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MHT CET · Physics · Center of Mass Momentum and Collision

A particle of mass ' \(\mathrm{m}\) ' collides with another stationary particle of mass ' \(M\) '. Particle of mass ' \(m\) ' stops just after collision. The coefficient of restitution is

  1. A \(\frac{M}{m}\)
  2. B \(\frac{\mathrm{m}+\mathrm{M}}{\mathrm{M}}\)
  3. C \(\frac{M-m}{M+m}\)
  4. D \(\frac{\mathrm{m}}{\mathrm{M}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\mathrm{m}}{\mathrm{M}}\)

Step-by-step Solution

Detailed explanation

Let \(v\) be the velocity of mass \(m\) and \(v\) ' be the velocity of mass \(M\) after collision.'
By law of conservation of momentum
\(\begin{aligned}& \mathrm{mv}=\mathrm{Mv} \\& \therefore \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}}\end{aligned}\)
Coefficient of restitutions,
\(\mathrm{e}=\frac{\text { Re lative velocity after collision }}{\text { Relative velocity before collision }}=\frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}}\)