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 $\text{B}$:

$J\cos \theta =Mv......\left(1\right)$

And for ball $\text{C}:$

$-2J\sin \theta =0-M{v}_0$

$\Rightarrow 2J\sin \theta =M{v}_0.......\left(2\right)$

From Eqs.$\left(1\right)$ and $\left(2\right)$,

$\tan \theta =\frac{v_0}{2v}.........\left(3\right)$

From $\Delta \text{AOC}$,

$\text{OC}=\sqrt{{\text{AC}}^2-{\text{AO}}^2}=\sqrt{\frac{9}{4}{R}^2-R^2}=\frac{\sqrt{5}}{2}R$

$\therefore \tan \theta =\frac{\text{OC}}{\text{AO}}=\frac{\sqrt{5}R/2}{R}=\frac{\sqrt{5}}{2}$

From equation $\left(3\right)$,

$\frac{\sqrt{5}}{2}=\frac{v_0}{2v}\Rightarrow v=\frac{v_0}{\sqrt{5}}$

Hence, option $\left(\text{C}\right)$ is correct.