Question

Two balls $X(2kg)$ and $Y(4kg)$ approach each equal speeds of $10m\cdot s^{-1}$. 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, $m_1=2kg,m_2=4kg$
$v_1=v_2=10m{s}^{-1}$
In perfectly elastic collision, momentum and kinetic energy is conserved.
$\therefore$ By conservation of momentum,
$m_1\times 10+m_2\times (-10)=m_1{v}_{1f}+m_2{v}_{2f}$
$\Rightarrow 2\times 10+4\times (-10)=2v_{1f}+4v_{2f}$
$\Rightarrow -20=2\left(v_{1f}+2v_{2f}\right)$
$\Rightarrow v_{1f}+2v_{2f}=-\mathrm{10...}(i)$
In perfectly elastic collision,
Velocity of separation = Velocity of approach
$\Rightarrow v_{2f}-v_{1f}=10-(-10)$
$\Rightarrow v_{2f}-v_{1f}=\mathrm{20...}(ii)$
Adding Eqs. (i) and (ii), we get
$3v_{2f}=10$
$v_{2f}=\frac{10}{3}m{s}^{-1}$
From Eqs. (ii), we get
$\frac{10}{3}-v_{1f}=20$
$\Rightarrow -v_{1f}=20-\frac{10}{3}=\frac{50}{3}$
$\Rightarrow v_{1f}=\frac{-50}{3}m{s}^{-1}$