Question

A body P strikes head-on with another body Q of mass that is 'p' times that of body P and moving with a velocity that is $\frac{1}{q}$ of the velocity of body P. If the body P comes to rest, the coefficient of restitution is

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Given $m_Q=P{m}_p$ and $v_Q\frac{v_p}{q}$ from the principle of conservation of momentum, we have (since body P comes to rest after collision)
$m_p{v}_p+m_Q{v}_Q=m_{Q}v$.
Where v is velocity of body Q after collision.
Thus, $m_p{v}_p+p{m}_p\frac{v_p}{q}=p{m}_{p}v.$
Which gives $\frac{v}{v_p}=\frac{p+q}{pq}\left(i\right)$
$e=\frac{v}{v_p-v_Q}=\frac{v}{v_p-\frac{v_p}{q}}$
which gives $\frac{v}{v_p}=\frac{e}{q}(q-1)\left(ii\right)$
Equating (i) and (ii), we get $\frac{p+q}{p(q-1)}$ which is choice (d).