Question 305: Centre of Mass and Rotational Motion

Question

An object of mass ' $m$ ' moving with velocity ' $u$ ' collides with another stationary object of mass ' $M$ ' and stops just after the collision. The coefficient of restitution is

Accepted Answer

Step 1: Write momentum conservation
Collision is assumed to be in 1D and no external force acts.
Initial momentum:
$ p_i = m u + M \cdot 0 = m u $
Final momentum (since mass $m$ stops after collision, its velocity = 0; velocity of $M = V$):
$ p_f = m \cdot 0 + M V = M V $
By conservation of momentum:
$ m u = M V \quad \Rightarrow \quad V = \frac{m}{M} u $
Step 2: Coefficient of restitution formula
Coefficient of restitution $e$ is defined as:
$ e = \frac{	ext{relative velocity after}}{	ext{relative velocity before}} $
Relative velocity before collision:
$ u_{	ext{rel,before}} = u - 0 = u $
Relative velocity after collision:
Object $m$ has velocity = 0, object $M$ has velocity $V = 	frac{m}{M}u$.
So relative velocity after collision:
$ u_{	ext{rel,after}} = V - 0 = \frac{m}{M}u $
Thus,
$ e = \frac{	frac{m}{M} u}{u} = \frac{m}{M} $
Final Answer:
The coefficient of restitution is
$ \boxed{	frac{m}{M}} $
Correct option: C

Answer

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

Answer

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

Answer

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

Answer

1

Centre of Mass and Rotational Motion

An object of mass ' m ' moving with velocity ' u ' collides with another stationary object of mass ' M ' and stops just after the collision. The coefficient of restitution is

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