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: Understanding the Question:
A moving particle of mass \(m\) collides with a stationary particle of mass \(M\). After the collision, the particle of mass \(m\) comes to rest. We need the coefficient of restitution \(e\) for this collision.

Step 2: Key Formula or Approach:
Coefficient of restitution is defined as: \[ e = \frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}. \] We also use conservation of linear momentum because no external forces act.

Step 3: Detailed Explanation:
Let the initial velocity of mass \(m\) be \(u\) (towards \(M\)). Mass \(M\) is initially at rest. After the collision, \(m\) stops, so its final velocity is \(0\). Let the final velocity of \(M\) be \(v\).
From conservation of momentum: \[ m u + M \cdot 0 = m \cdot 0 + M v \quad \Rightarrow \quad m u = M v \quad \Rightarrow \quad v = \frac{m}{M} u. \] Relative velocity of approach (before collision): \(u - 0 = u\) (since they move towards each other, but here only one moves).
Relative velocity of separation (after collision): \(v - 0 = v\) (both move in the same direction after? Actually after collision, \(m\) stops, \(M\) moves forward; they separate with velocity \(v\) relative to each other).
Thus: \[ e = \frac{v}{u} = \frac{\frac{m}{M} u}{u} = \frac{m}{M}. \]

Step 4: Final Answer:
The coefficient of restitution is \(\frac{m}{M}\), which is option (D).