In a head-on one dimensional inelastic collision, the velocities of two balls of masses $m_{1} {and m}_{2}$ are $+ u_{1} a n d + u_{2}$ before collision and $+ v_{1} a n d + v_{2}$ after collision respectively. It is observed that the loss of kinetic energy of the first ball during the collision is the same as the loss of kinetic energy of the second ball. Which of the following is correct relation?

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$u_{1} + u_{2} - v_{1} - v_{2} = 0$

$u_{1} + u_{2} + v_{1} + v_{2} = 0$

$u_{1} + u_{2} + v_{1} - v_{2} = 0$

$u_{1} + u_{2} - v_{1} + v_{2} = 0$

answer is B.

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Detailed Solution

As each loses same amount of KE,

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

And as momentum is conserved

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

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

Using (1) and (2).

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

$\Rightarrow u_{1} + u_{2} + v_{1} + v_{2} = 0$

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