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प्रश्न
A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001s, the force that the batsman had to apply to hold the bat firmly at its place would be ______.
पर्याय
10.5 N
21 N
1.05 × 104 N
2.1 × 104 N
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उत्तर
A cricket ball of mass 150 g moving with a speed of 126 km/h hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001 s, the force that the batsman had to apply to hold the bat firmly at its place would be `underline(1.05 xx 10^4 N)`.
Explanation:
Given, m = 150 g = `150/1000` kg = `3/20` kg
Δt = time of contact = 0.001 s
u = 126 km/h = `(126 xx 1000)/(60 xx 60)` m/s = 35 m/s
v = – 126 km/h = – 35 m/s
Change in the momentum of the ball
Δp = m(v – u) = `3/20(- 35 - 35)` kg-m/s
= `3/20 (-70)`
= `- 21/2`
We know that force F = `(Δp)/(Δt)`
= `(-21/2)/0.001` N
= – 1.05 × 104 N
Here, the – ve sign shows that force will be opposite to the direction of movement of the ball before hitting.
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