Question

A block $A$ of mass $2\text{kg}$ moving with speed of $15\text{m/s}$, collides with a stationary block $B$ of mass $5\text{kg}$. If the coefficient of restitution is $0.6$, find the magnitude of velocity of blocks $A$ and $B$ after the collision respectively.

• A. $5.2\text{m/s}$ and $10.5\text{m/s}$
• B. $3.5\text{m/s}$ and $6.85\text{m/s}$
• C. $10.5\text{m/s}$ and $5.2\text{m/s}$
• D. $2.15\text{m/s}$ and $6.85\text{m/s}$

Answer 1

The correct option is D $2.15\text{m/s}$ and $6.85\text{m/s}$

Mass of block A, $m_A=2\text{kg}$
Mass of block B, $m_B=5\text{kg}$
Let $u_A\&u_B$ be the velocity before collision of block $A$ and block $B$.
$u_A=15\text{m/s},u_B=0\text{m/s}$
Let $v_A\&v_B$ be the velocity after collision of block $A$ and block $B$.
Applying conservation of linear momentum before and after collision.
$m_A{u}_A+m_B{u}_B=m_A{v}_A+m_B{v}_B$
$2\times 15+5\times 0=2\times v_A+5\times v_B$
$2v_A+5v_B=30$ ... (1)
Coefficient of restitution $\left(e\right)=\frac{\text{velocity of separation}}{\text{velocity of approach}}$
$0.6=\frac{v_B-v_A}{u_A-u_B}\Rightarrow 0.6=\frac{v_B-v_A}{15-0}$
$0.6\times 15=v_B-v_A$
$v_B-v_A=9$ ... (2)
Multiplying by 2 in equation (1)
$2v_B-2v_A=18$ ... (3)
Adding equation (1) and equation (3), we get
$7v_B=48$
$v_B=6.85\text{m/s}$
$v_B=6.85\text{m/s}$ put in equation (2)
$v_A=-2.15\text{m/s}$
$\therefore$ Magnitude of velocities of ball $A$ and ball $B$ after collision is $2.15\text{m/s}\&6.85\text{m/s}$
The correct option is (d).