Question

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

• 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)

The correct options are
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 $\frac{5}{16}m{u}^2$

Applying conservation of linear momentum in general for collision between a body at rest and other one moving with initial velocity $u$,
$mu=m{v}_2+m{v}_1$
Using coefficient of restitution
$v_2-v_1=eu$
$\begin{array}{}⟹v_2=\frac{u}{2}(1+e)& \text{(1)}\end{array}$
$\begin{array}{}{v}_1=\frac{u}{2}(1-e)& \text{(2)}\end{array}$

Final velocity of B after collision with A using equation $\left(1\right)$ will be,$v_B=\frac{u}{2}(1+\frac{1}{2})=\frac{3u}{4}$
Final velocity of C after collision with B using equation $\left(1\right)$ will be, $v_C=\frac{1}{2}\left(\frac{3u}{4}\right)(1+\frac{1}{3})=\frac{u}{2}$ Hence option 3 is correct.
Final velocity of A after collision with B using equation $\left(2\right)$ will be, $v_A=\frac{u}{2}(1-\frac{1}{2})=\frac{u}{4}$ Hence option 1 is correct.
Final velocity of B after collision with C using equation $\left(2\right)$ will be, $v_B^{\prime }=\frac{1}{2}\left(\frac{3u}{4}\right)(1-\frac{1}{3})=\frac{u}{4}$ Hence option 2 is correct.
Total numbers of collisions = 2
Energy loss $\Delta E$ for collision between A and B
$\Delta E_1=\frac{1}{2}m{u}_A^2-\frac{1}{2}m[v_A^2+v_B^2]$
$\Delta E_1=\frac{1}{2}m[u^2-(\frac{u}{4}{)}^2-\left(\frac{3u}{4}{)}^2\right]=\frac{6}{32}m{u}^2$
Energy loss $\Delta E$ for collision between B and C
$\Delta E_2=\frac{1}{2}m{u}_B^2-\frac{1}{2}m[v_A^2+v_B^2]$
$\Delta E_2=\frac{1}{2}m\left[\right(\frac{3u}{4}{)}^2-(\frac{u}{2}{)}^2-(\frac{u}{4}{)}^2]=\frac{4}{32}m{u}^2$
Total energy loss,
$\Delta E=\Delta E_1+\Delta E_2=\frac{5}{16}m{u}^2$
Hence option 4 is also correct