Two balls $A a n d B$ having masses $m k g a n d 2 m k g$ , moving with speeds $21 m / s a n d 4 m / s$ respectively in opposite directions, collide head-on. After collision, $A$ moves with a speed of $1 m / s$ in the same direction. Then :

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the loss of kinetic energy due to collision is $200 m J$

the velocity of $B$ after collision is $6 m / s$ opposite to the direction before collision

the impulse of the force between the two balls is $20 m N s .$

the coefficient of restitution is $0.2$

answer is A, B, D.

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Detailed Solution

$m \times 21 - 2 m \times 4 = m \times 1 + 2 m \times v_{2} ∴ v_{2} = 6 m / s$

$e = - [\frac{v_{2} - v_{1}}{u_{2} - u_{1}}] = - [\frac{6 - 1}{- 4 - 21}] = 0.2 J = m (v_{1} - u_{1}) = m [21 - 1] = 20 m N - s .$

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