JEE Advanced 2017 Paper 1

Net external force acting on the system along the x-axis is zero.∴Along the x-axis Momentum is conserved$\Rightarrow mv=MV$From conservation of energyLoss in GPE of particle of mass m = Gain in kinetic energy of both the masses $\Rightarrow mgh=\frac{1}{2}mv^2+\frac{1}{2}MV^2\Rightarrow mgh$ $=\frac{1}{2}mv^2+\frac{1}{2}M\left(\frac{mv}{M}\right)^2$$\Rightarrow 2gh=\left(1+\frac{m}{M}\right)v^2\Rightarrow v$ $=\sqrt{\frac{2gh}{1+\frac{m}{M}}}$ hence option (A) is correct.Hence option [B] is incorrect.Since the location of center of mass does not change along the x-axis$m\Delta x_{m/G}+M\Delta x_{M/G}=0$ $\Rightarrow m(R-x)-Mx=0f$$x=\frac{mR}{m+M}.$$\Delta x_{M/G}=\frac{-mR}{m+R}$Hence option (D) is correct$\Delta x_{m/G}=R-x=\frac{MR}{M+m}$Final position of $m=-x-\frac{mR}{M+m}$Hence option (C) is incorrect.