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.
A particle of mass $m$ moving with speed $u$ hits a still particle of mass $M$. After the hit, the first particle ($m$) stops dead. We want the coefficient of restitution $e$.

Step 2: Use momentum conservation.
No outside push acts, so total momentum stays the same: $mu + 0 = m(0) + Mv$, where $v$ is the speed of $M$ after the hit.

Step 3: Solve for the speed of $M$.
From $mu = Mv$ we get $v = \dfrac{m}{M}u$.

Step 4: Recall what $e$ means.
The coefficient of restitution is the speed of separation divided by the speed of approach: $e = \dfrac{\text{separation speed}}{\text{approach speed}}$.

Step 5: Find both speeds.
Before the hit they approach at speed $u$ (since $M$ was at rest). After the hit $m$ is still and $M$ moves at $v$, so they separate at speed $v$.

Step 6: Compute $e$.
$e = \dfrac{v}{u} = \dfrac{(m/M)u}{u} = \dfrac{m}{M}$, which is option (D).
\[ \boxed{e = \frac{m}{M}} \]