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Momentum, Work, Energy Question

  1. Mar 17, 2008 #1
    1. The problem statement, all variables and given/known data
    A verbatim copy from the problem set:

    Two children wearing skates on ice push apart from each other in opposite directions. The child of mass 30kg slides 15m before stopping. How far will the smaller child of mass 20kg travel before stopping? Assume they experience the same frictional force by the ice. However, the frictional force is negligible during the time they push each other away. (This question involves energy and momentum).

    2. Relevant equations

    The "This question involves energy and momentum" through me off a bit because I was originally looking at just momentum. I suppose that the equations involved are:

    [tex]\sum[/tex]P(initial) = [tex]\sum[/tex]P(final)
    E(kinetic) = [tex]\frac{1}{2}[/tex]mv[tex]^{2}[/tex]
    [tex]\sum[/tex]F = [tex]\frac{dP}{dt}[/tex] = m[tex]\frac{dv}{dt}[/tex]

    3. The attempt at a solution
    The absense of any time makes the above equation with dt in it a bit hard.
    Is the acceleration constant (slowing each skater down)? I would have thought so because the kinetic friction force is generally constant.

    I've drawn out the basic situation in my book but this addition of energy to the problem is confusing me. We've only had a small talk about energy so far in the lectures so all I have to go on is my memory of first semester of last years high school physics which is now a fair time ago.

    What would be the recommended starting point? I was thinking that looking at the initial vs final momentum (both being zero) wouldn't be of much use as neither of them take into account the displacement. I know that this all sounds like a pretty poor attempt at the solution but I really am baffled and it has come as the hardest of the 6 questions on the problem set to other people I've talked to (it is question 6).
  2. jcsd
  3. Mar 17, 2008 #2
    To add my query.

    Should I be looking at this from a force perspective and try to relate it to momentum? It's the absense of time that's really bothering me and the suggestion of using energy has also left me a bit stunned. Some starting points and explanations would be greatly appreciated.
  4. Mar 17, 2008 #3

    Doc Al

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    Hint: Given that momentum is conserved, how do the initial speeds of the skaters compare? How do their kinetic energies compare?
  5. Mar 17, 2008 #4
    Isnt it wrong saying that the friction force is constant?? The force differ, but the friction constant is the same.
  6. Mar 17, 2008 #5

    Doc Al

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    That would seem to be a more realistic model. But I would just solve the problem as given.
  7. Mar 17, 2008 #6
    Well they should both have the same momentum which would give them different velocities. This would give the lighter of the two skaters more kinetic energy.

    I just can't quite incorperate distance into it. I know that Work = Force x Displacement but I don't have a frictional force here. That also means that I can't work out (or at least I don't think I can) the decceleration so using straight line motion equations doesn't work therefore.
  8. Mar 17, 2008 #7


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    It seems like you could work out the deceleration from F=ma.
  9. Mar 17, 2008 #8
    Well I have m but what's F? I don't have any frictional values to work with, nor do I have any values of time to work with dP/dt

    I really need a basic plan of attack for solving this problem because at the moment I'm just throwing equations at it and trying to somehow get my way to an answer but it isn't working yet.
  10. Mar 17, 2008 #9

    Doc Al

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    Right. Now find the exact relationship between the two speeds.
    Using what you find about the speeds, find the exact relationship between the two energies.

    You don't need the force. You know that whatever it is, it was enough to dissipate all of the energy of one child in the given distance.

    No need for that. Focus on the energy-work connection.
  11. Mar 17, 2008 #10


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    Assembling equations is a perfectly good plan of attack. In the end, of course, you would like to have as many equations as variables. Go ahead and call F a variable. I think you will find that it will cancel out in the end.
  12. Mar 17, 2008 #11
    Okay. I'm calling V1 the velocity of the 30kg kid and V2 the velocity of the 20kg kid.

    V1 x m1 = V2 x m2
    Add in numbers
    30V1 = 20V2
    V1 = (2/3)V2

    Now for energy.

    E1 = (1/2) x 30 x (V1)^2 Sub in V1 = (2/3)V2 to get E1 = (1/2) x 30 x ((2/3)V2)^2
    E2 = (1/2) x 20 x (V2)^2

    Rearrange to get (V2)^2 on its own.

    (V2)^2 = 3E1/20
    (V2)^2 = E2/10

    Equate the two.

    3E1/20 = E2/10

    E1 = (2/3)E2

    That's the same relationship that we had for the velocities. Does this all sound correct so far?

    How do I incorperate displacement now?
  13. Mar 17, 2008 #12
    Work = Force x Displacement

    The work for each skater will be equal to their kinetic energy because Work = Change in energy and they both lose all of their kinetic energy as they come to a complete stop (and gain no gravitational potential energy). That still leaves me with 2 unknowns though.

    EDIT: Wait. Simultanious equation?

    I'll do this and get back to you.
  14. Mar 17, 2008 #13

    Doc Al

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    Sounds perfect.

    So far, so good.
  15. Mar 17, 2008 #14


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    … energy is always positive …

    Hi David! :smile:
    I agree - so far as the initial velocities are concerned, I don't see where energy comes into it.
    No … E1/E2 = V1/V2 is impossible … as I'm sure you've worked out! :smile:

    The velocity calculation was valid because momentum is always conserved in a collision (well, this is the opposite of a collision - but the principle's the same!).

    Momentum can be conserved because if one skater's momentum is positive, then the other's is negative.

    But energy is always positive - so you can't add the two skaters' energies and expect them to be zero! :frown:

    You can assume the deceleration of the skaters is the same - it's the same ice!

    You know V1 = (2/3)V2.

    Let the acceleration be -a.

    How does stopping distance depend on initial velocity?

    (It's probably linear, or square, or square-root …)

    Then … that's it! :smile:

    But I don't see where energy come into it … :frown:
  16. Mar 17, 2008 #15

    Doc Al

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    Sure about that? :wink:

    No one added any energies to get zero. Where did you see that?

    Ordinarily a reasonable assumption, but the problem tells us to assume that each experiences the same frictional force.
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