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

Answer:

Mass of the ball=70 g

Initial Velocity(u)=0.5 m/s

Final velocity(v)=0

Change in momentum=m(v-u)=-70×0.5×10⁻³=0.035 kgm/s

Time taken=0.5 s

∴Force applied to stop the ball=change in momemtum/time taken=0.07 N

Answer 2

Given:

mass of ball =m=70g=0.070kg

Initial velocity=u= 0.5 m/s

Final velocity =v=0m/s time=t= 0.5 seca

a= v-u/t

- 0-0.5/0.5

1 m/s2

F=m*a

=0.07x-1

-0.07 N