Question

A body of mass m is dropped from a height h on a hard smooth horizontal surface it bounces back to same height the magnitude of change in momentum of body is

Answer 1

Answer:

Charge in momentum

Explanation:

(P) = mass × velocity

but, v = square root of 2gh

therefore, change in P = m× square root of 2gh - (- square root of 2gh)

therefore, change in P = 2m× square root of 2gh