A cricket ball of mass $0.25 k g$ with speed $10 m / s$ collides with a bat and returns with same speed within $0.01 s$ . The force acted on bat is

25 N

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50 N

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250 N

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500 N

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Solution

The correct option is D $500 N$
Given: $v = 10$ $m s^{- 1}$ ; $u = - 10$ $m s^{- 1}$ ; $m = 0.25 k g$ ; $Δ t = 0.01 s$
Change in momentum, $Δ P = m v - m u$
$\Rightarrow Δ P = 0.25 (10) - 0.25 (- 10) = 5$ $k g m s^{- 1}$
Force acting on bat, $F = \frac{Δ P}{Δ t} = \frac{5}{0.01} = 500 N$

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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

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A cricket ball of mass 0.25kg with speed 10m/s collides with a bat and returns with same speed within 0.01s. The force acted on bat is

The correct option is D 500 NGiven: v=10 ms−1; u=−10 ms−1; m=0.25kg; Δt=0.01sChange in momentum, ΔP=mv−mu⇒ΔP=0.25(10)−0.25(−10)=5 kgms−1Force acting on bat, F=ΔPΔt=50.01=500N

A cricket ball of mass $0.25 k g$ with speed $10 m / s$ collides with a bat and returns with same speed within $0.01 s$ . The force acted on bat is

The correct option is D $500 N$
Given: $v = 10$ $m s^{- 1}$ ; $u = - 10$ $m s^{- 1}$ ; $m = 0.25 k g$ ; $Δ t = 0.01 s$
Change in momentum, $Δ P = m v - m u$
$\Rightarrow Δ P = 0.25 (10) - 0.25 (- 10) = 5$ $k g m s^{- 1}$
Force acting on bat, $F = \frac{Δ P}{Δ t} = \frac{5}{0.01} = 500 N$