Question

A ball of mass $m$ is suspended by a massless string of length $l=10\text{m}$ from a fixed point as shown in the figure below. A ball of mass $2m$ strikes at an angle $\theta ={45}^{\circ }$ from the horizontal and sticks to it. If the system deflects by angle $\varphi =\frac{\pi }{2}$, then the velocity of the system just after the collision and the velocity of $2m$ particle before collision are, respectively $\left(\text{in m/s}\right)$

• A. $10\sqrt{2},30$
• B. $10,30\sqrt{2}$
• C. $10,30$
• D. $10\sqrt{2},30\sqrt{2}$

Answer 1

The correct option is A $10\sqrt{2},30$

Assume the velocity of mass $2m$ initially is $u$ and just after the collision, common velocity of the system is $v$.

Immediately after collsion, vertical velocity of the system ($m+2m$) will be zero due to the string.
By momentum conservation along horizontal direction :-
$\left(2m\cos {45}^{\circ }\right)u=(m+2m)v$
$2u\times \frac{1}{\sqrt{2}}=3v$
or $u=\frac{3}{\sqrt{2}}v$ ...(1)

By energy conservation after the collision,
Initial Kinetic Energy = Final Potential Energy
$\frac{1}{2}\left(3m\right)v^2=3mgl$
(since system deflects by angle $\frac{\pi }{2}$, final height achieved is $l$)

$\Rightarrow v=\sqrt{2gl}=\sqrt{2\times 10\times 10}$
$\Rightarrow v=10\sqrt{2}\text{m/s}$

$\therefore$ Velocity of system just after collision
$v=10\sqrt{2}\text{m/s}$ in horizontal direction

and velocity of $2m$ mass particle initially, will be
$u=\frac{3}{\sqrt{2}}\times v=\frac{3}{\sqrt{2}}\times 10\sqrt{2}=30\text{m/s}$
(from (1))