Question

A bowling ball of mass 5 kg hits a wall going 7 m/s and rebounds at a speed of 2 m/s. What was the impulse applied to the bowling ball

Answer 1

The impulse applied to the bowling ball is 45 kg m/s

Answer 2

Explanation :

Given that,

Mass of the ball, m = 5 kg

Initial velocity of the ball, u = 7 m/s

Final speed of the ball, v = 2 m/s

We know that the impulse applied to an object is equal to the change in momentum.

`J=\Delta p`

`J=p_f-p_i`

`J=m(v-u)`

`J=5\ kg(2\ m/s-7\ m/s)`

J = -25 kgm/s

Hence, this is the required solution.