29. A block of mass 'M' is placed on thetop of a wedge of mass 4M'. Al

1.	29. A block of mass 'M' is placed on thetop of a wedge of mass 4M'. All thesurfaces are frictionless. The system isreleased from rest. The distance movedby the wedge at the instant, the blockreaches the bottom will be1) 0.2 m2) 0.4 m3) 0.8 m4) Zero
Answer» Answer: Option (3) 0.8 m Explanation: The horizontal DISTANCE travelled by the block of MASS M = 4 m Since there is no force in the horizontal direction Therefore, the MOMENTUM in the horizontal direction will be conserved If the velocities of block and the wedge in the horizontal direction are $v_B$ and $v_W$ then $Mv_B+4Mv_A=0$ if the distance travelled in the horizontal direction by the block and the wedge are $x_B$ and $x_W$ then $M\frac{dx_B}{dt}+4M\frac{dx_W}{dt}=0$ $\implies Mdx_B+4Mdx_W=0$ $\implies M\Delta x_B+4M\Delta x_W=0$ If the wedge travels distance x then $\Delta x_W=x$ The block has a displacement of 4 m with RESPECT to the wedge but the wedge also has a displacement of x m in the same time Therefore, displacement of block with respect to the earth = $4+x$ Thus, $M\times (4+x)+4M\times x=0$ $\implies 4+x+4x=0$ $\implies 5x=-4$ $\implies x=-0.8$ m Thus, the wedge travels a distance of 0.8 m Hope this HELPS.