Question

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

• A. $\text{5.6}ms^{- 1}$
• B. $6 \, ms^{- 1}$
• C. $8 \, ms^{- 1}$
• D. $10 \, ms^{- 1}$

Answer 1

Answer: (A)

For collision between blocks $A$ and $B$ ,
$e=\frac{v_{B} - v_{A}}{u_{A} - u_{B}}=\frac{v_{B} - v_{A}}{10 - 0}=\frac{v_{B} - v_{A}}{10}$
$\therefore v_{B}-v_{A}=10e=10\times 0.5=5\ldots .\left(\right.i\left.\right)$
From principle of momentum conservation,
$m_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}$
$m\times 10+0=mv_{A}+mv_{B}$
$\therefore v_{A}+v_{B}=10 \, \ldots .\left(\right.ii\left.\right)$
Adding eqs. (i) and (ii), we get
$v_{B}=\text{7.5} \, ms^{- 1} \, \ldots \left(\right.iii\left.\right)$
Similarly for collision between $B$ and $C$ ,
$v_{C}-v_{B}=7.5e=7.5\times 0.5=3.75$
$\therefore v_{C}-v_{B}=3.75ms^{- 1}$ ...(iv)
Adding Eqs. (iii) and (iv) we get
$2v_{C}=\text{11.25}$
$\therefore v_{C}=\frac{11 .25}{2}=\text{5.6}ms^{- 1}$