Question

A block of mass $m$ is released from the top of a wedge of mass $M$ as shown in the figure. Find the displacement of the wedge on the horizontal ground, when the block reaches the bottom of the wedge.

Neglect friction everywhere.

• A. $\frac{mh\tan \theta }{M+m}$
• B. $\frac{mh\tan \theta }{M-m}$
• C. $\frac{mh\cot \theta }{M+m}$
• D. $\frac{mh\cot \theta }{M-m}$

Answer 1

The correct option is C $\frac{mh\cot \theta }{M+m}$


Here, the system is wedge $+$ block.

Net force on the system in horizontal direction $(i.ex$-direction$)$ is zero, therefore, the centre of mass of the system will not move in $x-$direction.

We can apply,

$x_R{m}_R=x_L{m}_L...\left(i\right)$

Let,

$x_L=$ displacement of wedge towards left $=x$

$m_L=$ mass of wedge moving towards left $=M$

$x_R=$ displacement of block with respect to ground towards right $=(h\cot \theta -x)$

and,

$m_R=$ mass of block moving towards right $=m$

Substituting all these values in equation $\left(i\right)$, we get,

$m(h\cot \theta -x)=xM$

$\Rightarrow x=\frac{mh\cot \theta }{M+m}$

Hence, $\left(C\right)$ is the correct option.