Question

A cricket ball of mass 150 g moving with a speed

of 30 ms–1 is brought to rest by a player in 0.03 s.

Find (i) the change in momentum of the ball

(ii) the average force applied by the player.
Guys pls answer step by step ​

Answer 1

Explanation:

change in momentum = mv-mu

Delta p = m(v-u)

as the object comes to rest so v = 0

u = 30 m/s²

m= 150g = 0.150 kg

p= 0.150× (0-30)

-4.5 kgm/s

-ve sign indicate the breaks are applied which is friction force.

f= ma

m(v-u) /t

= 0.150(0-30)/0.03

= -4.5/0.03

= -150N

Answer 2

Answer:

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