First slide

Coefficient of restitution

Question

A ball collides impinges directly on a similar ball at rest. The first ball is brought to rest after the impact. If half of the kinetic energy is lost by impact, the value of coefficient of restitution (e) is

Moderate

A

$\frac{1}{\sqrt{3}}$
B

$\frac{\sqrt{3}}{2}$
C

$\frac{1}{2 \sqrt{2}}$
D

$\frac{1}{\sqrt{2}}$

Solution

$∴ Initial KE, K_{i} = \frac{1}{2} {mu}_{1}^{2} + \frac{1}{2} {mu}^{2}_{2} = \frac{1}{2} {mu}^{2}_{1}$
and final KE, $K_{f} = \frac{1}{2} {mv}^{2}_{1} + \frac{1}{2} {mv}_{2}^{2} = \frac{1}{2} {mv}^{2}_{2}$
Loss in KE, $∆ K = K_{i} - k_{f} = \frac{1}{2} {mu}^{2}_{1} - \frac{1}{2} {mu}^{2}_{2}$
It is given $\frac{1}{2} (\frac{1}{2} {mu}^{2}_{1}) = \frac{1}{2} {mu}^{2}_{1} - \frac{1}{2} {mv}^{2}_{2}$
( $∴ Half of its KE is lost by impact$ )
or ${u^{2}}_{1} = 2 {v^{2}}_{2} or v_{2} = \frac{u_{1}}{\sqrt{2}}$
Hence, coefficient of restitution, $e = | \frac{v_{2} - v_{1}}{u_{1} - u_{2}} | = \frac{v_{2}}{u_{1}} = \frac{1}{\sqrt{2}}$

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