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Momentum collision difficulty

  1. Dec 9, 2004 #1
    Please help me!
    I have been thinking on this problem

    A car of mass 450kg travels east at 4.5m/s, and collides with a 550kg truck infront of it traveling east at 3.7m/s. What are their velocity after the elastic collision.

    I know the energy formula and the momentum formula, and I know that when the energy formular is divided by the momentum formula, it came out that:

    the ' sign means prime/final

    But I can't figure it out anyway... due to my stupidity... because V'2 and V'1 is unknown.
    Would anyone help me? If so, please explain to me how you would do it using the energy and momentum formula.
    Last edited: Dec 9, 2004
  2. jcsd
  3. Dec 9, 2004 #2


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    Staff Emeritus
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    Conservation of momentum

    [tex]m_1v_1 + m_2v_2 = m_1v'_1 +m_2v'_2[/tex]

    Conservation of energy
    [tex]m_1{v_1}^2 + m_2{v_2}^2 = m_1{v'_1}^2 +m_2{v'_2}^2[/tex]

    2 equations, 2 unkowns.
  4. Dec 10, 2004 #3
    But there must be answer since its my textbook problem...
  5. Dec 10, 2004 #4

    Doc Al

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    Staff: Mentor

    Of course there's an answer. Just solve those two equations that Astronuc provided.
  6. Dec 10, 2004 #5
    HOW you solve it?
    V'2 and V'1 is unknown...

    I found a link http://physics.bu.edu/~duffy/py105.html where it has a similar problem situation under momentum and collision section in the Energy and Momentum section, but it didn't clear explain how to do it... can somobody help me!!!..........
    Last edited: Dec 10, 2004
  7. Dec 10, 2004 #6


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    [tex]1/2{m_1}{v_1}^2 + 1/2{m_2}{v_2}^2 = 1/2{m_1}{v'_1}^2 +1/2{m_2}{v'_2}^2[/tex]
    Conservation of KE has a coefficient of 1/2.

    Also just a hint, In an elastic collision the relative speeds of the two objects will be the same before and after the collision. Just opposite directions. :p And mass does not matter, unless your dealing with objects of considerable mass difference.
  8. Dec 10, 2004 #7

    Doc Al

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    Staff: Mentor

    Sure [itex]v'_1[/itex] and [itex]v'_2[/itex] are unknown--they are what you are trying to find. Luckily everything else is known and you have two equations.

    One way to solve the problem is to solve the two conservation equations together (as Astronuc was suggesting). In the momentum equation, solve for [itex]v'_2[/itex] in terms of [itex]v'_1[/itex]. Then plug that into the second equation and solve the quadratic for [itex]v'_1[/itex]. It's a bit tedious, but you can do it.

    An easier way is to make use of that equation for relative speeds that you posted: [itex]v_1 - v_2 = v'_2 - v'_1[/itex]. (This is what Tjl was suggesting, but you already knew it.) Combine that equation with the conservation of momentum equation. Two equations and two unknowns again, but no quadratic this time.
    Last edited: Dec 10, 2004
  9. Dec 10, 2004 #8
    thanx alot!!! Now I m enlightened by you people thanks aloT!!!!

    THANK YOU!!!!
    THANK YOU!!!!
    actually my teacher went over this... but i wasn't listening... and... i dont wana get a lecture... DANX ALOT!!!!
    Last edited: Dec 10, 2004
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