1. The problem statement, all variables and given/known data A wedge of mass m and angle θ is placed on a smooth ground. As shown in the diagram A is a point on ground. A particle also having mass m is dropped from a height h at a horizontal distance x from A . It finally touches the ground at point B . Find √3AB in metres. Details and Assumptions Take m=5kg ,h=2√2 metre ,x=1metre.tanθ=1/√2 The collision of the ball with the wedge is elastic and the wedge is free to move. THIS IS THE FIGURE https://d3pq38zxuosm5i.cloudfront.net/solvable/2676130fa9.e594e044b6.ZQmdPV.jpg 2. Relevant equations AS the collision is elastic so momentum of system is conserved and kinetic energy is conserved. 3. The attempt at a solution given tanθ=1/sq.root(2) using this we can find the the initial distance of ball above point A(v3)=3/√2. so velocity of ball when it collided wedge=√(3√2)g. since K.E. is conserved so 3√2g=v1^2+v2^2...............(1)(V1 is velocity of ball after collision and v2 is velocity of wedge after collision) now tanθ=1/sq.root(2) so θ=35.26 applying law of conservation of momentum in Y-direction 3/√2g=-v1 sin2θ+0...................(2)(considering upward direction -ve). Are my two equations correct. Answer of this question is 1.268.