A cricket ball of mass 70 g moving with a velocity of 0.5 m/s is stopped by a player in 0.5 s. What is the force applied by the player to stop the ball?
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Answer:
mass of ball=70g=0.07 kg
velocity of ball=0.5 m/s
impulse=rate of change of momentum=m(v-u)=0.07(0-0.5)=0.07×-0.5=-0.035 Ns
impulse =force ×time
-0.035=force ×0.5
-0.035÷0.5=force applied
-0.07 N=force applied
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