Two masses $2 m$ and $m$ are connected by an inextensible light string. the string is passing over a light frictionless pulley. The mass $2 m$ is resting on a surface and mass $m$ is hanging in air as shown in figure. A particle of mass $m$ strikes the mass $m$ from below in case (I) with a velocity $v_{0}$ and in case (II) strikes mass $m$ with a velocity $v_{0}$ from top and sticks to it.

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The conservation of linear momentum can be applied in both the cases just before and just after collision

The conservation of linear momentum can be applied in case I but cannot be applied in case II just before and just after collision

The ratio of velocities of mass $m$ just after collision in first and second cases is 1/2

The ratio of velocities of mass $m$ just after collision in first and second case is 2

answer is B, D.

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Detailed Solution

In the first case: $(m + m) v_{1} = m v_{0} \Rightarrow v_{1} = \frac{v_{0}}{2}$

In the second case: $- \int T d t = (m + m) v - m v_{0}, \int T d t = 2 m v$

From Eqs. (i) and (ii), $4 m v = m v_{0} \Rightarrow v = \frac{v_{0}}{4}$ and required ratio = 2

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