A ball of mass $m_{1}$ experiences a perfectly elastic collision with a stationary ball of mass $m_{2}$ . What fraction of the K.E. does the striking ball lose, if it recoils at right angle to the original direction of motion.

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Solution

Mass $m_{1}$ undergoes elastic collision mass $m_{2}$ is at rest $4$ let the recoiling angle be denoted by A.
Momentum:
$m_{1} \times 4 = m_{2} \times v_{2} \times cos A$
$m_{2} \times v_{1} = m_{2} \times v_{2} \times sin A$
Squaring $4$ adding $m_{1}^{2} (u^{2} + v_{1}^{2}) = m_{2}^{2} \times v_{2}^{2}$
square initial $4$ final K.E
$1 / 2 \times m_{1} \times u^{2} = 1 / 2 \times m_{1} \times v_{1}^{2} + 1 / 2 m_{2} v_{2}^{2}$
$m_{2} (u^{2} - v_{1}^{2}) = m_{1} (u^{2} + v_{1}^{2}) (v_{1} / u)^{2} = (m_{2} - m_{1}) (m_{2} + m_{1})$
fraction of kinetic energy lost
$(0.5 \times m_{1} \times u^{2} - 0.5 \times m_{1} v_{1}^{2}) / 0.5 \times m_{1} \times u_{2} 1 - (v_{1} / u)^{2}$
Therefore, $1 - (v_{1} / u)^{2} = 1 (m_{2} - m_{1} / m_{2} + m_{1})$
fraction of KE lost $= 2 m_{1} / (m_{1} + m_{2})$ .

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