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Homework Help: Conservation of Momentum and Kinetic Energy

  1. Nov 20, 2011 #1
    There was a question in my textbook that howed a graph of momentum (a straight line because it is constant) and a graph of kinetic energy before and after a collision. After the collision the energy was less than the kinetic energy it stared out with. Doesn't that mean that speed has decreased and that momentum too had decreased since the speed of one or both objects have decreased?
  2. jcsd
  3. Nov 20, 2011 #2
    Note that momentum depends on BOTH the speed and the mass.

    One can give better help if more details of the graph or problem are given.
  4. Nov 20, 2011 #3

    But if no mass is lost, then it would stay constant right? Then if the speed decreases, so does momentum ... ? :?
  5. Nov 20, 2011 #4
    The momentum should always be the same before and after a collision, its a vector quantity, so it also has direction. If you are working in 1D, this means it can have a positive as well as negative value. Before the collision, the initial momentum should be pi=p1+p2, where p1 and p2 are the momentum of the 1st and 2nd object respectively. After the collision the final momentum should be pf=p1'+p2', where p1' is the momentum of the first object and p2' is the momentum of the second object, after the collision. Since momentum is conserved you have pi=pf=p1+p2=p1'+p2'. So, the momentum of the individual particles 1 and 2 are different before and after the collision, but the total momentum is always the same.

    The Kinetic energy is different. It is a scalar quantity and always positive. It can change depending on whether you have an elastic collision (energy is the same before and after) or if you have an inelastic collision (energy is different before and after).
  6. Nov 20, 2011 #5
    Ok then lets consider an inelastic collision. Kinetic energy is not conserved, so the total kinetic energy before is different after. Wouldn't that mean there is a different speed before and after. And if the masses stay the same, and the velocities change, then how would momentum be conserved? (I don't want to say that momentum isn't conserved because that's a general law).
  7. Nov 20, 2011 #6
    Although the masses of the two trolleys are constant, yet the distribution of speeds amongst them before and after will keep the total momentum constant.For example the bigger trolley may decrease its speed but the smaller one may increase it just enough to keep the total momentum constant.
  8. Nov 20, 2011 #7
    ^ But on the graph, the total kinetic energy of the system drops after the collision.
  9. Nov 20, 2011 #8
    If the two things your are colliding stick together before and after the collision, then you would have [itex]KE_i=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2[/itex] while after [itex]KE_f=\frac{1}{2}(m_1+m_2)v_f^2[/itex] while the momentum would be [itex]p=p_i=m_1v_1+m_2v_2=p_f=(m_1+m_2)v_f[/itex] so you would get [itex]v_f=\frac{p}{m_1+m_2}=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/itex] and inserting this into [itex]KE_f[/itex] we get [itex]KE_f=\frac{1}{2}(m_1+m_2)\frac{(m_1v_1+m_2v_2)^2}{(m_1+m_2)^2}=\frac{1}{2}\frac{(m_1v_1+m_2v_2)^2}{m_1+m_2}[/itex] which is clearly not equal to [itex]KE_i[/itex] in general.

    If I take for example [itex]m_1=m_2=1 kg[/itex] and [itex]v_1=1 m/s, v_2=0 m/s[/itex] then we get [itex]p= 1 kg\cdot m/sec[/itex], [itex]KE_i=1/2 J[/itex] and [itex]v_f=1/2 m/sec[/itex] so that we finally get [itex]KE_f=\frac{1}{2}(2)\left(\frac{1}{2}\right)^2=1/4 J[/itex]. So [itex]KE_f<KE_i[/itex]
    Last edited: Nov 20, 2011
  10. Nov 20, 2011 #9
    And this is only true for completely inelastic collisions. But what about if it was just inelastic, so the two masses don't stick together. They bounce off each other but the total kinetic energy after the collision is not equal to the total kinetic energy before?
  11. Nov 20, 2011 #10
    Let me give an example.
    A big mass M with initial speed u hits a stationary smaller mass m so that after the impact the two masses move together with speed v in the same direction as u.

    Then by conservation of momentum: Mu = (M+m)v.

    [itex]\frac{KE_{f}}{KE_{i}}[/itex] = [itex]\frac{\frac{1}{2}(M+m)v^{2}}{\frac{1}{2}Mu^{2}}[/itex]

    But v = Mu/(M+m)


    [itex]\frac{KE_{f}}{KE_{i}}[/itex] = [itex]\frac{\frac{1}{2}(M+m)((Mu)/(M+m))^{2}}{\frac{1}{2}Mu^{2}}[/itex] = M/(M+m) which is less than 1.

