Question

A ball of mass $m$ is suspended by the string of length $l=2\text{m}$. Another ball of mass $2m$ collides horizontally with ball of mass $m$ with a speed of $u$ so that it completes vertical circle as depicted in the below figure. Find the initial speed of the ball of mass $2m$ before collision.
[Consider, collision is perfectly elastic, take $g=10{\text{ms}}^{-2}$]

• A. $5\text{m/s}$
• B. $\sqrt{5}\text{m/s}$
• C. $7.5\text{m/s}$
• D. $\sqrt{15}\text{m/s}$

Answer 1

The correct option is C $7.5\text{m/s}$

Let the initial speed of the ball of mass $2m$ be $u$


As we know that the speed required to complete a verticle circle is $\sqrt{5gl}$
Just after collision;


Since, no horizontal force is present, momentum will be conserved along horizontal direction.
$\Rightarrow 2m\left(u\right)+m\left(0\right)=2m\left(v_1\right)+m\left(\sqrt{5gl}\right)$
$\Rightarrow 2u=2v_1+\sqrt{5gl}.........\left(1\right)$
Also, collision is elastic hence, $e=1$
$\Rightarrow e=\frac{\text{velocity of seperation}}{\text{velocity of approach}}=1$
$\Rightarrow 1=\frac{\sqrt{5gl}-v_1}{u}\Rightarrow \sqrt{5gl}-v_1=u..........\left(2\right)$
from eq. $\left(1\right)$ & $\left(2\right)$ we have,
$u=\frac{3}{4}\sqrt{5gl}$
$u=\frac{3}{4}\times \sqrt{5\times 10\times 2}\Rightarrow u=7.5\text{m/s}$