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

Explanation:

mass, m = 1kg

height, s = 20m

initial velocity of ball, u = 0m/s

acceleration, a = 10m/s

2

Using, v

2

= u

2

+ 2as

= 0 + 2×10 ×20

= 400

v = 20m/s

Now, let us take upward direction as positive and downward direction as negative.

Initial momentum of the ball (before striking the ground) = m×v

= 1×(-20) (velocity is downward, hence negative)

= -20kgm/s

Now, when the ball strikes and rebounds, its velocity becomes +20m/s (after striking, it moves upward, hence positive velocity)

Final momentum (after striking the ground) = m×v

= 1×(+20)

= +20kgm/s

Change in momentum = Final momentum - Initial momentum

= 20kgm/s - (-20kgm/s)

= 40kgm/s in upward direction.