Question

A body ${}^{\prime }{A}^{\prime }$ experiences perfectly elastic collision with a stationary body ${}^{\prime }{B}^{\prime }$. If after collision the bodies fly apart in the opposite directions with equal velocities, the mass ratio of ${}^{\prime }{A}^{\prime }$ and ${}^{\prime }{B}^{\prime }$ is

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{5}$

Answer 1

The correct option is B $\frac{1}{3}$

Let the velocity of particle A before collision $=v_1$

After collision: Velocity of each particle $=$v

Applying conservation of momentum we get:

$m_A{v}_1+0=m_{A}v-m_{B}v$(initial momentum before collision $=$ final momentum after collision)

$\Rightarrow v_1=\frac{(m_A-m_B)v}{m_A}\to \left(1\right)$

By applying conservation of kinetic energy we get:

$\frac{1}{2}{m}_A{v}_1^2=\frac{1}{2}{m}_A{v}^2+\frac{1}{2}{m}_B{v}^2$

$\Rightarrow m_A\frac{(m_A-m_B{)}^2{v}^2}{m_A^2}=m_A{v}^2+m_B{v}^2$

$\Rightarrow m_A^2+m_B^2-2m_A{m}_B=m_A^2+m_A{m}_B$

$\Rightarrow m_B^2=3m_A{m}_B$

$\Rightarrow \frac{m_A}{m_B}=\frac{1}{3}$.