Question

A ball of mass $3kg$ moving with a velocity of $3m/s$ collides with another ball of mass $1kg$ 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, $m_1=3$, $u_1=3$,
$m_2=1$, $u_2=-u$
$v_1=0$, $e=\frac{2}{7}$
$\therefore 3\cdot 3-1\cdot u=3\cdot 0+1\cdot v_2$
$\Rightarrow 9-u=v_2$
$\Rightarrow v_2+u=9\dots \left(i\right)$
and $v_2-0=\frac{2}{7}\left(3+u\right)$
$\Rightarrow v_2=\frac{2}{7}\left(3+u\right)\dots \left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get
$9=u+\frac{2}{7}(3+u)$
$\Rightarrow 9-\frac{6}{7}=u+\frac{2}{7}u$
$\Rightarrow \frac{9}{7}u=\frac{57}{7}$
$\Rightarrow u=\frac{57}{9}$
$\Rightarrow u=\frac{19}{3}$