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

Question

Infinity Learn · Accepted Answer

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)

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

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