Question

A ball of mass m is released from top of an inclined plane.... Help

Answer 1

Answer:

energy of the ball when strikes to a rigid surface={ I - 3I/4} + KE

= mgI/4 + mv^2/2

following conservation of energy, mgl = mgl/4 + mv^2/2

solve it n u will get v=$\sqrt{1.5gl}$

impulse=mv - (-mv) = 2mv = m$\sqrt{6gl}$