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Partially Inelastic Collision

A particle of...

Question

A particle of mass $m_{1}$ moving with velocity v in a positive direction collides elastically with a mass $m_{2}$ moving in opposite direction also at velocity v. If $m_{2}$ >> $m_{1}$ , then :

the velocity of

m_{1}

immediately after collision is nearly 3v

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the change in momentum of

m_{1}

is nearly

4 m_{1} v

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the change in kinetic energy of

m_{1}

is nearly

4 m v_{2}

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all of the above

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Solution

The correct option is D all of the above
Let the velocity of mass $m_{1}$ after the collision be $v_{1}$ in positive direction.
Given : $u_{1} = v$ $u_{u} = - v$ $e = 1$ $m_{2} >> m_{1}$

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

$v_{1} = \frac{(m_{1} - (1) m_{2}) v + (1 + 1) m_{2} (- v)}{m_{1} + m_{2}} = \frac{m_{1} v - 3 m_{2} v}{m_{1} + m_{2}}$
For $m_{2} >> m_{1}$ $⟹ v_{1} = - 3 v$ (-ve sign represents $m_{1}$ recoils)
Now change in momentum $Δ P = m_{1} v_{2} - m_{1} v$
$Δ P = m_{1} (- 3 v) - m_{1} v = - 4 m_{1} v$ $⟹ | Δ P | = 4 m_{1} v$

Change in kinetic energy of $m_{1}$ $Δ E = \frac{1}{2} m_{1} v_{1}^{2} - \frac{1}{2} m_{1} v^{2}$
$Δ E = \frac{1}{2} m_{1} (3 v)^{2} - \frac{1}{2} m_{1} v^{2} = 4 m_{1} v^{2}$

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A particle of mass m1 moving with velocity v in a positive direction collides elastically with a mass m2 moving in opposite direction also at velocity v. If m2 >> m1, then :

A particle of mass $m_{1}$ moving with velocity v in a positive direction collides elastically with a mass $m_{2}$ moving in opposite direction also at velocity v. If $m_{2}$ >> $m_{1}$ , then :