Question

A cricket ball of mass 0.5 kg strikes a bat normally with a velocity of 30 M per second and rebounds with a velocity of 20 m per second in the opposite direction what is the Impulse of force exerted by the ball on the bat

Answer 1

The solution can be done easily using vectors

Answer 2

Answer:

Magnitude of impulse, J = 25 kg-m/s

Explanation:

Given that,

Mass of the ball, m = 0.5 kg

Initial velocity of the ball, u = 30 m/s

Final velocity of the ball, v = -20 m/s (as it rebounds)

We need to find the impulse of force exerted by the ball on the bat. Let it is equal to J. The change in momentum is equal to the impulse. Mathematically,

$J=m(v-u)$

$J=0.5\times (-20-30)$

J = -25 kg-m/s

So, the impulse exerted by the ball on the bat is 25 kg-m/s. Hence, this is the required solution.