An object of mass 2kg fall a distance of 5m on a horizontal surface and rebounds to a vertical height of 3m. Calculate the change in momentum.

Answer 1

starting m g h = 2 *g * 5 Joules
so
(1/2) 2 v^2 = 10 g
v^2 = 10 g
if g = 10 m/s^2
v = -sqrt 100 = -10 m/s
momentum = -20 kgm/s
no w upward
m g h = 2 * 10 * 3 = 60
so
(1/2) 2 v^2 = 60
v = +sqrt 60
m v = 2 sqrt 60 = 15.5 kg m/s up
change = 15.5 - (-20) = 35.5 kg m/s

Answer 2

no change in momentum.
It had zero momentum at the start, and at the top of the bounce it is zero again.

Answer 3

LOL - true, but usually they mean the impulse at the turn.