    That is althought the total KE decreased yet the momentum reamined constant.
  12. Nov 20, 2011 #11
    Those are both examples of completely inelastic collisions. What about just inelastic where the two masses don't stick together ... ?
  13. Nov 20, 2011 #12
    I'm beginning to understand completely inelastic with the math ... thanks guys.
  14. Nov 20, 2011 #13
    Then that would be an elastic collision, elastic collisions conserve both momentum and energy (they don't stick together), while inelastic collisions conserve momentum, but not energy. Doesn't this answer your problem? a.) the first Blue line is momentum, since its constant. b.) its an inelastic collision because the KE is different before than after.
  15. Nov 20, 2011 #14
    "When the tennis balls collide, the total kinetic energy of the system after the collision
    is not equal to the total kinetic energy of the system before the collision. This is an
    inelastic collision."

    "When two objects stick together during a collision, as is the case with the putty
    balls, we have a completely inelastic collision. The decrease in total kinetic energy
    in a completely inelastic collision is the maximum possible."

    So those mathematical processes you implemented above work for both kinds of inelastic collisions?
  16. Nov 20, 2011 #15
    Well, whether they stick together or not, you have heat generated in the process that takes away the remainder of the energy (which you are not accounting for), and also the sound they make will to a much lesser degree also remove some energy of the system
  17. Nov 20, 2011 #16

    D H

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    Only elastic collisions conserve mechanical energy. Any collision that does not conserve mechanical energy is, by definition, an inelastic collision.
  18. Nov 20, 2011 #17
    So if kinetic energy is lost, how does it make sense that m1v1 + m2v2 = m1v1' + m2v2' if the speed of both of these two objects or at least of of them has decreased? The transfer has caused some transfer and some loss, so the speeds won't match up in the equation? I'm so confused :O :S :( :$.
  19. Nov 20, 2011 #18

    D H

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    Energy and momentum are conserved in an isolated system. Collisions can be modeled as an isolated system because there is little time for the colliding objects to transfer energy or momentum to the environment.

    Anticipating your question, so why isn't kinetic energy conserved? The answer is that the law of conservation of energy says that energy is conserved. There is no law of conservation of kinetic energy. There are lots of other forms of energy besides the kinetic energy of the colliding objects. The crack of the bat when a baseball player hits the ball: That sound is energy. The collision between ball and bat is not an elastic collision. The crunching and bending that occurs during an automobile accident indicate that that collision also is not elastic. The sounds, the bending and breaking, and the heat generated by the collision are all examples of kinetic energy being converted to some other form of energy.

    Momentum, on the other hand, is momentum.
  20. Nov 20, 2011 #19
    Assume all velocities mentioned are in the same direction.
    Let mass M with velocity 2u hit mass m which was moving with velocity u. Now let us choose the ratio of the masses M and m so that after impact mass M moves with velocity 1.5u and the other mass m moves with velocity v.

    By conservation of momentum one can find v interms of M, m and u.

    Then I suggest that the OP will try to find the ratio final KE/initial KE.
  21. Nov 20, 2011 #20


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    Say you have two equal masses approaching with speed V. They collide with a loud smack and bounce off each other. By symmetry, you know they will have the same speeds. Let's call it v. The sound wave from the collision carries off some energy, so the masses had to lose some energy. It's an inelastic collision.

    Before the collision, the total momentum was 0 = mV + m(-V). After the collision, the momentum is also 0 = mv + m(-v). Momentum is conserved, but kinetic energy isn't.
  22. Nov 20, 2011 #21
    Maybe better to use that after the collision the momentum is 0=mv'+m(-v'), since if the velocity v was the same before and after, how could KE not be conserved?
  23. Nov 20, 2011 #22


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    V and v aren't the same variable.
  24. Nov 20, 2011 #23
    Oh, I am completely sorry. I did not notice you capitalized v in the first part. I thought you just rewrote the same thing
  25. Nov 20, 2011 #24


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    No worries. It probably would have been less prone to misunderstanding if I had used primes, but I just don't like using them for some reason. :wink:
  26. Nov 20, 2011 #25
    I think it makes sense to me now; I'm beginning to understand a little. Perhaps I need to re-read the entire thread the strengthen my understanding. Thanks everyone!
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