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Standard IX

Physics

Momentum

A cricket bal...

Question

A cricket ball of mass 150 g 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 the blow acts for 0.001 & on the ball. Find that force?

4800 N

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5200 N

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6000 N

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7000 N

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Solution

The correct option is A 4800 N
$I m p u l s e = F \times t =$ Change in Momentum
$\Rightarrow$ Change in Momentum $= m v - m u = 0.15 (12) - 0.15 (- 20) = 4.8$
$\Rightarrow F \times 0.001 = 4.8$
$\Rightarrow F = 4800 N$
Therefore, A is correct option.

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Similar questions

Q. A cricket ball of mass

150 g

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

0.01 s

on the ball. The average force exerted by the bat on the ball is (in

N

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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 cricket ball of mass 120 g is moving with a velocity of 12 m

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when hit by a bat. Find the impulse if the force acts on the ball for 0.02 s.

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.

Q. A cricket ball of mass 120 g is moving with a velocity of

12 m s^{- 1}

, it turns back and move with a velocity of

20 m s^{- 1}

when hit by a bat. Find the impulse and the force, if the force acts for 0.02 s.

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A cricket ball of mass 150 g 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 the blow acts for 0.001 & on the ball. Find that force?

The correct option is A 4800 NImpulse=F×t= Change in Momentum⇒ Change in Momentum =mv−mu=0.15(12)−0.15(−20)=4.8⇒F×0.001=4.8 ⇒F=4800N Therefore, A is correct option.

The correct option is A 4800 N
$I m p u l s e = F \times t =$ Change in Momentum
$\Rightarrow$ Change in Momentum $= m v - m u = 0.15 (12) - 0.15 (- 20) = 4.8$
$\Rightarrow F \times 0.001 = 4.8$
$\Rightarrow F = 4800 N$
Therefore, A is correct option.