Q110 / 150 · MHT CET 2024 (15 May Shift 2) · Physics

A particle of mass $m$ collides with another stationary particle of mass M . The particle m stops just after collision. The coefficient of restitution is .

  • Correct answer
    A
    $\frac{m}{M}$
  • B
    $\frac{M-m}{M+m}$
  • C
    1
  • D
    $\frac{\mathrm{m}}{\mathrm{M}+\mathrm{m}}$

Solution

Let v be the velocity of mass m and $\mathrm{v}^{\prime}$ be the velocity of mass M after collision. By law of conservation of momentum, $\mathrm{mv}=\mathrm{Mv}^{\prime}$ $\therefore \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}}$
Coefficient of restitution $=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}$ Here, $\mathrm{v}_1=0$ and $\mathrm{u}_2=0$ $\therefore \quad e=\frac{v^{\prime}}{v}=\frac{m}{M}$

Hence, the correct answer is option: 'A'.

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