Question

. 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?

Answer 1

m=70g=7/100kg
u=0.5m/s
t=0.5s
v=0m/s
F=∆p/t
=m(v-u)/t
=7/100(0-0.5) /0.5
=7/100(-0.5) /0/5
=7/100(-1) = -7/100N=0.07N
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Answer 2

the force applied by player to stop the ball is equal to 0.07N