CM Punk

The force that act on the block of mass $m$ is the weight force $W_m$ ; the normal component of weight force $W_{mn}=mg\cos\theta$ is equilibrate by the reaction of the vinculus (ramp), hence the nect force directed towards the ramp is: $W_{mp}=mg\sin\theta;$ If $(O,x,y)$ is a system of cartesian coordinates with $x$ towards the floor, $y$ perpendicular at the floor and $r$ the direction parallel to the ramp of the wedge, the force acting on the block is by 3rd law of dynamics the orizzontal component of the reaction (the vertical component is blocked by the vinculus of the floor) is: $F_x=mg\sin\theta\cos\theta;\,(1)$ The acceleration of the wedge is: $a_x=\frac{F_x}{M+m};\,(2)$ The kinematic law of the motion with constant acceleration $a_x$ for the wedge leads to: $x=\frac 1 2 a_x t^2=\frac 1 2 \frac{m}{M+m}\sin\theta\cos\theta t^2;\,(3)$ The space covered by the block in the same time is: $S=\frac H {\sin\theta} = \frac 1 2 a_r t^2;\,a_r =\frac{mg\sin\theta} m = g\sin\theta;\,(4)$ $\frac H {\sin\theta}=\frac 1 2 g\sin\theta t^2;\,(5)$ Equating $t^2$ from $(3)$ and $(5)$ , we get: $t^2=2\frac{M+m}{mg\sin\theta\cos\theta}x=2\frac H{g{\sin}^2\theta};$ $x=\frac m {M+m} H \frac{\cos\theta}{\sin\theta};$ $x=\frac m {M+m} H \cot\theta$

Considering the block and the wedge as my system .. since there is no external force the position of center of mass should not change. ... so as the block reaches the end it has moved hcot (theta) with respect to wedge ..... assuming wedge moved x..... movement of block with respect to ground should be equal to that of wege ... since center of mass does not change.... mH(HCot (theta)-x)=Mx

we will consider movement in only x axis taking tan $\theta$ = $\frac{H}{base}$ we get base as H $\cot\theta$ we are concerned with the base is the distance traveled by small block in the x axis with respect to the wedge so the actual distance traveled by the block will be $\boxed{\text{distance traveled by block with respect wedge + distance traveled by wedge}}$ as we took the distance traveled by wedge as X but wedge moves in the opposite direction to block so apparent distance = H $\cot\theta$ - X now come to c\point of center of mass since there is no force for my system in x axis the position of center of mass does not change F.Y.I we actually don't need to know the actual position of the center of mass just assume it to be at origin {where it may be} but as even after block reaches bottom it is remains at origin so m(h $\cot\theta$ - X )=M X

we have to consider the movements in x axis only .... the block did come down via the hypotenuse but the displacement( of block with respect to wedge ) in the x axis is THE BASE. the actual distance traveled by the block can be understood as the displacement the block was supposed to be moving ( if the wedge was stationary ) - displacement by wedge ( we are subtracting it as the movement of wedge decreases the supposed displacement of the block ) .. the displacement of block gets X less ; than it was supposed to be ( if the wedge was stationary ) i.e it would have moved H $\cot\theta$ but was pulled back by X (cause wedge moved) apparent distance = H $\cot\theta$ - X

Any Solution Any one !!!!!!!!!!!!!!