A particle of mass m is kept on a smooth cube of mass M and side L as shown in figure. Cube starts moving with a constant velocity v . Displacement of the center of mass along the horizontal direction when particle hits the ground is

Time taken by particle to hit the ground, t=t1+t2 here t1 is the time to leave the block by the particle, and t2 is the time to hit the ground after leaving the block.Using 2nd equation of motiont1=Lv,Also, h=12gt22⇒t2=2hg=2Lg∴t=Lv+2LgNow, in horizontal direction, Vcm=Mvm+MThe distance moved by the center of mass in this time,xcm=vcmt∴xcm=Mvm+MLv+2LgTherefore, the correct answer is (C)