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Homework Help: Laws of motion

  1. Jan 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A girl weighing 30kg is standing motionless on a 2-kg skateboard holding a 7-kg bowling ball. She throws the ball with an avg force of 75N

    a.) what is the magnitude of her acceleration?
    b.) what is the magnitude of the acceleration of the bowling ball?

    2. Relevant equations

    3. The attempt at a solution
    b.) 75/2
  2. jcsd
  3. Jan 3, 2008 #2


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    The choice of the skateboard that she is standing on is to assure that the motion to follow will take place (essentially) frictionlessly. This is a problem applying Newton's Third Law. If she is applying 75N to the ball when she throws it away, what force does the ball apply to her? Then answer the questions in that light. Keep in mind that during the throw, she and the skateboard will move together.
  4. Jan 3, 2008 #3
    My book doesnt even explain how to do the problem. There arent any equaitons for the 3rd law ( in my book). Do yo know how to do it
  5. Jan 3, 2008 #4
    Newtons Third Law:
    [tex] \vec{F_{A on B}}=\vec{-F_{B on A}}[/tex]
  6. Jan 3, 2008 #5


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    The applied force of the girl against the ball equals the force of the ball against the girl's hands. (This is what the equation Winzer posted says.) So the force acting on her is...?
  7. Jan 3, 2008 #6
    I have never seen that equation before in my life.

    Could you tell me what to plug in ? Maybe I can take it from there
  8. Jan 3, 2008 #7
    Don't think of it as just an equation, think of it as common sense as well.
    Suppose: you are facing a wall. You chose to push on the wall applying a Force(your feet are firmly planted); however this wall exerts an opposing Force->-F, no? If not, then your hand would go stright through the wall.

    Now consider the case you have listed above.
    This person exerts a force on the bowling ball. Then this ball would have to provide an opposing force or the girl's hand would go through the ball. However the girl is on a skateboard and that forced gets translated through the motion of the skateboard .
  9. Jan 3, 2008 #8
    I am lost.. maybe it is because it is late

    I dont know

    I'm about to go to bed. Could you just tell me the answer and I'll come back on (tomorrow afternoon) and you could explain it to me?
    Last edited: Jan 3, 2008
  10. Jan 3, 2008 #9
    Ok, lets do this:
    [tex] \vec{F_AonB}=\vec{-F_BonA}[/tex] right? Let A be the ball and B be the skateboard and girl.{The minus sign will depend on orientation}
    By Newtons second law: F=ma
    So.. [tex] m_b\vec{a_b}=-(m_g+m_s)\vec{a}[/tex] b=ball, g=girl, s=skateboard
    [tex] -\frac{a_b m_b}{m_g+m_s}=a[/tex] This is the acceleration of the girl+skateboard
    Do you see it? Can you explain to me why this is?
    Last edited: Jan 3, 2008
  11. Jan 3, 2008 #10
    No, we never learned that.
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