Question

A ball of mass m.Is released from the top pf a inclined plane of inclination theta as shown.It strikes a rigid surface at a distance 3l/4 from top elastically.Impulse imparted to ball by the rigid surface is

Answer 1

initial energy of ball = potential energy at height h of ball = mgl

energy of the ball when it strikes the rigid surface = potential energy at ( l - 3l/4) + kinetic energy
= mgl/4 + 1/2mv²

from conservation of energy,

initial energy = final energy

mgl = mgl/4 + 1/2mv²

3mgl/2 = mv²

v² = 1.5gl

v = √(1.5gl)

now, impulse = change in momentum = mv - (-mv) = 2mv = 2m√(1.5gl) = m√(6gl)

hence, Impulse imparted to ball by the rigid surface is m√(6gl)

Answer 2

Explanation:

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