Question

In the figure shown below, two identical balls of mass $M$ and radius $R$ each, are placed in contact with each other on the frictionless horizontal surface. The third ball of mass $M$ and radius $R/2$, is coming down vertically and has a velocity $v_0$ when it hits the two balls symmetrically and itself comes to rest. Then, each of the two bigger balls will move after collision with a speed equal to

• A. $\frac{4v_0}{\sqrt{3}}$
• B. $\frac{2v_0}{\sqrt{3}}$
• C. $\frac{v_0}{\sqrt{3}}$
• D. $\sqrt{3}{v}_0$

Answer 1

The correct option is C $\frac{v_0}{\sqrt{3}}$

Assuming that the impulse imparted to each ball on the surface, during the collision is $J$. Since the situation is symmetric, the two balls on the ground will move with the same velocity.


For ball $B$:
$J\cos \theta =Mv\dots \left(i\right)$
& for ball $C$
$-2J\sin \theta =0-M{v}_0$
$\Rightarrow 2J\sin \theta =M{v}_0\dots \left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$,
$\tan \theta =\frac{v_0}{2v}\dots \left(iii\right)$

From $\Delta AOC$
$\tan \theta =\frac{OC}{AO}=\frac{\sqrt{5}\left(R/2\right)}{R}=\frac{\sqrt{5}}{2}$
From $\left(iii\right)$,
$\frac{\sqrt{5}}{2}=\frac{v_0}{2v}\Rightarrow v=\frac{v_0}{\sqrt{5}}$.