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Bocce ball momentum

  1. Jun 16, 2011 #1
    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
     
  2. jcsd
  3. Jun 16, 2011 #2

    gneill

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    Staff: Mentor

    "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?
     
  4. Jun 16, 2011 #3
    i dont know, thats just what the problem says... its not necessarily realistic
     
  5. Jun 16, 2011 #4

    gneill

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    Staff: Mentor

    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?
     
  6. Jun 16, 2011 #5
    i dont 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.
     
  7. Jun 17, 2011 #6

    gneill

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    Staff: Mentor

    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.
     
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