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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
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B
(Mm+M)l
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C
(mM+m)l
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D
(M+mM)l
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Solution

The correct option is C (mM+m)l
Initially 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(lX)
X=mlm+M

Physics

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