Question

A block released from the state of rest from a wedge as shown. The block and the wedge are of the same mass. If all surfaces are smooth, the total horizontal displacement of the block with respect to earth before it falls on the earth is given by $xR$. Then $x=$
Take $\sqrt{2}=1.41$

Answer 1

Since, x-coordinate of center of mass remains same.
$x_i=\frac{m\left(a\right)+m\left(a_1\right)}{2m}=\frac{a+a_1}{2}$
Let $x$ is the displacement of wedge when block reaches the bottom.
$x_f=\frac{m(a-x)+m(a_1+R-x)}{2m}=\frac{a+a_1+R-2x}{2}$

$\therefore x_i=x_{f}i.e.x=\frac{R}{2}=0.5R$
After that particle will follow the parabolic path
$R=ut+\frac{1}{2}g{t}^2$
where,$u=0m/s$
so,
$R=\frac{1}{2}g{t}^2$
$\therefore t=\sqrt{\frac{2R}{g}}$

Let $x^{\prime }$ is the displacement of the block when it reaches the surface and $V_m$ is the initial velocity of the block.
$x^{\prime }=vt=\sqrt{\frac{2R}{g}}.V_m$

From momentum conservation
$m{V}_m=m{V}_M$ (where $V_m$ and $V_M$ are velocity's of the block and the wedge when the block just leaves the wedge)
Hence $V_m=V_M$

Using conservation of energy:
$mgR=\frac{1}{2}m{V}_m^2+\frac{1}{2}m{V}_m^2$
$mgR=m{V}_m^2$
$V_m=\sqrt{gR}$
Using eq. (1)
$x^{\prime }=\sqrt{\frac{2R}{g}}\times \sqrt{gR}$
$\therefore x^{\prime }=\sqrt{2}R$
$\therefore \text{Total displacement}=x+x^{\prime }=0.5\text{R}+\sqrt{2}\text{R}$
$=(0.5+1.41)R=1.91R$