Question

A U shaped tube of mass '2m' is placed on a smooth horizontal surface. Two identical spherical balls each of mass 'm' and of diameter slightly less than the inner diameter of tube enters into the tube with a velocity u as shown. (Assume no loss of energy anywhere and all collisions to be elastic).

Speed of each spherical ball, just before their collision is

• A. $\frac{u}{\sqrt{3}},\frac{u}{\sqrt{3}}$
• B. $\frac{2u}{3\sqrt{3}},\frac{u}{\sqrt{3}}$
• C. $\frac{\sqrt{3}u}{2},\frac{\sqrt{3}u}{2}$
• D. $\frac{u}{2},\frac{\sqrt{3}u}{2}$

Answer 1

The correct option is C

$\frac{\sqrt{3}u}{2},\frac{\sqrt{3}u}{2}$

Let, velocity of tube becomes V along x direction just before collision, then velocities of balls will also be

V in this direction along with components of velocities perpendicular to this direction just before

collision. Apply conservation of momentum along X direction.

$2mu$ = $2mV+mV+mV$ $\Rightarrow V$ = $\frac{u}{2}$

Apply Conservation of energy:

$\left(\frac{1}{2}m{u}^2\right)\times 2$ = $\frac{1}{2}\left(2m\right)V^2+2\left[\frac{1}{2}m\right(V^2+V_y^2\left)\right]$

$\Rightarrow V_y$ = $\frac{u}{\sqrt{2}}$

$V_{net}$ = $\sqrt{(\frac{u}{2}{)}^2+(\frac{u}{\sqrt{2}}{)}^2}$ = $\frac{u\sqrt{3}}{2}$