Two similar balls $P$ and $Q$ having velocities of 0.5 ms $^{- 1}$ and -0.3 ms $^{- 1}$ respectively collide elastically. The velocities of $P$ and $Q$ after the collision will be respectively :

0.3 ms

^{- 1}

and 0.5 ms

^{- 1}

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-0.5 ms

^{- 1}

and 0.3 ms

^{- 1}

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0.5 ms

^{- 1}

and 0.3 ms

^{- 1}

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-0.3 ms

^{- 1}

and 0.5 ms

^{- 1}

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Solution

The correct option is D -0.3 ms $^{- 1}$ and 0.5 ms $^{- 1}$
Let, $m_{1}$ and $m_{2}$ be the masses of two balls and $u_{1}$ and $u_{2}$ be the velocities of two balls before collision. Also, $v_{1}$ and $v_{2}$ be the velocities of two balls after collision.
Here, Two similar balls P and Q collide elastically hence the momentum will be conserved
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2})$ ...................(1)
and the conservation of the total kinetic energy is given by
$\frac{1}{2} m_{1} (u_{1})^{2} + \frac{1}{2} m_{2} (u_{2})^{2} = \frac{1}{2} m_{1} (v_{1})^{2} + \frac{1}{2} m_{2} (v_{2})^{2}$
$m_{1} (u_{1})^{2} + m_{2} (u_{2})^{2} = m_{1} (v_{1})^{2} + m_{2} (v_{2})^{2}$
$m_{1} (u_{1})^{2} - m_{1} (v_{1})^{2} = m_{2} (v_{2})^{2} - m_{2} (u_{2})^{2}$
$m_{1} [(u_{1})^{2} - (v_{1})^{2}] = m_{2} [(v_{2})^{2} - (u_{2})^{2}]$
$m_{1} [(u_{1} - v_{1}) (u_{1} + v_{1})] = m_{2} [(v_{2} - u_{2}) (v_{2} + u_{2})]$ .....................(2)
Dividing (2) by (1), we get
$u_{1} + v_{1} = v_{2} + u_{2}$
$v_{1} - v_{2} = u_{2} - u_{1}$
$\Rightarrow$ the relative velocity of one ball with respect to the other is reversed by the collision. Hence, after collision the velocities of $P$ and $Q$ will be $- 0.3 m s^{- 1}$ and $0.5 m s^{- 1}$ respectively.

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