Question

Two balls fastened to light string of same length $l$ and of different masses $m$ & $2m$. The other end of the strings are fixed at $O$. If both the masses collide with same speed (just before collision). Find the ratio of heights raised by the balls.
[Consider, collision is elastic.]

• A. $1:\sqrt{5}$
• B. $1:5$
• C. $1:2$
• D. $1:25$

Answer 1

The correct option is D $1:25$

Given, just before collision, both the masses have same speed, let's say $v$
Just After the collision, let the condition be as shown in the below figure.


There is no external force present in the horizontal direction.
Hence, momentum will be conserved along horizontal direction.
$\Rightarrow 2mv-mv=2m{v}_1+m{v}_2$
$\Rightarrow v=2v_1+v_2------\left(1\right)$
Collision is elastic hence, $e=1$
$\Rightarrow e=\frac{\text{speed of seperation}}{\text{speed of approach}}=\frac{v_2-v_1}{2v}=1$
$\Rightarrow 2v=v_2-v_1-----\left(2\right)$
from $\left(1\right)$ & $\left(2\right)$, we have
$v_2=\frac{5v}{3};v_1=\frac{-v}{3}$
Here, $-ve$ represents that the assumed direction of $v_1$ is wrong, it will be opposite to the assumed direction.
So, the situation of the masses are as shown in the below figure.


Here, assumed height attained by the mass $2m$ is $h_1$ and by $m$ is $h_2$.
From Work-Energy theorm - (Just after collision to maximum height attained by the balls) we have,
For $2m$ mass ball
$\frac{1}{2}\left(2m\right)v_1^2+0=2mg{h}_1+0$
$\Rightarrow v_1=\sqrt{2g{h}_1}-----\left(3\right)$
Similarly, for $m$ mass ball
$\frac{1}{2}m{v}_2^2+0=mg{h}_2+0$
$\Rightarrow v_2=\sqrt{2g{h}_2}------\left(4\right)$
from eq $\left(3\right)$ & $\left(4\right)$, we have
$\frac{v_1}{v_2}=\sqrt{\frac{2g{h}_1}{2g{h}_2}}$
$\Rightarrow \left(\frac{\frac{v}{3}}{\frac{5v}{3}}\right)=\sqrt{\frac{h_1}{h_2}}$
$\Rightarrow {\left(\frac{1}{5}\right)}^2=\frac{h_1}{h_2}\Rightarrow \frac{h_1}{h_2}=\frac{1}{25}$