Question

Two balls $A$ and $B$ of masses $2\text{kg}$ each are moving with speeds $1\text{m/s}$ and $2\text{m/s}$ on a frictionless surface. After colliding, the ball $A$ returns back with speed $0.5\text{m/s}$, then the impulse of reformation of ball $B$ is

• A. $-2\text{Ns}$
• B. $0\text{Ns}$
• C. $4\text{Ns}$
• D. $-4\text{Ns}$

Answer 1

The correct option is B $0\text{Ns}$

Given,
$m_A=2\text{kg};m_B=2\text{kg}$
$u_A=1\text{m/s};u_B=-2\text{m/s}$
$v_A=-0.5\text{m/s}$
(taking rightward +ve)
Applying $PCLM$,
$m_A{u}_A+m_B{u}_B=m_A{v}_A+m_B{v}_B$
$2\left(1\right)+2(-2)=2(-0.5)+2\left(v_B\right)$
$\Rightarrow v_B=-0.5\text{m/s}$

Common velocity of balls during collision, $v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}=\frac{2\left(1\right)+2(-2)}{4}$
$=-\frac{1}{2}\text{m/s}$

Impulse of reformation of ball $B$
$J_r=m_B(v_B-v)$
$=2(-\frac{1}{2}+\frac{1}{2})$
= $0\text{Ns}$

This is because the balls $A$ and $B$ after collision move with the same velocity i.e perfectly inelastic collision.