A cricket ball of mass 0.5 kg strikes a bat normally with a velocity of 30 m/s and rebound with a velocity of 20 m/s in the opposite direction. The impulse of the force exerted by the ball on the bat is
(A) 0.5 Ns
(B) 1.0 Ns
(C) 25 Ns
(D) 50 Ns
Spamming will be reported immediately. Best answer will be marked as Brainliest.
(A) 0.5 Ns
(B) 1.0 Ns
(C) 25 Ns
(D) 50 Ns
Spamming will be reported immediately. Best answer will be marked as Brainliest.
Answers & Comments
Answer:
25 ns is the perfect answer
Verified answer
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 × (-20-30)
J = -25 kg-m/s
So, the impulse exerted by the ball on the bat is 25 kg-m/s.