Question

A ball of mass $ 3 \,kg $ moving with a velocity of $ 3 \,m/s $ collides with another ball of mass $ 1 \,kg $ moving with velocity $ u $ in the opposite direction. If the first ball comes to rest after the impact and $ e=\frac{2}{7} $ , then $ u $ is in $ m/s $ , is

• A. $ \frac{13}{3} $
• B. $ \frac{17}{3} $
• C. $ \frac{19}{3} $
• D. $ \frac{23}{3} $

Answer 1

Answer: (C)

Law of conservation
$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$
and $v_{2}-v_{1}=e\left(u_{1}-u_{2}\right)$
Here, $m_{1}=3$, $u_{1}=3$,
$m_{2}=1$, $u_{2}=-u$
$v_{1}=0$, $e=\frac{2}{7}$
$\therefore 3\cdot3-1\cdot u=3\cdot0+1\cdot v_{2}$
$\Rightarrow 9-u=v_{2}$
$\Rightarrow v_{2}+u=9\quad\ldots\left(i\right)$
and $v_{2}-0=\frac{2}{7}\left(3+u\right)$
$\Rightarrow v_{2}=\frac{2}{7}\left(3+u\right)\quad\ldots\left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$ 9 =u+\frac{2}{7}(3+u) $
$ \Rightarrow \, 9-\frac{6}{7} =u+\frac{2}{7} u $
$ \Rightarrow \, \frac{9}{7} \,u =\frac{57}{7} $
$ \Rightarrow \, u =\frac{57}{9} $
$ \Rightarrow \, u =\frac{19}{3} $