Question

Three balls A, B and C each of mass $m$ and same size, are placed along the same line on a smooth horizontal surface. A is given a velocity $u$ towards B as shown. If coefficient of restitution for the collision between A and B is $\frac{1}{2}$ and between B and C is $\frac{1}{3}$, choose the correct options: [Assume all the collisions to be head-on]

• A. Final velocity of A is $\frac{u}{4}$.
• B. Final velocity of B is $\frac{u}{4}$.
• C. Final velocity of C is $\frac{u}{2}$.
• D. Total energy loss is $\frac{5}{16}m{u}^2$.

Answer 1

Answer: (A), (B), (C), (D)

Total energy loss is $\frac{5}{16}m{u}^2$.

Just before collision between A and B:


Applying conservation of linear momentum,
$mu=m{v}_2$+ $m{v}_1$
and $v_2$ – $v_1=eu$
where $e\to$ coefficient of restitution
$\Rightarrow v_2=\frac{u}{2}(1+e)$
& $v_1=\frac{u}{2}(1-e)$

For collision between A and B:
$v_A=\frac{u}{2}(1-\frac{1}{2})=\frac{u}{4}$
$v_B=\frac{u}{2}(1+\frac{1}{2})=\frac{3u}{4}$

For collision between B and C:
$v_C=\frac{1}{2}\left(\frac{3u}{4}\right)(1+\frac{1}{3})=\frac{u}{2}$
& $v_B^{\prime }=\frac{1}{2}\left(\frac{3u}{4}\right)(1-\frac{1}{3})=\frac{u}{4}$

Total number of collisions $=2$

Loss in kinetic energy in each collision is given by
$\Delta KE=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}{u}_1^2(1-e^2)$

Therefore, total loss in K.E
$=\frac{1}{2}\left(\frac{m}{2}\right)u^2(1-\frac{1}{4})+\frac{1}{2}\frac{m}{2}{\left(\frac{3u}{4}\right)}^2(1-\frac{1}{9})$

$=\frac{1}{2}m{u}^2\left(\frac{3}{4}\right)+\frac{1}{4}m{u}^2\left(\frac{9}{16}\right)\left(\frac{8}{9}\right)$

$=\frac{3m{u}^2}{16}+\frac{1}{8}m{u}^2=\frac{5}{16}m{u}^2$

Hence, all options are correct.