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Show that there are three collisions and find all find velocities

  1. Oct 17, 2007 #1
    There are the 3 spheres in a line, of masses m , m and M. The 2 spheres on the right are slightly seperated and initially at rest, the left sphere is incident with speed Vo. Assuming head on elastic collisions:
    a) if M =< m , show that there are two collisions and find all the final velocities
    b) if M > m, show that there are three collisions and find all find velocities


    if M =< m , there will be two collisions because m will hit m which will hit M and all will continue in the +x direction

    if M > m, there will be 3 collisions, as m will hit m which will hit M and bounce back hitting the first m again.

    mv1i + mv2i = mv1f + mv2f
    mVo + 0 = mv1f + mv2f
    Vo = v1f + v2f
    v1f = Vo - v2f
    v2f = Vo - v1f

    and then

    mv2i + Mv3i = mv2f + Mv3f
    m(Vo - v1f) + 0 = mv2f + Mv3f

    I'm getting sortof lost here, is there an easier was to work this into 2 equations

    Thanks
     
  2. jcsd
  3. Oct 17, 2007 #2
    You have to PROVE these statements, you cannot just assert them.

    Conservation of linear momentum on its own is NOT enough to solve this problem. But you are told that the collision are ELASTIC (an elastic collision means kinetic energy is conserved). Aha! you should say.
    After each single collision, you have to use conservation of kinetic energy to find the exact solution (there is one, only one). You do this for each solution. You might have to use equations of motions to decide whether there are 2 or 3 collisions.
     
  4. Oct 17, 2007 #3
    " You might have to use equations of motions to decide whether there are 2 or 3 collisions."

    I'm not sure what you mean by equations of motion

    Do I want to use conservation on energy instead of momentum? Or momentum and then kinetic energy?

    KEi = KEf
    (1/2)mVo^2 + 0= (1/2)mv1^2 + (1/2)mv2^2
    Vo= sqrt(v1^2 + v2^2)

    or

    v1f = Vo - v2f
    v2f = Vo - v1f

    KEi = KEf
    (1/2)mv^2 = (1/2)mv^2
    (1/2)mVo^2 = (1/2)m(Vo - v2f)^2 + (1/2)m(Vo - v1f)^2
    Vo = sqrt[(Vo - v2f)^2 + (Vo - v1f)^2]

    thanks
     
  5. Oct 17, 2007 #4
    I'm still stuck on this one, why do I need to use Kinetic energy?
     
  6. Oct 18, 2007 #5
    oops, i meant to solve for final velocities

    KEi = KEf
    (1/2)mVo^2 + 0= (1/2)mv1f^2 + (1/2)mv2f^2
    V1f= sqrt(vo^2 - v2^2)

    or

    v1f = Vo - v2f
    v2f = Vo - v1f

    KEi = KEf
    (1/2)mv^2 = (1/2)mv^2
    (1/2)mVo^2 = (1/2)m(Vo - v2f)^2 + (1/2)m(Vo - v1f)^2
    v2f = sqrt[(Vo)^2 - (Vo - v1f)^2]

    thanks
     
  7. Oct 18, 2007 #6
    So you're all sorted?
     
  8. Oct 18, 2007 #7
    is that how i need to do it:

    v1f = Vo - v2f
    v2f = Vo - v1f

    KEi = KEf
    (1/2)mv^2 = (1/2)mv^2
    (1/2)mVo^2 = (1/2)m(Vo - v2f)^2 + (1/2)m(Vo - v1f)^2
    v2f = sqrt[(Vo)^2 - (Vo - v1f)^2]

    and then again for each following collision?
     
  9. Oct 18, 2007 #8
    Vo = v1f + v2f..................(1)

    KEi = 1/2m(Vo)^2
    KEf = 1/2m(v1f)^2 + 1/2m(v2f)^2
    KEi = KEf
    1/2m(Vo)^2 = 1/2m(v1f)^2 + 1/2m(v2f)^2
    Vo= sqrt[(v1f)^2 + (v2f)^2]
    Equate this with (1):
    sqrt[(v1f)^2 + (v2f)^2] = v1f + v2f
    (v1f)^2 + (v2f)^2 = (v1f)^2 + (v2f)^2 + 2(v1f)(v2f)

    So: 2(v1f)(v2f) = 0
    So: v1f = 0 or v2f = 0
    But how can v2f = 0? This means it collided with the first ball and remained stationary! So obviously:
    v1f = 0...........(2)
    Substitute (2) in (1):
    v2f = Vo
    So the balls m and m collide once, the first one comes to rest, the second one continues at Vo. This is what we could have guessed as they have the same mass and KE is cnserved.
    Do a similar analysis for the second collision.
     
    Last edited: Oct 18, 2007
  10. Oct 18, 2007 #9
    ok that makes sense
    i will be able to complete this problem,

    thanks :)
     
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