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 - 1)/(A + 1))^2`
B. `((A + 1)/(A - 1))^2`
C. `((A)/(A + 1))^2`
D. `((A)/(A - 1))^2`

Answer

A

Explanation

Mass of neutron `(m_1) =1` unit. Mass of nucleus `(m_2)`=A units. Here`u_1 = u` and `u_2 = 0`.Therefore, the velocity of the neutron after the collision is
`v_1 = [(m_1 - m_2)/(m_1 + m_2)] u = [(1 - A)/(1 + A)] u`
KE of neutron after collision = `1/2 m_1 v_1^2 = 1/2 xx 1 xx [(1 - A)/(1 + A)]^2 u^2`
KE of neutron before collision is `= 1/2 m u^2 = 1/2 xx 1 xx u^2 = 1/2 u^2 , ` the ratio is `[(1 - A)/(1 + A)]^2`