A ball of mass $200 g$ is thrown with a speed $20 m s^{- 1}$ . The ball strikes a bat and rebounds along the same line at a speed $40 m s^{- 1}$ Variation of the interaction force, as long as the ball remains in contact with the bat, is as shown in Fig. Average force exerted by the bat on the ball is:

5000 N

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2000 N

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2500 N

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6000 N

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Solution

The correct option is A $2000 N$
Given,

Mass, $m = 0.2 k g$

Final velocity, $v_{f} = 40 m s^{- 1}$

Initial velocity, $v_{i} = - 20 m s^{- 1}$

Change in momentum, $Δ P = m (v_{f} - v_{i}) = 0.2 [40 - (- 20)] = 12 k g m s^{- 1}$

Average force, $F_{a v} = \frac{Δ P}{Δ t} = \frac{12}{6 \times 10^{- 3}} = 2000 N$

Hence, average force is $2000 N$

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A ball of mass 200g is thrown with a speed 20ms−1. The ball strikes a bat and rebounds along the same line at a speed 40ms−1 Variation of the interaction force, as long as the ball remains in contact with the bat, is as shown in Fig. Average force exerted by the bat on the ball is:

The correct option is A 2000NGiven, Mass, m=0.2kg Final velocity, vf=40ms−1 Initial velocity, vi=−20ms−1 Change in momentum, ΔP=m(vf−vi)=0.2[40−(−20)]=12kgms−1 Average force, Fav=ΔPΔt=126×10−3=2000N Hence, average force is 2000N