Collision and the phase space - 2

Conservation of momentum tells us that $A$ has velocity $\tfrac12v$ , while $B$ has velocity $\tfrac{M}{2m}v$ , after the collision. Since energy is not created, the speed of separation cannot exceed the speed of approach of the two particles, and hence $\tfrac{M}{2m}v - \tfrac{1}{2}v\; \le \; v$ and hence $\tfrac{M}{m} \le 3$ . Thus $\boxed{2}$ of the given options are possible.