A ball of mass $10 k g$ is moving with a velocity of $10 m / s$ . It strikes another ball of mass $5 k g$ , which is moving in the same direction with a velocity of $4 m / s$ . If the collision is elastic, their respective velocities after the collision will be

12 m / s, 6 m / s

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12 m / s, 25 m / s

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6 m / s, 12 m / s

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8 m / s, 20 m / s

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Solution

The correct option is C $6 m / s, 12 m / s$
Let their velocities after the collision be $v_{1} a n d v_{2}$ . As we know for elastic collision, relative velocity of approach = relative velocity of separation
$10 - 4 = v_{2} - v_{1} \Rightarrow 6 = v_{2} - v_{1}$
$\Rightarrow v_{1} = v_{2} - 6$
Applying conservation of momentum,
$10 \times 10 + 5 \times 4 = 10 v_{1} + 5 v_{2}$
$120 = 10 v_{1} + 5 v_{2}$
$120 = 10 (v_{2} - 6) + 5 v_{2} = 15 v_{2} - 60$
$15 v_{2} = 180 \Rightarrow v_{2} = 12 m / s,$
$v_{1} = 6 m / s$
$\Rightarrow v_{2} = 12 m / s$

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