Question

A particle of mass 'm' collides with another stationary particle of mass 'M'. A particle of mass 'm' stops just after collision. The coefficient of restitution is

• A. \(\frac{M}{m}\)
• B. \(\frac{m+M}{M}\)
• C. \(\frac{M-m}{M+m}\)
• D. \(\frac{m}{M}\)

Hint:

In a collision where one particle stops, the coefficient of restitution equals the ratio of the masses only if the target was initially at rest. Remember: \(e = \frac{\text{velocity gained by target}}{\text{initial velocity of projectile}}\) when the projectile stops.

Answer 1

The Correct Option is D

Step 1: Set up the collision.
Let $m$ move with speed $u$ and hit the resting mass $M$. After impact $m$ stops, and $M$ moves with speed $v$.

Step 2: Conserve momentum.
$mu = Mv$, so $v = \tfrac{m}{M}u$.

Step 3: Form the restitution ratio.
The speed of approach is $u$ and the speed of separation is $v$, so \[ e = \frac{v}{u} = \frac{(m/M)u}{u}. \]

Step 4: Simplify.
\[ \boxed{e = \frac{m}{M}} \]