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A ball of mass $m_{1}$ , collides elastically and head on with ball of mass $m_{2}$ at rest. Then

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If $m_{1} > >$ $m_{2}, v e l o c i t y o f t h e s e c o n d b a l l i s 4 u_{1}$

The transfer of kinetic energy to the second ball is maximum when $m_{1} = m_{2}$

The velocity of the second ball is maximum, when $m_{1} ≫ m_{2} .$

The change of momentum of first ball is maximum, when $m_{1} ≪ m_{2}$

answer is A, B, C.

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Detailed Solution

Velocity of first ball after the collision is

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

If $m_{1} = m_{2}$ , then $v_{2} = u_{1} a n d v_{1} = u_{2}$

In this case whole of the $K . E .$ is transferred to the second ball after the collision.

If $m_{1} ≪ m_{2}; v_{1} = - u_{1}$ and $v_{2} = 0$

i.e., momentum of first ball is reversed, thus for maximum change in the momentum of the first ball.

If $m_{1} ≫ m_{2}$ and $v_{2} = 2 u_{1}$

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