A neutron moving at a speed v undergoes a head-on elastic collision with a nucleus of mass number A at rest. The ratio of the kinetic energies of the neutron after and before collision is

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Solution

The correct option is A
Mass of neutron $(m_{1}) = 1 u n i t$ . Mass of nucleus $(m_{2}) = A u n i t s$ .
The velocity of the neutron after the collision is
$v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u = (\frac{1 - A}{1 + A}) u$
KE of neutron after collision = $\frac{1}{2} m_{1} v_{1}^{2} = \frac{1}{2} \times 1 \times {(\frac{1 - A}{1 + A})}^{2} u^{2}$
KE of neutron before collision = $\frac{1}{2} m u^{2} = \frac{1}{2} \times 1 \times u^{2} = \frac{1}{2} u^{2}$ .
Their ratio is ${(\frac{1 - A}{1 + A})}^{2}$ , which is choice (a).

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