A $2 \mathrm{~kg}$ ball moving at $24 \mathrm{~ms}^{-1}$ undergoes inelastic head-on collision with a $4…

A $2 \mathrm{~kg}$ ball moving at $24 \mathrm{~ms}^{-1}$ undergoes inelastic head-on collision with a $4 \mathrm{~kg}$ ball moving in the opposite direction at $48 \mathrm{~ms}^{-1}$. If the coefficient of restitution is $2 / 3$, their velocities in $\mathrm{ms}^{-1}$ after impact are
  1. $-56,-8$
  2. $-28,-4$
  3. $-14,-2$
  4. $-7,-1$

Solution

$\begin{aligned} & \begin{aligned} & v_1^{\prime}= {\left[\frac{m_1-e m_2}{m_1+m_2}\right] v_1-\left[\frac{(1+e) m_2}{m_1+m_2}\right] v_2 } \\ &=\left[\frac{2-\left(\frac{2}{3}\right) 4}{2+4}\right] 24-\left[\frac{\left(1+\frac{2}{3}\right) 4}{2+4}\right] 48 \\ &=-56 \mathrm{~m} / \mathrm{s} \\ & \text { and } v_2^{\prime}=\left[\frac{(1+e) m_1}{m_1+m_2}\right] v_1-\left[\frac{m_2-e m_1}{m_1+m_2}\right] v_2 \\ &=\left[\frac{\left(1+\frac{2}{3}\right) 2}{2+4}\right] 24-\left[\frac{4-\left(\frac{2}{3}\right) 2}{2+4}\right] 48 \\ &=-8 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}$

Asked in: AP EAMCET 2004

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