Dear Student
Let the inclined plane be at an angle phi with horizontal and the ball is of mass m.
For a ball falling from a vertical height h the velocity is (2gh)^(1/2).
[Initial velocity of the ball is zero at height h
v^2-u^2=2*g*h, From Equation of motion
v^2-0=2*g*h
v=(2gh)^(1/2)]
If the ball rebounds elastically with the coefficient of restitution e =1 , momentum will be conserved along the horizontal direction of the inclined plane. If the ball rebounds with velocity u at an angle theta, then
m*(2gh)^(1/2) *sin (phi) = m*e*u*sin (theta)
(2gh)^(1/2)*sin (phi)= u*sin (theta).....(1) (cancelling m and e=1)
along vertical direction of the plane the velocity components must be same
(2gh)^(1/2)*cos (phi) = u*cos (theta).......(2)
From (1) and (2) we get theta=phi and u =(2gh)^(1/2)......(3)
Thus impulse = rate of change of momentum
Along horizontal direction the velocity components are in the same direction so the rate of change of momentum is zero. Along vertical they are in opposite direction so they will add up.
Thus m*(2gh)^(1/2)*cos(phi)+m*u*cos(theta)
=2*m*(2gh)^(1/2)*cos(phi) (From (3))
2*mv*cos θ = {2m(cos θ)}*{√(2gh)}
Regards
Arun (askIITians forum expert)
