Question

A ball of mass 2 kg is dropped from a height of 5 m. On striking the ground it rebounds with same speed

with which it struck the ground. Find the magnitude of net change in its momentum just after striking the

ground. [Take, g = 10 m/s2
]​

Answer 1

Answer:

no

Explanation:

th speed will be change

Answer 2

Explanation:

initial velocity =0m/s^2

height=5m

g=10m/s^2

So Final velocity=√(u^2 +2gh)

=√(2×10×5)

=10m/s^2

Now initial momentum =m×v= 2×10=20kgm/s^2

Final momentum =m×-v(direction is opposite)

=2×-10= -20kgm/s^2

So net change in momentum=

final momentum-initial momentum

= -20-(20) = -40 kgm/s^2

Thus the net change in momentum is 40kgm/s^2