Two identical balls A and B are kept on a smooth table as shown. B collides with A with speed v. For different conditions mentioned in List I, match with speed of A after collision given in List II.

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Solution

In one dimensional elastic collision between two particles,
${v_{1}}^{'} = (\frac{m_{1} - e m_{2}}{m_{1} + m_{2}}) v_{1} + (\frac{m_{2} + e m_{2}}{m_{1} + m_{2}}) v_{2}$ .....(1)
${v_{2}}^{'} = (\frac{m_{2} - e m_{1}}{m_{1} + m_{2}}) v_{1} + (\frac{m_{1} + e m_{1}}{m_{1} + m_{2}}) v_{2}$ .....(2)
$m_{1} = m_{2} = m$ ;
$v_{1} = v$ ;
$v_{2} = 0$ ;
Substituting the values $m_{1}$ , $m_{2}$ , $v_{1}$ and $v_{2}$ in eqns (1) and (2)
${v_{1}}^{'} = (\frac{1 - e}{2}) v_{1}$ ; ....(3)
${v_{2}}^{'} = (\frac{1 + e}{2}) v_{1}$ ; .....(4)
Elastic collision: velocities get exchanged, hence A will move with velocity v, B will come to rest.
Perfectly inelastic collision: A & B will stick together. Since both the balls have the same mass, velocity of the combined object will be v/2.
Inelastic collision with e=1/2:
Substituting $e = \frac{1}{2}$ in eqn (4)
${v_{2}}^{'} = (\frac{3}{4}) v$ ;
Inelastic collision with e=1/4:
Substituting $e = \frac{1}{4}$ in eqn (4)
${v_{2}}^{'} = (\frac{5}{8}) v$ ;

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