizontal force is present so momentum along the horizontal direction will be conserved.
Just after collision:


Assumed just after collision velocity of $2m$ & $m$ masses are $v_1$ & $v_2$ respectively.
Fom momentum conservation we have
$2mv-mv=2m{v}_1+m{v}_2$
$v=2v_1+v_2-----\left(1\right)$
Collision is elastic, hence, $e=1$
$\Rightarrow e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}=\frac{v_2-v_1}{v+v}=1$
$2v=v_2-v_1$ ------(II)
from eq (I) & (II)
$v_2=\frac{5}{3}v{v}_1=\frac{-v}{3}$
(-ve means opposite in direction.)
$\Rightarrow \frac{v_2}{v_1}=\frac{\frac{5}{3}v}{\frac{v}{3}}=\frac{5}{1}$
Hence, the ratio of their velocities is $5:1$