Question

Three identical blocks A, B, C are placed on horizontal frictionless surface. The blocks B and C are at rest. But A is approaching towards B with a speed $10m{s}^{-1}$. The coefficient of restitution for all collisions is $0.5$. The speed of the block C just after collision is:

• A. $5.6m{s}^{-1}$
• B. $6m{s}^{-1}$
• C. $8m{s}^{-1}$
• D. $10m{s}^{-1}$

Answer 1

The correct option is A $5.6m{s}^{-1}$

Here the collision is not perfectly elastic so, there is some loss in speed

$ma=mb=mc=mAcc$

To linear momentum,

$\begin{array}{c}UA=VA+\frac{VB}{10}\\ =VA+VB.......\left(1\right)\end{array}$

Now $e=VB-\frac{VA}{UA}05UA=VB-VA........\left(2\right)$

from(1) &(2)

$\begin{array}{c}2VB=15VB\\ =\frac{15}{2}\end{array}$

Now collision between B &C $\begin{array}{c}UB=VB+VC.\frac{15}{2}\\ =VB+VC..........\left(3\right)\end{array}$

or $\begin{array}{c}e=\frac{VC-VB}{0.5UB}0.5UB\\ =VC-VB.........\left(4\right)\end{array}$

from (3) & (4)

$VC=\frac{15}{2}+0.5\times \frac{15}{2}$

on solving equation

$\begin{array}{c}UVC=\frac{225}{40}VC\\ =5.6m/s\end{array}$

So the option $A$ is the correct answer