Center of Mass, Momentum & Collision

JEE Main level, single correct, medium

90 of 130

Question 90 of 130

As per the given figure, a small ball PP slides down the quadrant of a circle and hits the other ball QQ of equal mass which is initially at rest. Neglecting the effect of friction and assume the collision to be elastic, the velocity of ball QQ after collision will be : (g=10ms−2)(g=10m{s}^{-2})

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Answer: (C) 2ms−12m{s}^{-1}

Let vp{v}_{p} be the velocity of ball PP just before collision. Therefore, applying conservation of energy for PP (between the moment it is released and the moment just before collision),
mgl+0=12mvp2mgl+0=\dfrac{1}{2}m{{v}_{p}}^{2}
⇒vp=2gl\Rightarrow {v}_{p}=\sqrt{2gl}
=2×10×0.2=\sqrt{2\times 10\times 0.2}
=2ms−1=2m{s}^{-1}
Since both balls have equal masses and the collision is perfectly elastic, the velocities of both the balls will get interchanged.
Therefore, the velocity of ball QQ just after the collision is 2ms−12m{s}^{-1}.

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