A ball of mass 0.2 kg moving with a velocity 10 m/s hits a bat normally and returns with a velocity 20 m/s. If the duration of impact is 0.01 s, find the average force exerted by the bat on the ball.

60 N

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500 N

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600 N

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240 N

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Solution

The correct option is C 600 N
Impulse P=change in momentum .
Now $V_{f} = - 20, V_{1} = 10$ because the direction has reversed .
P= $m (V_{f} - V_{1}) = - 0.2 \times 30 = - 6 N s$
Average force = $\frac{I m p u l s e}{t i m e} = \frac{6}{.01} = 600 N$

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A cricket ball of mass $150 g$ is moving with a velocity $12 m s^{- 1}$ and is hit by a bat so that the ball is turned back with a velocity of $20 m s^{- 1}$ . If the duration of contact between the ball and bat is $0.01 s$ , find the impulse and the average force exerted on the ball by the bat.

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and is hit by a bat so that the ball is turned back with a velocity of

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A cricket ball of mass 150 gm is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of 20 m/s. The force of blow acts for 0.01s on the ball. The average force exerted by the bat on the ball is

Q. A 150 g cricket ball moving at a speed of

12 m s^{- 1}

is hit by a bat and turned back at a speed of

20 m s^{- 1}

. The duration of impact is 0.01 s. What is the average force exerted on the ball by the bat?

Q. A cricket ball of mass 500 gm strikes a bat normally with a velocity 30 m/s and rebounds with a velocity 20 m/s in the opposite direction. The impulse of the force exerted by the ball on the bat is:

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A ball of mass 0.2 kg moving with a velocity 10 m/s hits a bat normally and returns with a velocity 20 m/s. If the duration of impact is 0.01 s, find the average force exerted by the bat on the ball.

The correct option is C 600 NImpulse P=change in momentum .Now Vf=−20 ,V1=10 because the direction has reversed .P=m(Vf−V1)=−0.2×30=−6NsAverage force =Impulsetime=6.01=600N

The correct option is C 600 N
Impulse P=change in momentum .
Now $V_{f} = - 20, V_{1} = 10$ because the direction has reversed .
P= $m (V_{f} - V_{1}) = - 0.2 \times 30 = - 6 N s$
Average force = $\frac{I m p u l s e}{t i m e} = \frac{6}{.01} = 600 N$