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Q.
A neutron makes a head on elastic collision with a stationary deuteron. The fractional energy loss of the neutron in the collision is :-
NTA AbhyasNTA Abhyas 2020
Solution:
$u=v_{1}+2v_{2}...\left(1\right)$
$e=1=\frac{- \left(v_{1} - v_{2}\right)}{u}$
$\therefore u=v_{2}-v_{1}...\left(2\right)$
solving $v_{1}=\frac{- u}{3},v_{2}=\frac{2 u}{3}$
fraction energy loss of neutron
$\frac{\frac{1}{2} m u^{2} - \frac{1}{2} m \left(- \frac{u}{3}\right)^{2}}{\frac{1}{2} m u^{2}}=\frac{8}{9}$
