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Steel ball dropped onto a plate

  1. Feb 21, 2006 #1
    A 5.60 kg steel ball is dropped onto a copper plate from a height of 10.0 m. If the ball leaves a dent 3.10 mm deep, what is the average force exerted by the plate on the ball during the impact?

    Ok so if you calculate out the speed of the ball you should get the ball is going 14 m/s when it hits the plate. Do you use Work = Force * Distance? With the distance being the amount the dent went (3.1 mm) and the force would be equal to the energy the ball has before it hits the plate? Thanks for any help/advice
  2. jcsd
  3. Feb 22, 2006 #2
    In my opinion, I'll solve this problem in this way.
    Consider [tex]\Delta W=W_i-W_f[/tex]
    I use the equation

    [tex]\Delta W=(F-P)\Delta x\Longrightarrow F=\frac{\Delta W}{\Delta x}+P[/tex]
  4. Feb 22, 2006 #3
    My attack would be to use the impulse-momentum theorem. You know how fast the ball is going when it hits the copper. This allows you to calculate the momentum of the ball before impact. The momentum of the ball when it reaches the bottom of the dent is zero, so you know the change in momentum. So you can figure the impulse. Now, impulse is average force times time. You can find the time by looking at the speed of the ball at impact and the depth of the dent by kinematics. Use [tex] y=y_0 +(1/2)(v_0+v)t [/tex].

    Typically, if a problem is looking for average force, the impulse-momentum theorem is a good starting point.

  5. Feb 22, 2006 #4


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    Forces and energies are completely different quantities with different units. You can not compare or equate them.
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