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Homework Help: Completely Inelastic collision

  1. Apr 8, 2005 #1
    Please help.

    Two particles move perpendicular to each other until they collide. View Figure Particle 1 has mass m and momentum of magnitude 2p, and particle 2 has mass 2m and momentum of magnitude p. Note: Magnitudes are not drawn to scale in any of the figures.

    a. Suppose that after the collision, the particles "trade" their momenta, as shown in the figure. That is, particle 1 now has magnitude of momentum p, and particle 2 has magnitude of momentum 2p; furthermore, each particle is now moving in the direction in which the other had been moving. How much kinetic energy, K_lost, is lost in the collision?

    b.Consider an alternative situation: This time the particles collide completely inelastically. How much kinetic energy K_lost is lost in this case?

    I've already found the answer to part a, which is 3p^2/4m, and as I was trying to work out the answer to b, I ended up with the asme answer, and I followed all the "instructions" in my textbook. Please help.

    Attached Files:

  2. jcsd
  3. Apr 8, 2005 #2


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    Your answer for part (a) was correct (although I had to calculate it twice to get the right answer!).

    In part (b), the particals collide "completely inelastically" which means that the "stick together" and have the same after-collision speed.

    Set up your coordinate system so that partical A is moving in the positive direction along the x-axis (so its momentum is <2p, o>) and partical B is moving in the negative direction along the y-axis (so its momentum is <0, p>). Before the collision, their total momentum is <2p, p>. After the collision, the move off together: let's call their velocity vector <vx, vy>. Since their total mass is now 3m we have <2p, p>= <3mvx,3mvy> so we must have vx= 2p/(3m) and vy= p/(3m). v2 is now (3p/3m)2+ (p/3m)2= (10p2)/(9m2) and the kinetic energy is
    (1/2)(3m)(10p2)/9m2)= (150/81)p2/m.

    Of course, the total initial kinetic energy was still (9/4)p2/m so the energy lost is (9/4- 150/81)p2/m= ((720-600)/324)p2/m = (30/81)p2/m.
  4. Apr 9, 2005 #3
    Ok...thanks for your help, but that answer is incorrect.
  5. Apr 11, 2005 #4
    Does anyone else know. Thank you.
  6. Apr 12, 2005 #5
    I just keep on ending up with the same answer.
  7. Apr 12, 2005 #6

    hmmm why did i get 3p^2/2m
  8. Apr 12, 2005 #7
    That is what I got as the final KE for the second question, but subtracting it from the initial KE to get the KE lost gave me the same answer as the first queston.
  9. Apr 12, 2005 #8
    Okay, we have established that the initial kinetic energy is

    [itex]K_1 = \frac{9p^2}{4m}.[/itex]

    The initial x-component of momentum is [itex]P_{1x} = 2p[/itex] and the y-component is [itex]P_{1y} = p[/itex]. After collision [itex]P_{2x} = 3mv_x[/itex] and [itex]P_{2y} = 3mv_y[/itex]. Conservation of momentum states that

    \vec{P}_1 & = \vec{P}_2 \\
    |\vec{P}_1| & = |\vec{P}_2| \\
    4p^2 + p^2 & = 9m^2v_x^2 + 9m^2v_y^2 \\
    5p^2 & = 9mv^2 \\
    v ^2 = \frac{5p^2}{9m^2}.

    Substituting into the kinetic energy equation you get

    [itex]K_2 = \frac{5p^2}{6m}[/itex]

    and therefore

    [itex]K_\mathrm{lost} = \frac{17p^2}{12m}.[/itex]
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