A cricket ball of mass 120 g is moving with a velocity of 12 m $s^{- 1}$ . It turns back and moves with a velocity of 20 m $s^{- 1}$ when hit by a bat. Find the impulse if the force acts on the ball for 0.02 s.

4 N s

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8 N s

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32 N s

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192 N s

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Solution

The correct option is D 192 N s
Initial velocity of ball $u = - 12 m / s$
Final velocity of ball $v = 20 m / s$
Magnitude of change in momentum $| Δ P | = m (v - u) = 0.12 \times (20 - (- 12)) = 3.84 k g m / s$
Time of action $Δ t = 0.02 s$
Thus impulsive force $F = \frac{Δ P}{Δ t} = \frac{3.84}{0.02} = 192 N$

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A cricket ball of mass 120 g is moving with a velocity of 12 m s−1. It turns back and moves with a velocity of 20 m s−1 when hit by a bat. Find the impulse if the force acts on the ball for 0.02 s.

The correct option is D 192 N sInitial velocity of ball u=−12 m/sFinal velocity of ball v=20 m/sMagnitude of change in momentum |ΔP|=m(v−u)=0.12×(20−(−12))=3.84 kg m/sTime of action Δt=0.02 sThus impulsive force F=ΔPΔt=3.840.02=192 N

A cricket ball of mass 120 g is moving with a velocity of 12 m $s^{- 1}$ . It turns back and moves with a velocity of 20 m $s^{- 1}$ when hit by a bat. Find the impulse if the force acts on the ball for 0.02 s.