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Momentum conserved but Kinetic Energy not conserved?

  1. Oct 15, 2014 #1
    I know that momentum is conserved when no "net" external forces such as Friction, Gravity, and air resistnace normal force; However, during inelastic collisions it says kinetic energy is not conserved because of external forces as well.

    1. Are the external forces different for mechanical energy and conservation of momentum?
    2. During inelastic collisions, KE is not conserved, therefore the velocity's will have values to make KEi not equal to KEf, How does momentum stay conserved if velocity is different for KineticInitial and Kineticfinal?

    Maybe the momentum takes into account the velocity change from mechanical energy and also has different external forces and therefore momentum stays conserved despite mechanical energy not being conserved....
     
  2. jcsd
  3. Oct 16, 2014 #2
    In inelastic collisions, work will be done deforming the objects. This will cause the sums of the initial KEs and final KEs to be different. Another way of writing KE is p^2/2m, and from that you can see that if the sums of the KEs are different, so will be the momenta.

    does that help?
     
  4. Oct 16, 2014 #3

    A.T.

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    Look at the definitions of KE and momentum. Calculate some examples.
     
  5. Oct 16, 2014 #4

    ehild

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    The sum of all forces acting on a system is equal to the sum of the external forces as the internal forces cancel each other. But the total energy of the system changes by the work of the external forces and also by the work of the internal ones.

    Think of two people A an B, on ice, pushing each other. If the force from the ice is negligible, the forces are FBA and FAB , the force A pushes B and the force B pushes A. According to Newton's Third Law these forces are equal in magnitude and of opposite direction.
    The total change of momentum in dT time is the net force times dt. The net force is zero, the net momentum does not change.
    But the separate momenta of the persons do change. Assuming both persons in rest initially, B is pushed by A with force F acquires PA=mAVA momentum in dT time. A is pushed by B, with force -F and it gains PB= mBVB=-Fdt momentum. PB=-PA, but the kinetic energies are KEA=PA2/(2mA) and KEB= PB2/(2mB ) The kinetic energy of the whole system is KEA+KEB, different from the initial one.

    ehild
     
  6. Oct 16, 2014 #5

    bobie

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    When p and KE are both conserved there is only one possible solution. When only p is conserved And ##E_i > E_f## there are many possible solutions.

    A change of KE without a change of momentum is not only possible but very frequent, because p = mv momentum varies linearly and KE quadratically. You can get the same product by a wide range of factors: 6 = 6*1, = 3*2, = 2*3, = 1*6, = 0.5*12, etc., different factors give same momentum

    All these factors give same values for m*v, but as the figure for v must be squared, you get all different values between momentum and energy, therefore the same factors give momentum = 6, but KE =3, =6, =9, =18, =72, etc, same momentum corresponds to many different values of KE
     
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