Question

A wedge of mass M and a cube of mass m are
shown in figure. The system is released.
Considering no frictional force between any two
surfaces, the distance moved by the wedge,
when the cube just reaches on the ground is​

Answer 1

Answer:

The distance move by the wedge is $\frac{2m}{M+m}$ m

Explanation:

The concept used is that the horizontal displacement of the centre of mass will be zero

Let the wedge moves a distance x in the +x direction

Therefore, the displacement of the com of wedge

$d_{W/E}=x$

If w.r.t. the wedge, the mass m moves a distance x' then

Displacement of block w.r.t. the wedge = -x' (since it will move in the opposite direction to that of the wedge)

$d_{B/W}=-x'$

But w.r.t. the wedge (i.e. is we assume the wedge to be in rest), the horizontal distance covered by the block is 2m

Therefore, x' = 2m

Thus

$d_{B/W}=-2$

Displacement of the block w.r.t. the earth

$d_{B/E}=d_{B/W}+d_{W/E}$

$\implies d_{B/E}=-2+x$

Now the shift of the centre of mass will be zero

Hence using

$X_{cm}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$\implies 0=\frac{m\times (x-2)+M\times x}{m+M}$

$\implies (M+m)x=2m$

$\implies x=\frac{2m}{M+m}$

Therefore, the distance move by the wedge is $\frac{2m}{M+m}$ m

Hope this helps.