(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A bocce ball (A) with a mass of 3 kg is thrown with a force of 40 N. It collides with another ball (B) that is at rest. The contact time is .5 seconds.

a. What is the Impulse given to ball B by ball A?

b. If ball B was at rest before being hit, with what velocity does ball B now have?

c. ball B collides with ball C (also 3kg) that was at rest. Ball B keeps moving at 4m/s, how fast does ball C move?

d. What is its kinetic energy?

e. After hitting C, B comes to rest in 1.8 seconds. How much force did the grass apply to the ball to stop it?

f. What is the coefficient of friction of the grass and ball?

2. Relevant equations

impulse=ft

ft=mv

(m1v1)+(m2v2)=(m1v1)'+(m2v2)'

KE=.5mv^2

p=mv

3. The attempt at a solution

heres what i have so far:

a. impulse= 20

b. V=6.666 m/s

c. V=2.666 m/s

d. KE=10.66 J

e. p=mv?

f. Ff=(mu) Fn?

are the answers right? i need help with E and F

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# Bocce ball momentum

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