Question

A block of mass $m=0.5\text{kg}$ is released from the top of a wedge of mass $M=9.5\text{kg}$ as shown in the figure. The inclination angle of the wedge is $\theta ={45}^{\circ }$ and its height is $h=2\text{m}$. The displacement of the wedge on the horizontal ground, when the block reaches the bottom of the wedge, is (Neglect friction everywhere.)

Answer 1

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.

Thus, we can apply,

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

Where,
$x_L=$ 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,

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

$\Rightarrow x=\frac{mh\cot \theta }{M+m}=\frac{0.5\times 2\times \cot {45}^{\circ }}{9.5+0.5}=0.1\text{m}$