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(C) For a head-on collision between two identical masses $m$ where one is at rest,the loss in kinetic energy $\Delta KE$ is given by: $\Delta KE = \fr

Vedclass · Accepted Answer

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

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(C) For a head-on collision between two identical masses $m$ where one is at rest,the loss in kinetic energy $\Delta KE$ is given by: $\Delta KE = \frac{1}{2} \frac{m_1 m_2}{m_1 + m_2} (1 - e^2) (u_1 - u_2)^2$ Since $m_1 = m_2 = m$,$u_1 = u$,and $u_2 = 0$: $\Delta KE = \frac{1}{2} \frac{m^2}{2m} (1 - e^2) u^2 = \frac{1}{4} m u^2 (1 - e^2)$ Given that $\Delta KE = \frac{1}{4} KE_{initial} = \frac{1}{4} (\frac{1}{2} m u^2)$: $\frac{1}{4} m u^2 (1 - e^2) = \frac{1}{4} (\frac{1}{2} m u^2)$ $(1 - e^2) = \frac{1}{2}$ $e^2 = 1 - \frac{1}{2} = \frac{1}{2}$ $e = \frac{1}{\sqrt{2}}$

$A$ moving ball collides head-on with a similar ball at rest. If $\frac{1}{4}$ of the kinetic energy is lost during the impact,the value of the coefficient of restitution is:

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