Question

A disc of mass $m$ is moving with constant speed $v_0$ on a smooth horizontal table. Another disc of same mass is placed on the table at rest as show in the figure. If collision is elastic then find the velocity of the discs after collision.

[Consider $r<<R$]$(\theta =ta{n}^{-1}\frac{h}{\sqrt{R^2-h^2}})$.

• A. $0,v_{0}sin\theta$
• B. $0,v_{0}cos\theta$
• C. $v_{0}sin\theta ,v_{0}cos\theta$
• D. $v_0,0$

Answer 1

The correct option is C

$v_{0}sin\theta ,v_{0}cos\theta$

since OO' is the common normal and in the direction of common normal there is no external impulse, hence momentum of the system will be conserved.Also momentum of the body along line PQ (common tangents) does not change.Let velocities of discs along the common normal after collision are v₁ and v₂ respectively.

Therefore, $\frac{v_2-v_1}{0-v_{0}cos\theta }$ = $-e$

Where $cos\theta$ = $\frac{\sqrt{(R+r{)}^2-h^2}}{R+r}$ = $\frac{\sqrt{R^2-h^2}}{R}$(as $r<<R$)

$\Rightarrow v_2-v_1$ = $v_{0}cos\theta$, (i) as $e$ = 1 for elastic collision.

From conservation of momentum of the system along $O{O}^{\prime }$

$M{v}_{0}cos\theta$ = $m(v_1+v_2)$ (ii)

Solving (i) and (ii), we get $v_1$ = 0, $v_2$ = $v_{0}cos\theta$

Velocity of the disc1 along $PQ$ after collision will be $v_{0}sin\theta$ only and that of the disc 2 will be zero.

Therefore, velocity of disc1 after collision = $v_{0}sin\theta$ and velocity of disc 2 after collision = $v_{0}cos\theta$