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Average force and a ball of mass

  1. Mar 30, 2006 #1
    Could someone help me out with this one, I cant seem to get on track here.

    A ball of mass 3.5 kg is thrown upward. It leaves the thrower's hand with a velocity of 12 m/s. The ball is in the air for 2.4 seconds. Final velocity before the ball returns to the hand in -12 m/s. The change in momentum in the ball is -84 N*s. The implus calculated from the change in momentum is -84 N*s. How would I calculate the average force acting on the ball?

    I though it should be -84/2.4 (change in momentum/change in time). -35 however is not correct.
     
  2. jcsd
  3. Mar 30, 2006 #2
    How did you get that -84 N*s for change in momentum?
     
  4. Mar 30, 2006 #3
    (-12*3.5)-(12*3.5)=-84 N*s
     
  5. Mar 30, 2006 #4

    Tide

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    HINT:

    [tex]\bar F = \frac { \int_{t}^{t + \Delta t} F dt}{\Delta t}[/tex]

    and

    [tex]F = \frac {dp}{dt}[/tex]
     
  6. Mar 30, 2006 #5
    [tex]F = \frac {dp}{dt}[/tex]
    is that not the formula I used and got the incorrect answer

    I though it should be -84/2.4 (change in momentum/change in time). -35 however is not correct.

    Could you explain the other formula, I'm not sure what your trying to tell me. I'm not familiar with the squiggly symbol. It's whatever multiplied by the force multiplied by the change in time all divided by the change in time?
     
    Last edited: Mar 30, 2006
  7. Mar 31, 2006 #6

    Hootenanny

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    Are you sure you've copied the question correctly?
     
  8. Mar 31, 2006 #7

    Doc Al

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    If the question is "What's the average force acting on the ball while its in the air?", then that answer is correct. (Realize that the only force acting on the ball, when in the air, is its weight.)

    Or are they asking: "What's the average force that the thrower's hand exerts on the ball?" A different problem.

    Why don't you transcribe the problem exactly as it was given to you.
     
  9. Mar 31, 2006 #8
    1. A. A ball of mass 3.5 kg is thrown upward. It leaves the thrower's hand with a velocity of 12 m/s. The following questions refer to the motion after the ball leaves the thrower's hand. Assume that the upward direction is positive. Show all calculations.
    a. How long does it take for the ball to return to the thrower's hand?

    0=12+(-9.8)T
    12=-9.8t
    t=1.2
    1.2*2= 2.4s


    b. What is the final velocity of the ball just before it reaches the hand?

    v=0+-9.8(1.2)
    v=-12 m/s

    c. What is the change in momentum of the ball?

    (-12*3.5)-(12*3.5)=-84 N*s

    d. What is the impulse calculated from the change in momentum?
    -84/2.4=-35
    -35*2.4=-84 N*s

    e. What is the average force acting on the ball?
     
  10. Mar 31, 2006 #9
    -34 is the correct answer.....rounding error.....I suppose that what I get for trying to work on homework at midnight...
     
  11. Mar 31, 2006 #10

    Doc Al

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    Looks good to me. In the context of that problem, the "average force" means the average force on the ball while the ball is in flight, which equals mg, as you calculated.

    (This isn't one of those computer graded exercises, is it? In that case be careful how you round off the answer and the sign of your answer. For example: While you correctly found the average force to be -35N, they may be looking for just the magnitude of the force, thus an answer of 35N.)

    [Edit: Looks like you figured it out! :smile: ]
     
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