Question

a body of mass 10 kg is dropped from a height of 1m on a hard smooth surface it bounce back to same height the magnitude of the change in momentum (kgm/s)

Answer 1

velocity of the object
$= \sqrt{2gh}$
therefore,
$\sqrt[]{2 0} = 4.4721$
change in momentum is
mv-(-mv)
=2mv=2*10*4.471

Answer 2

2gh=v^2-u^2
2*10*1=v^2-0
v^2=20
v=√20m/s or2√5

momentum=mass*velocity
=10*2√5
=20√5