Two masses undergo a perfectly elastic head-on collision. The first mass $^{'} m^{'}$ move with inertia velocity of $50 m / s$ in the $+ v e$ $x$ direction. The second mass $^{'} 2 m^{'}$ moves with initial velocity of $40 m / s$ in the $- v e$ $x$ direction.

(i)Calculate their final velocities (ii) What is the ratio of the final $K . E .$ to the initial $K . E$ for mass $m$ ? (iii) Calculate the change in momentum of masses $m$ and $2 m$ if $m = 1 k g$ . (iv) What is the change in momentum of the system?

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Solution

Final velocity of first body (of mass $m$ ) is

$V_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1} + (\frac{2 m_{1}}{m_{1} + m_{2}}) u_{2}$

$= (\frac{m - 2 m}{m + 2 m}) 50 + (\frac{2 \times 2 m}{m + 2 m}) (- 40)$

$= - \frac{50}{3} - \frac{160}{3}$

$= - \frac{210}{3}$

$= - 70 m / s$

Final velocity of second body of mass $2 m$ is

$V_{2} = (\frac{2 m}{m + 2 m}) (50) + (\frac{2 m - m}{m + 2 m}) (- 40)$

$= \frac{2}{3} (50) + \frac{1}{3} (- 40)$

$= \frac{100}{3} - \frac{40}{3}$

$= \frac{60}{3}$

$= 20 m / s$

(ii) Ratio of final to initial kinetic energies for body of mass mis

$\frac{\frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{2} v_{1}^{2}} = \frac{v_{1}^{2}}{u^{2} u_{1}} = \frac{(- 70)^{2}}{(50)^{2}} = \frac{4900}{2500} = \frac{49}{25}$

$= A n t [log 49 - log 25]$

$= A n t (1.6902 - 1.3979)$

$= A n t (0.2923)$

$= 1.96$

(iii) $m_{1} = 1 k g$

$m_{2} = 2 k g$

$∴$ Change in momentum of first body is $m_{1} v_{1} - m_{1} u_{1} = m_{1} (v_{1} - u_{1})$

$= 1 (- 70 - 50) = - 120 k g m / s$

Change in momentum of second body is

$m_{2} v_{2} - m_{2} u_{2} = m_{2} (v_{2} - u_{2}) = 2 (20 (- 40))$

$= + 120 k g m / s$

(iv) Momentum of the system will be conserve so change will be zero.

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