Question

A block of mass m is placed at rest on the top of a smooth wedge of mass M, which in turn is placed at rest on a smooth horizontal surface as shown in figure. Then the distance moved by the wedge as the block reaches the foot of the wedge is

A
(M+mm)l
B
(Mm+M)l
C
(mM+m)l
D
(M+mM)l

Solution

The correct option is C (mM+m)lInitially both wedge and block are at rest. There is no external force in horizontal direction of the wedge block system. Thus acceleration, ⇒acom in x direction is zero. Therefore, x− coordinate of the COM of the system will remain at rest even when block slides down. Assume that wedge moves X when  block reaches the ground.  As xcom is not changing. m1x1=m2x2  ⇒MX=m(l−X) X=mlm+MPhysics

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