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Homework Help: Impulse and Average Force. Ball by the Wall.

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    A ball of mass m= .25kg is moving with an initial velocity of [itex]\vec{v}[/itex]=(8.0[itex]\hat{x}[/itex]+8.0[itex]\hat{y}[/itex])(m/s). It rebounds from a wall and has a final velocity (after impact) of [itex]\vec{v}[/itex]= (-5.0[itex]\hat{x}[/itex]+6.0[itex]\hat{y}[/itex])(m/s)
    a.) Determine the impulse exerted on the ball by the wall.
    b.) Calculate the average force exerted on the ball by the wall.
    c.) Is the momentum conserved in this collision? Explain.

    2. Relevant equations

    3. The attempt at a solution
    a.) The forces for this part I solved for in part b. I looked at the question closer, and it only wants one impulse, and does not mention anything about finding them for both the x and y coordinates so that is why I think I did it wrong.
    Jx= 21.7N (.15s) = 3.25J ([itex]\hat{x}[/itex])
    Jy= 3.3N (.15s) = .49J ([itex]\hat{y}[/itex])
    b.) Calculate the average force exerted on the ball by the wall.
    Favg = (deltaP)/(deltat)
    deltax=-5-8 = -13 m/s
    deltay= 6-8 = -2 m/s
    deltaPx = (.25 kg)-13 m/s = -3.25 N ([itex]\hat{x}[/itex])
    deltaPy = (.25 kg)-2 m/s = -.5 N ([itex]\hat{y}[/itex])
    Favgx = (-3.25 ([itex]\hat{x}[/itex]))/.15s = -21.7N ([itex]\hat{x}[/itex])
    Favgy = (-.5N ([itex]\hat{y}[/itex]))/.15s = -3.3N ([itex]\hat{y}[/itex])
    As so for this I think I solved for the force exerted on the wall by the ball. I took the opposites of the two answers I just got to make them the force exerted on the ball by the wall.
    c.) by this point I was pretty lost, but I do know that In an isolated system, the total momentum of the system is constant.

    The more I look at the the more I think I am very wrong. I am pretty confused so if you could explain in as much detail as possible please that would be very helpful. I am kinda frustrated with it. Thank you very much for your time.
  2. jcsd
  3. Dec 8, 2013 #2
    Should the question give a time for the collision of 0.15 seconds? Are you aware of how impulse and the change in momentum are related?
  4. Dec 9, 2013 #3
    Oh yes I am sorry for leaving that out. I know the equations, but no I am not entirely sure on how they relate to one another. I read through my notes and am only confused.
  5. Dec 9, 2013 #4
    The impulse is equal to the change in momentum.

    Ft = mv - mu

    Any object with momentum is going to be hard to stop. To stop such an object, it is necessary to apply a force against its motion for a given period of time.
  6. Dec 9, 2013 #5
    Awesome that helps a lot! I know how to do it now, other than I am unsure about how to find the velocity before and after. I have never seen the velocities in x and y coordinates like that before. Would it be vx= -5-8= -13 m/s ; vy= 6-8= -2 m/s for a Δv (-13-(-2))/2 = -5.5 m/s ?
    Last edited: Dec 9, 2013
  7. Dec 9, 2013 #6
    When finding the resultant vector from components you use pythagoras (you will use this a lot)

    v^2 = vx^2 + vy^2
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