A neutron moving at a speed 'v' undergoes a head-on elastic collision

Question

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

Options:

A . (A−1A+1)2

B . (A+1A−1)2

C . (AA+1)2

D . (AA−1)2

Answer: Option A
:
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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