Question

An object of mass $1kg$ is dropped from a height of $20m$. It hits the ground and rebounds with the same speed. Find the change in momentum. (Take $g=10m/s^2)$

Answer 1

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$\times$10 $\times$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$\times$v

= 1$\times$(-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$\times$v

= 1$\times$(+20)

= +20kgm/s

Change in momentum = Final momentum - Initial momentum

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

= 40kgm/s in upward direction.