Question

Two balls $X(2 kg )$ and $Y(4 kg )$ approach each equal speeds of $10\, m \cdot s ^{-1}$. If the collision is elastic, the new velocities of $X$ and $Y$ balls respectively

• A. $\frac{50}{3} m \cdot s^{-1}, \frac{-10}{3} m \cdot s^{-1}$
• B. $\frac{-50}{3} m \cdot s^{-1}, \frac{-10}{3} m \cdot s^{-1}$
• C. $\frac{-50}{3} m \cdot s^{-1}, \frac{10}{3} m \cdot s^{-1}$
• D. $\frac{50}{3} m \cdot s^{-1}, \frac{10}{3} m \cdot s^{-1}$

Answer 1

Answer: (C)

Given, $m_{1}=2\, kg , m_{2}=4\, kg$
$v_{1}=v_{2}=10\, ms ^{-1}$
In perfectly elastic collision, momentum and kinetic energy is conserved.
$\therefore $ By conservation of momentum,
$m_{1} \times 10+m_{2} \times(-10)=m_{1} v_{1 f}+m_{2} v_{2 f}$
$\Rightarrow 2 \times 10+4 \times(-10)=2 v_{1 f}+4 v_{2 f}$
$\Rightarrow -20=2\left(v_{1 f}+2 v_{2 f}\right)$
$\Rightarrow v_{1 f}+2 v_{2 f}=-10 ...(i)$
In perfectly elastic collision,
Velocity of separation = Velocity of approach
$\Rightarrow v_{2 f}-v_{1 f}=10-(-10)$
$\Rightarrow v_{2 f}-v_{1 f}=20 ...(ii)$
Adding Eqs. (i) and (ii), we get
$3 v_{2 f}=10$
$v_{2 f}=\frac{10}{3} m s^{-1}$
From Eqs. (ii), we get
$\frac{10}{3}-v_{1 f}=20$
$\Rightarrow -v_{1 f}=20-\frac{10}{3}=\frac{50}{3}$
$\Rightarrow v_{1 f}=\frac{-50}{3} m s^{-1}$