Question 313: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

Let $v$ be the velocity of mass $m$ and $v^{\prime}$ be the velocity of mass M after collision.
By law of conservation of momentum,
$\mathrm{mv}=\mathrm{Mv}^{\prime}$
$	herefore \quad \frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}}$
Coefficient of restitution $=\frac{v_2-v_1}{u_2-u_1}$
Here, $\mathrm{v}_1=0$ and $\mathrm{u}_2=0$
$	herefore \quad \mathrm{e}=\frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\frac{\mathrm{m}}{\mathrm{M}}$

Answer

$\frac{m}{M}$

Answer

$\frac{M-m}{M+m}$

Answer

1

Answer

$\frac{\mathrm{m}}{\mathrm{M}+\mathrm{m}}$

Centre of Mass and Rotational Motion

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

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