A small ball of mass m is placed on top of the “super ball” of mass M and the two ball are dropped to the floor from height h. Height reached by small ball after the collision is Nh, then ‘N’ is: (Assume that the collision with the super ball are elastic, and the m<<M)

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answer is 9.

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Detailed Solution

If initial velocities are: $u_{1} = \sqrt{2 g h} (↓), u_{2} = \sqrt{2 g h} (↑)$
By using conservation of momentum and equation of e & m <<M.
Then final velocities: $v_{1} = 3 \sqrt{2 g h}, v_{2} = \sqrt{2 g h}$

So height reached by m will be 9h.

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