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Conservation of Momentum

A particle of...

Question

A particle of mass $m_{1}$ moves with speed $u_{1}$ and collide head on with a stationary particle of mass $m_{2}$ . After the collision, the velocities of the particles are $v_{1}$ and $v_{2}$ , and $e$ is coefficient of restitution for the collision. If $v_{1}$ is to be positive, i.e., for the first particle to continue moving in same direction then:

A

\frac{m_{2}}{m_{1}} > e

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B

\frac{m_{1}}{m_{2}} > e

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C

\frac{m_{1} + m_{2}}{m_{2}} > e

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D

\frac{m_{1} - m_{2}}{m_{2}} > e

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Solution

The correct option is B $\frac{m_{1}}{m_{2}} > e$
Draw a labelled diagram.

Diagram

Find the ratio of masses.
Formula used:
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$e = \frac{relative velocity of separation}{relative velocity of approach}$

Applying momentum conservation,
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} \dots (i)$

Apply coefficient of restitution
$e = \frac{- (v_{1} - v_{2})}{u_{1} - u_{2}}$

$\Rightarrow e = \frac{- (v_{1} - v_{2})}{u_{1}}$

$∵ v_{2} - v_{1} = e u_{1} \dots (i i)$

From $(i)$ & $(i i)$ ,

$v_{1} = v_{2} - e u_{1}$

$m_{1} u_{1} = m_{1} (v_{2} - e u_{1}) + m_{2} v_{2}$

$m_{1} u_{1} = m_{1} v_{2} - m_{1} e u_{1} + m_{2} v_{2}$

$m_{1} u_{1} + m_{1} e u_{1} = m_{1} v_{2} + m_{2} v_{2}$

$v_{2} = \frac{m_{1} u_{1} (1 + e)}{m_{1} + m_{2}}$

and
$v_{1} = \frac{(m_{1} - e m_{2}) u_{1}}{m_{1} - m_{2}}$

If $v_{1}$ is to be positive,

$∵ v_{1} > 0$

$m_{1} - e m_{2} > 0$

$\frac{m_{1}}{m_{2}} > e$
Final answer: (b)

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