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Particles colliding

  1. Oct 6, 2007 #1
    A neutron (m= 1.6 x10-27) moving with speed Vo strikes a stationary particle of unknown mass and rebounds elastically straight back along its path with a velocity of .7Vo. What is the mass of the struck particle?

    2. Relevant equations
    P initial = P final
    Ke initial = Ke final

    3. The attempt at a solution
    alright i tried saying that the velocity imparted on the stationary particle was .3Vo but that gave me an answer of a mass of a neutron when the actual answer is 9.46 x 10-27.

    i know: v1i = Vo
    v2i = 0
    v1f = .7Vo

    but i am unsure as how to continue on from here. Please help.
  2. jcsd
  3. Oct 6, 2007 #2
    Let mass of the stationary particle be m and its velocity after collision be V. Write down the equations for conservation of momentum and kinetic energy (keep in mind the +/- sign of the velocities). Solve these to obtain m. [If you are able to write down the equations but do not know how to solve them to get m, post again writing down upto the step you are able to reach]
  4. Oct 7, 2007 #3
    yeah i understand what your saying:

    1/2m1Vo^s = 1/2m1v1f^s+ 1/2m2 .7Vo^s

    m1Vo = m1v1f + m2 .7Vo

    what i have done is rearrange it for one value and then substitute it into the other equation but this still gives me the wrong answer. any ideas?
  5. Oct 7, 2007 #4


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    You should recall that momentum is a vector quantity and hence, direction is important.
  6. Oct 7, 2007 #5
    1/2m1Vo^s = 1/2m1v1f^s - 1/2m2 .7Vo^s

    ahhh well a negative in place of a positive might help :P
    thanks hootenanny!
  7. Oct 7, 2007 #6


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    Careful though, I'm sure 1/2m2(0.7v02) should be 1/2m1(0.7v02). Assuming that m1 is your known mass
  8. Oct 7, 2007 #7
    argh its these small errors that really throw me off
    thanks again :)
  9. Oct 7, 2007 #8


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    No problem :smile:
  10. Oct 8, 2007 #9
    alright so i used

    m1Vo^2 = -m1 .7Vo^2 + m2 V2f

    v2f = m1Vo^2 + m1 .7Vo^2/m2

    then i subs the v2f into the kinetic formula equation and ended up with this:

    m1Vo^2 + m1 .7Vo^2/m2 = (m1Vo^2 + m1 .7Vo^2/m2)^2

    0 = m1Vo^2 + m1 .7Vo^2/m2

    alright, have i gone wrong somewhere and if so, where?
  11. Oct 8, 2007 #10


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    I'm not sure what you've done there, I think you may have got your equations mixed up. Kinetic energy is a scalar quantity; i.e. direction doesn't matter. And make sure you square the whole velocity. I'll start you off;

    [tex]m_1v_0 = m_1(-0.7\times v_0) + m_2v_2[/tex]

    Kinetic Energy
    [tex]m_1v_0^2 =m_1(0.7\times v_0)^2 + m_2v_2^2[/tex]
  12. Oct 8, 2007 #11
    hmm i see the directions with ke did stuff me up

    if the .7Vo velocity is negative, could you not just say that m1.7Vo as a whole is negative, and then rearrange as such?

    i try to rearrange it but keep stuffing it up.

    what should i do, rearrange on for v2f and then substitute it in the other? if so i need some more guidance

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