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How are Energy and Momentum related?

  1. May 8, 2007 #1
    I know that both energy and momentum are transferred from one object to another in collisions. But I'm not use to relating energy to collisions, rather I use momentum mostly. How are Energy and momentum related? I know that Kinetic energy has the same variables (mv) as momentum except that it has the extra (1/2) and squared sign on the mass.
  2. jcsd
  3. May 8, 2007 #2
    e= p^2 / 2m ....here e is energy.....p is momentum...and m is mass(either of teh system or the body)......
  4. May 8, 2007 #3
    so in order to get KE can you just mutiply momentum by 1/2m?? I know that E=pc.. but I haven't really looked into relativity yet. lol
  5. May 8, 2007 #4
    It is actually the velocity that is squared, not the mass as in 1/2mv^2.
    It is no longer a vector quantity unlike momentum. Generally, momentum is always conserved in the absence of an external force, where as energy may or may not be. So in the case of a simple collision thats elastic, energy is conserved, whereas in one that is inelastic, it is not.

    In some simple collisions such as where objects stick together, it is enough to use conservation of momentum. In others, even fairly simple ones, such as the following:

    a small block of mass m moving at velocity, v, collides with a heavier block of
    mass 3m, find the final velocities of each block.
    Using conservation of momentum,


    Two unknowns, one equation. A dead end. Thats where conservation of energy comes to the rescue, it gives us another equation so that both v1 and v2 can be solved for. That help at all?
  6. May 8, 2007 #5
    Thanks for your explanation. =] So can you just multiply the momentum of an object by (1/2)v to get its scalar Kinetic energy? Does this multiplying/dividing subtracting method work for all equations?
  7. May 9, 2007 #6
    Maybe looking at the relationships between the different quantities may help.

    F=dP/dt so P=integral of force wrt time, p=F*t

    Work=Integral (force*cos(theta)dx), under "ideal" conditions= F*distance

    Work (in absence of dissipation such as friction and change in potential energy)=kinetic energy

    so P*distance/time=KE

    So there are a lot of constraints. But yes, if you wanted to know the kinetic energy of a constant mass with a particular momentum, you could multiply by v/2. Was there a particular example of a problem you had in mind?
    That might be more hepful.
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