First slide

Coefficient of restitution

Question

A steel sphere A is at rest on a frictionless horizontal surface. Another identical sphere B moving with velocity 2m/s collides head on with A. If coefficient of restitution e = 0.5, fraction of kinetic energy of B lost in the process is

Moderate

A

15/16
B

3/4
C

1/8
D

1/4

Solution

Conserving momentum: mv= $m v_{1} + m v_{2}$ where $v_{1}$ is the final velocity of B while $v_{2}$ is the final velocity of A
$v_{2} - v_{1} = \frac{v}{2}$
Putting above value in first equation, we get
$v_{2} = 3 v_{1}$
$v_{1} = \frac{v}{4} v_{2} = \frac{3 v}{4}$
Change in kinetic energy of B = $\frac{15}{32} m v^{2}$
Thus fraction of KE lost= $\frac{15}{16}$

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