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 $10m{s}^{-1}$ . The coefficient of restitution for all collisions is $\text{0.5}$ . The speed of the block $C$ just after the collision is

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

Answer 1

Answer: (A)

For collision between blocks $A$ and $B$ ,
$e=\frac{v_B-v_A}{u_A-u_B}=\frac{v_B-v_A}{10-0}=\frac{v_B-v_A}{10}$
$\therefore v_B-v_A=10e=10\times 0.5=5\dots .(i)$
From principle of momentum conservation,
$m_A{u}_A+m_B{u}_B=m_A{v}_A+m_B{v}_B$
$m\times 10+0=m{v}_A+m{v}_B$
$\therefore v_A+v_B=10\dots .(ii)$
Adding eqs. (i) and (ii), we get
$v_B=\text{7.5}m{s}^{-1}\dots (iii)$
Similarly for collision between $B$ and $C$ ,
$v_C-v_B=7.5e=7.5\times 0.5=3.75$
$\therefore v_C-v_B=3.75m{s}^{-1}$ ...(iv)
Adding Eqs. (iii) and (iv) we get
$2v_C=\text{11.25}$
$\therefore v_C=\frac{11.25}{2}=\text{5.6}m{s}^{-1}$