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Collision Example II

A ball of mas...

Question

A ball of mass $m_{1}$ in motion hits directly another ball of mass $m_{2}$ at rest and sticks to it, the total kinetic energy after collision is 2/3 of their total K.E. before the collision. Determine the ratio of $m_{1} : m_{2}$ .

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Solution

Velocity after collision is $v_{1}$ .

By momentum conservation,

$m_{1} v = (m_{1} + m_{2}) v_{1} v_{1} = \frac{m_{1}}{(m_{1} + m_{2})} v$

K.E after collision= $\frac{2}{3}$ K.E before collision.
$\frac{1}{2} (m_{1} + m_{2}) {v_{1}}^{2} = \frac{2}{3} (m_{1} v^{2}) \frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{1}}{m_{1} + m_{2}})}^{2} {v_{1}}^{2} = \frac{2}{3} \times \frac{1}{2} m_{1} v^{2} \frac{{m_{1}}^{2}}{m_{1} + m_{2}} = \frac{2}{3} m_{1} 3 m_{1} = {2 m}_{1} + 2 m_{2} \frac{m_{1}}{m_{2}} = 2$

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