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Kinetic Energies vs. Momentum

  1. Mar 3, 2005 #1
    A simple question:

    A small truck and a large truck have the same kinetic energies. Which truck has the greater momentum? Justify your answer.
  2. jcsd
  3. Mar 3, 2005 #2
    Smells like homework...

    Show us an attempt at solving the problem. Start with the equations you know for kinetic energy and momentum as functions of velocity, and relate them through that variable.
  4. Mar 3, 2005 #3

    Andrew Mason

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    Homework Helper

    What is the expression for kinetic energies (ie in terms of ms, vs and ml and vl)? What is the expression for momenta of the two trucks? If the energies are equal, what is the ratio of the speeds vs/vl ?. Using that ratio, what is the ratio of their momenta?

  5. Mar 4, 2005 #4
    I thought I saw another post just like this...
  6. Mar 4, 2005 #5
    Maybe I didn't. Anyway, set up your problem using the equation for kinetic energy.
    Use m1 and m2... and to make things easy, let's say m2 = 1/2 m1.

    [tex]K.E. = \frac{1}{2}mv^2[/tex]

    Setting the two equation equal to each other...
    [tex]\frac{1}{2}m_{1}^{2}v_{1}^2 = \frac{1}{2}*\frac{1}{2}m_{1}v_{2}^2[/tex]

    The "m's" cancel and you get the velocities in terms of each other.
  7. Mar 5, 2005 #6
    yes there was a post exactly like this, i think he double posted it for one reason or another,

    but i simply said that whenever a mass is increased the momentum is also increased - P=MV

    of course i dont understand what they mean by larger truck, is it mass? or simply bigger?

    because you can get equal KEs but haveing one of them going slightly faster, and the other with more mass, that when plugged into the momentum equation, the larger one will have more

  8. Mar 5, 2005 #7
    You mean the math typesetting? That's Latex. This should get you started.
  9. Mar 5, 2005 #8


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    I assume you mean the larger truck has greater mass.

    Let :

    Larger truck : mass M, velocity V
    Smaller truck : mass m, velocity v


    [tex]\frac{1}{2}mv^2 = \frac{1}{2}MV^2[/tex]

    [tex]V = v\sqrt{\frac{m}{M}}[/tex] -- eqn (1)

    Momentum of the larger truck is MV, that of the smaller truck is mv.

    Now, using eqn (1)

    [tex]MV = Mv\sqrt{\frac{m}{M}} = v\sqrt{Mm} > v\sqrt{m^2} = mv[/tex]

    Hence the larger truck has the greater momentum.

    This doesn't apply in the trivial case when both trucks are at rest.
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