Question

A ball of mass 'm', moving with a velocity 'v' along X-axis, strikes another ball of mass '2m' kept at rest. The first ball comes to rest after collision and the other breaks into two equal pieces. One of the pieces starts moving along Y-axis with a speed $v_1$. What will be the velocity of the other piece?

• A. $\sqrt{v^2+v_1^2}$
• B. $\sqrt{2v^2+v_1^2}$
• C. $\sqrt{v^2-v_1^2}$
• D. $\sqrt{\sqrt{v^2+2v_1^2}}$

Answer 1

The correct option is A

$\sqrt{v^2+v_1^2}$

Can you recall,

${\overrightarrow{v}}_{COM}=\frac{m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}}{m_1+m_2}$

Considering both balls as one system Initially the velocity of COM of system

${\overrightarrow{v}}_{COM}=\frac{mv\hat{i}}{m+2m}=\frac{v}{3}\hat{i}$

After collision,

Let the velocity of other piece be $\overrightarrow{v_2}=v_x\hat{i}+v_y\hat{j}$

As there is no net external force, v will not change and fit ball of mass 'm' comes to rest.

$({\overrightarrow{v}}_{COM}{)}_x=\frac{m_1\left(v_1\right)x+m_2(v_2{)}_x}{m+m_1+m_2}$

$\frac{v}{3}=\frac{0+m{v}_k}{3m}$

$v_x=v$ ......... (1)

Similarly,

$({\overrightarrow{v}}_{COM}{)}_y=\frac{m_1\left(v_1\right)y+m_2(v_2{)}_y}{m+m_1+m_2}$

0 = $\frac{m{v}_1+m{v}_y}{(2m+m)}\Rightarrow v_y=v_1$

Therefore, $\overrightarrow{v_2}=v\hat{i}-v_1\hat{j}$

$|\overrightarrow{v_2}|=\sqrt{v^2+v_1^2}$