Question

An object of mass 'm' moving with velocity 'u' collides with another stationary object of mass 'M' and stops just after the collision. The coefficient of restitution is ______.

• A. $\frac{m}{M + m}$
• B. $\frac{M - m}{M + m}$
• C. $\frac{m}{M}$
• D. 1

Hint:

If a perfectly elastic collision ($e=1$) occurs between identical masses ($m=M$), they exchange velocities. Here, $m$ stops and $M$ moves. If $m=M$, then $e=1$, which perfectly aligns with our derived formula $e=m/M$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We use the Law of Conservation of Momentum and the definition of the coefficient of restitution ($e$).

Step 2: Formula Application:
Conservation of Momentum: $mu + M(0) = m(0) + Mv_2 \implies v_2 = \frac{mu}{M}$.
Coefficient of restitution: $e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2}$.

Step 3: Explanation:
Substituting the values: $v_1 = 0$ (m stops), $v_2 = \frac{mu}{M}$, $u_1 = u$, $u_2 = 0$.
$e = \frac{(mu/M) - 0}{u - 0} = \frac{mu}{Mu} = \frac{m}{M}$.

Step 4: Final Answer:
The coefficient of restitution is $m/M$.