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Physics momentum problem: units

  1. Mar 17, 2005 #1
    physics momentum problems

    Fred (mass 70.0kg) is running with the football at a speed of 4.70m/s when he is met head-on by Brutus (mass 130kg), who is moving at 8.10m/s. Brutus grabs Fred in a tight grip, and they fall to the ground. How far do they slide?
    Part A
    The coefficient of kinetic friction between football uniforms and Astroturf is 0.290.
    Give answer in cm

    I got an answerof 13.1m/s=5.684m/s^2(deltaX) so ultimately deltaX=2.3 but what? meters? and could someone remind me of how to go from m to cm again.
    Last edited: Mar 17, 2005
  2. jcsd
  3. Mar 17, 2005 #2
    to go from metres to centimetres you multiply by 100

    since your WHOLE CALCULATION is in metres what do you think the final answer will come out.
  4. Mar 17, 2005 #3
    well i know it should be meters, but assumptions always get you lost in some corner of physics you dont want to find yourself in. And since i cant find out how to get to meters, i asked here.

    Also, another problem is saying 9.8 meters per second per second. this is 9.8m/s^2, correct?

    same problem as the 9.8 question: if i have 39.2m^2/s^2+64m/s is the result 103.2m/s?
    Last edited: Mar 17, 2005
  5. Mar 17, 2005 #4
    No. It's a nonsense expression. You've made a mistake in your original question as well, if [itex] \Delta x[/tex] should be a distance (the mistake could just be keeping track of units wrong, of course).

    If you're ever doing a physics problem and the units don't work out, it means you've made a mistake.
  6. Mar 17, 2005 #5
    A girl of mass 60 kilograms springs from a trampoline with an initial upward velocity of 8 meters per second. At height 2 meters above the trampoline, the girl grabs a box of mass 15 kilograms.

    For this problem, use 9.8 meters per second per second for the magnitude of the acceleration due to gravity.

    What is the speed of the girl immediately before she grabs the box?
    Express your answer numerically, to two significant figures.

    i did 1/2Vf^2+60kg(9.8m/s^2)(2m)=1/2(8m/s)^2 I got this down to Vf^2+588J=16m/s

    the units dont jive if they want the answer in m/s
  7. Mar 17, 2005 #6
    I have absolutely no idea where you get

    from. Do you perhaps mean to use conservation of energy? In that case, your expression should be in the form

    \frac{1}{2}mv_f^2 + mgh = \frac{1}{2}mv_i^2

    in which case you will find the units work out perfectly.
  8. Mar 17, 2005 #7
    i cancelled the masses initially, why couldnt you do this?
  9. Mar 17, 2005 #8
    [tex] \frac{1}{2}mv_f^2 + mgh = \frac{1}{2}mv_i^2 \Longrightarrow \frac{1}{2}v_f^2 + gh = \frac{1}{2}v_i^2[/tex]

    the units will still work fine, of course (how could dividing both sides of an equation with balanced units by the same quantity affect the balance, anyways?).
  10. Mar 17, 2005 #9
    i find myself in the same place again. vf^2+ 39.2m^2/s^2=64m/s
  11. Mar 17, 2005 #10
    [tex] \frac{1}{2}v_f^2 + gh = \frac{1}{2}v_i^2[/tex]

    [tex]\Longrightarrow v_f^2 + 2\left(9.8\frac{\mbox{m}}{\mbox{s}^2}\right)(2\mbox{m}) = \left(8\frac{\mbox{m}}{\mbox{s}}\right)^2[/tex]

    [tex]\Longrightarrow v_f^2 = 64\frac{\mbox{m}^2}{\mbox{s}^2} - 39.2\frac{\mbox{m}^2}{\mbox{s}^2} = 24.8 \frac{\mbox{m}^2}{\mbox{s}^2}[/tex]

    [tex]\Longrightarrow v_f \approx 5 \frac{\mbox{m}}{\mbox{s}}[/tex]
    Last edited: Mar 17, 2005
  12. Mar 17, 2005 #11
    And if you didn't notice, your only error in your last calculation was not realising that [tex]\left(\frac{\mbox{m}}{\mbox{s}}\right)^2 = \frac{\mbox{m}^2}{\mbox{s}^2}[/tex].
  13. Mar 17, 2005 #12
    same question, what is maximum height of the girl to two decimal places.


    got answer of 3.395, i submitted answer of 3.40 which is incorrect.

    Last edited: Mar 18, 2005
  14. Mar 18, 2005 #13
    anyone know an answer?
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