Bocce Ball Impulse & Kinetic Energy Calculations

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Homework Statement


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?


Homework Equations


impulse=ft
ft=mv
(m1v1)+(m2v2)=(m1v1)'+(m2v2)'
KE=.5mv^2
p=mv

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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"A bocce ball (A) with a mass of 3 kg is thrown with a force of 40 N." This doesn't make much sense unless the time over which the 40N is impressed upon the ball (A) is given.

40N applied to a 3kg ball for a period of an hour would give the ball an impressive speed of almost 173,000 km/hr. :smile:

So, precisely how was the ball thrown?
 
gneill said:
"A bocce ball (A) with a mass of 3 kg is thrown with a force of 40 N." This doesn't make much sense unless the time over which the 40N is impressed upon the ball (A) is given.

40N applied to a 3kg ball for a period of an hour would give the ball an impressive speed of almost 173,000 km/hr. :smile:

So, precisely how was the ball thrown?

i don't know, that's just what the problem says... its not necessarily realistic
 
green123 said:
i don't know, that's just what the problem says... its not necessarily realistic

Was the problem translated from another language? Perhaps the context has different implications in another language. Is it possible that they meant that the ball is thrown in such a way that it imparts 40N over 0.5 seconds to the ball it strikes?
 
gneill said:
Was the problem translated from another language? Perhaps the context has different implications in another language. Is it possible that they meant that the ball is thrown in such a way that it imparts 40N over 0.5 seconds to the ball it strikes?

i don't believe so, all i need to know is the force it took for the grass to stop the ball in 1.8 seconds and the coefficient of friction between the grass and the ball.
 
green123 said:
i don't believe so, all i need to know is the force it took for the grass to stop the ball in 1.8 seconds and the coefficient of friction between the grass and the ball.

Okay, assuming that the ball was thrown with a force of 40N applied over a period of 0.5s, then your results for a and b look okay. How did you arrive at 2.666 m/s for the velocity of ball C in part (c)?

Keep in mind that momentum is a vector quantity, and if ball B has any speed after a collision with another ball of equal mass initially at rest, that means the balls are not going to be moving co-linearly after the impact -- it was not a head-on strike.