A moving particle of mass $m,$ makes a head on elastic collision with another particle of mass $2 m,$ which is initially at rest. The percentage loss in energy of the colliding particle on collision, is close to

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33%

67%

90%

10%

answer is C.

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Detailed Solution

$Conservation of total kinetic energy of the system during collision \frac{1}{2} m_{1} {(u_{1})}^{2} + \frac{1}{2} m_{2} {(u_{2})}^{2} = \frac{1}{2} m_{1} {(v_{1})}^{2} + \frac{1}{2} m_{2} {(v_{2})}^{2} o r v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1} + (\frac{2 m_{2} u_{2}}{m_{1} + m_{2}}) p u t \begin{aligned} m_{1} = m \\ m_{2} = 2 m \\ u_{2} = 0 \end{aligned} \Rightarrow v_{1} = - \frac{u_{1}}{3} Initial kinetic energy of the colliding body is given by, (K . E)_{i} = \frac{1}{2} m {(u_{1})}^{2} (K . E)_{f} = \frac{1}{2} m {(\frac{- u_{1}}{3})}^{2} Δ K . E = (K . E)_{i} - (K . E)_{f} % \frac{Δ K . E}{(K . E)_{i}} = \frac{8}{9} \times 100 = 89 % \approx 90 %$