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Standard IX

Physics

Second Law of Motion

A 150 g ten...

Question

A $150 g$ tennis ball coming at a speed of $40 m/s$ is hit straight back by a bat to a speed of $60 m/s$ . The magnitude of the average force $F$ on the ball, when it is constant for $5 ms$ , is:

2500 N

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3000 N

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3500 N

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4000 N

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Solution

The correct option is C $3000 N$

The change in momentum

$Δ p = {m(v}_{f} - v_{i})$

$= 0.150 (60 - (- 40))$

$= 0.150 \times 100 = 15 N s$

Thus, $F = \frac{Δ P}{Δ t} = \frac{15}{5 \times 10^{- 3}} = 3 \times 10^{3} N$

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A 150g tennis ball coming at a speed of 40m/s is hit straight back by a bat to a speed of 60m/s. The magnitude of the average force F on the ball, when it is constant for 5ms, is:

The correct option is C 3000N The change in momentum Δp=m(vf−vi) =0.150(60−(−40)) =0.150×100=15Ns Thus, F=ΔPΔt=155×10−3=3×103N

A $150 g$ tennis ball coming at a speed of $40 m/s$ is hit straight back by a bat to a speed of $60 m/s$ . The magnitude of the average force $F$ on the ball, when it is constant for $5 ms$ , is:

The correct option is C $3000 N$

The change in momentum

$Δ p = {m(v}_{f} - v_{i})$

$= 0.150 (60 - (- 40))$

$= 0.150 \times 100 = 15 N s$

Thus, $F = \frac{Δ P}{Δ t} = \frac{15}{5 \times 10^{- 3}} = 3 \times 10^{3} N$