Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
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
PhysicsCenter of Mass Momentum and CollisionMHT CETMHT CET 2021 (21 Sep Shift 1)
Options:
  • A $\frac{M}{m}$
  • B $\frac{\mathrm{m}+\mathrm{M}}{\mathrm{M}}$
  • C $\frac{M-m}{M+m}$
  • D $\frac{\mathrm{m}}{\mathrm{M}}$
Solution:
2145 Upvotes Verified Answer
The correct answer is: $\frac{\mathrm{m}}{\mathrm{M}}$
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,
$$
\mathrm{e}=\frac{\text { Re lative velocity after collision }}{\text { Relative velocity before collision }}=\frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.