Q: A neutron travelling with a velocity v and K.E. ‘E’ collides perfectly elastically head on with nucleus of an atom of mass number A at rest. The fraction of total energy retained by the neutron is :

(a) $ \displaystyle (\frac{A-1}{A+1})^2 $

(b) $ \displaystyle (\frac{A-1}{A})^2 $

(c) $\displaystyle (\frac{A+1}{A-1})^2 $

(d) $\displaystyle (\frac{A-1}{A})^2 $

Ans: (a)

velocity of body after collision is

$ \displaystyle v_1 = \frac{(m_1-m_2)u_1}{m_1 +m_2} $

$\displaystyle v_1 = \frac{(1-A)u_1}{1 + A} $

Fraction of K.E retained is

$ \displaystyle \frac{1/2 m v_1^2}{1/2 m u_1^2} = (\frac{v_1}{u_1})^2 $

$ \displaystyle = (\frac{(1-A)}{1 + A})^2 $