A $1 \mathrm{~kg}$ ball moving with a speed of $6 \mathrm{~ms}^{-1}$ collides headon with a $0.5…

A $1 \mathrm{~kg}$ ball moving with a speed of $6 \mathrm{~ms}^{-1}$ collides headon with a $0.5 \mathrm{~kg}$ ball moving in the opposite direction with a speed of $9 \mathrm{~ms}^{-1}$. If the coefficient of restitution is $\frac{1}{3}$, then the energy lost in the collision is
  1. $303.4\ J$
  2. $66.7\ J$
  3. $33.3\ J$
  4. $67.8\ J$

Solution

Using $\Delta \mathrm{KE}=\frac{m_1 m_2}{2\left(m_1+m_2\right)}\left(1-e^2\right)\left(u_1+u_2\right)^2$ Given: $m_1=1 \mathrm{~kg}$ $\begin{aligned} & m_2=\frac{1}{2} \mathrm{~kg} ; \\ & u_1=6 \mathrm{~ms}^{-1} ; \quad u_2=9 \mathrm{~ms}^{-1} \text { and } e=\frac{1}{3} \\ & \therefore \Delta \mathrm{KE}=\frac{1 \times \frac{1}{2}}{2\left(1+\frac{1}{2}\right)}\left[1-\left(\frac{1}{3}\right)^2\right](6+9)^2 \\ & =\frac{1}{2} \times \frac{2}{6} \times \frac{8}{9} \times(15)^2 \\ & =\frac{1}{6} \times \frac{8}{9} \times 225 \\ & =33.33 \mathrm{~J} \end{aligned}$

Asked in: AP EAMCET 2016

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