Question

A particle of mass $m_{1}$ experienced a perfectly elastic collision with a stationary particle of mass $m_{2}$ . What fraction of the kinetic energy does the striking particle lose, if

$(a)$ it recoils at right angles to its original motion direction,

$(b)$ the collision is a head-on one?

Accepted Answer

$(a)$ Assuming that the initial velocities of $m_{1}$ and $m_{2}$ are $\vec{u_{1}}, \vec{u_{2}}$ , and the final velocities are $\vec{v_{1}}, \vec{v_{2}}$ respectively. Since during the collision there is no external force on the system, then using the law of momentum conservation along with horizontal and vertical directions.

For horizontal direction,

$m_{1} u_{1} = m_{2} v_{2} \cos θ . . . . . . (1)$

For vertical direction,

and $m_{1} v_{1} = m_{2} v_{2} \sin θ . . . . . (2)$

Doing $(equation 1)^{2} + (equation 2)^{2}$ we get that,

$m_{2}^{2} v_{2}^{2} = m_{1}^{2} (u_{1}^{2} + v_{1}^{2})$ .

As we know, kinetic energy is always conserved during the elastic collision, hence,

initial kinetic energy $=$ final kinetic energy.

$\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

$\Rightarrow \frac{1}{2} (m_{1} u_{1}^{2} - m_{1} v_{1}^{2}) = \frac{1}{2} m_{2} v_{2}^{2} . . . . . . . (3)$ since ( $u_{2}$ $= 0$ )

Or $m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} v_{2}^{2}$

$m_{2} (u_{1}^{2} - v_{1}^{2}) = m_{1} (u_{1}^{2} + v_{1}^{2}) \Rightarrow m_{2} u_{1}^{2} - m_{2} v_{1}^{2} = m_{1} u_{1}^{2} + m_{1} v_{1}^{2} \Rightarrow m_{2} u_{1}^{2} - m_{1} u_{1}^{2} = m_{1} v_{1}^{2} + m_{2} v_{1}^{2} \Rightarrow (m_{2} - m_{1}) u_{1}^{2} = (m_{1} + m_{2}) v_{1}^{2}$ $[$ $Using (3)$ $]$

Or ${(\frac{v_{1}}{u_{1}})}^{2} = \frac{m_{2} - m_{1}}{m_{1} + m_{2}} . . . . . . (4)$

So, the fraction of kinetic energy lost by the particle of mass $m_{1}$ ,

$= \frac{\frac{1}{2} m_{1} u_{1}^{2} - \frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{1} u_{1}^{2}} = 1 - \frac{v_{1}^{2}}{u_{1}^{2}}$

$= 1 - \frac{m_{2} - m_{1}}{m_{1} + m_{2}} = \frac{2 m_{1}}{m_{1} + m_{2}} . . . . . (5)$

$(b)$ When the collision occurs head-on using the law of linear momentum conservation, the equation for masses,

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

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

And from the law of conservation of kinetic energy, plugging the value of $v_{2}$ ,

$\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} {[\frac{m_{1} {(u_{1} - v_{1})}^{2}}{m_{2}}]}^{2}$ .

Simplifying it more,

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

or $\frac{v_{1}}{u_{1}} = \frac{(\frac{m_{1}}{m_{2}} - 1)}{(1 + \frac{m_{1}}{m_{2}})}$ $= \frac{m_{1} - m_{2}}{m_{1} + m_{2}} . . . . . . (6)$

In this particular case, the fraction of kinetic energy lost by mass $m_{1}$ ,

$= 1 - \frac{v_{1}^{2}}{u_{1}^{2}} = 1 - {(\frac{m_{1} - m_{2}}{m_{1} + m_{2}})}^{2} = \frac{4 m_{1} m_{2}}{{(m_{1} + m_{2})}^{2}}$ .

Embibe Experts Solutions for Chapter: Centre of Mass, Momentum and Collisions, Exercise 4: Exercise - 4

Embibe Experts Physics Solutions for Exercise - Embibe Experts Solutions for Chapter: Centre of Mass, Momentum and Collisions, Exercise 4: Exercise - 4

Questions from Embibe Experts Solutions for Chapter: Centre of Mass, Momentum and Collisions, Exercise 4: Exercise - 4 with Hints & Solutions

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