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Ball collision Momentum Problem

  1. Oct 16, 2007 #1
    Two balls, A and B, having different but unknown masses, collide. A is initially at rest and B has a speed v. After collision, B has a speed v/2 and moves at right angles to its original motion.
    a) Find the direction in which ball A moves after the collision.
    b) Can you determine the speed of A from the information given? Explain

    So, conversation of momentum

    lets let ball A have mass m, and ball B have mass M

    mVai + Mvi = mVaf + Mvf

    0 + Mv = mVaf + (1/2)Mv
    (1/2)Mv = mVaf
    Vaf = (Mv)/2m

    It seems I need to solve B before solving A

    Since ball B moves at a right angle to its initial velocity, if we say it was moving only in the x direction initally, it now has no x coordinate of velocity, so all of its x momentum will be transferred to ball A

    so cosx = v/[(Mv)/2m]
    cosx = 2mv/Mv
    cosx = 2m/M
    x = arccos(2m/M)

    Have I done all of this right?

    Thanks
     
  2. jcsd
  3. Oct 16, 2007 #2
    It would be best to break the problem down into two perpendicular directions. One direction in the original motion of ball B, and the other perpendicular.
    In each of the two directions linear momentum must be conserved. So the linear momentum in each direction, after the collision, must equal the total momentum before the collision, but in the same direction; this is why your answer will not be correct. And you need to be careful with your signs.
     
  4. Oct 16, 2007 #3
    ok,
    in the x direction

    mVai + Mvi = mVaf + Mvf

    0 + Mv = mVaf + 0
    Vaf = Mv/m

    in the y direction

    mVai + Mvi = mVaf + Mvf

    I know Vf for B needs to be v/2
    but what can I use for initial momentum

    do i need to use sin/cos for initial velocity components?

    in the x direction

    mVai + Mvi = mVaf + Mvf

    0 + Mvcosx = mVaf + 0
    Vaf = Mvcosx/m

    in the y direction

    mVai + Mvi = mVaf + Mvf
    0 + Mvsinx = mVaf + Mv/2
    mVaf = Mvsinx - Mv/2
    mVaf = Mv(sinx - 1/2)
    Vaf = Mv(sinx - 1/2)/m

    is this better?

    thanks
     
  5. Oct 16, 2007 #4
    Did any of the balls have momentum in the y-direction to start with? It should be apparent what the initial momentum in the y-direction is.
     
  6. Oct 16, 2007 #5
    A initially at rest
    B with inititial velocity v

    initial momentums:
    x direction
    vcosx
    y direction
    vsinx

    so

    in the x direction

    mVai + Mvi = mVaf + Mvf

    0 + Mvcosx = mVaf + 0
    Vaf = Mvcosx/m

    in the y direction

    mVai + Mvi = mVaf + Mvf
    0 + Mvsinx = mVaf + Mv/2
    mVaf = Mvsinx - Mv/2
    mVaf = Mv(sinx - 1/2)
    Vaf = Mv(sinx - 1/2)/m
     
  7. Oct 16, 2007 #6
    B was the only ball moving initiallt, correct? And we defined the x-direction to be the ball B was moving in initially, so how can there be any initial momentum in the y-direction? You may have to read that a few times over before you get it.
     
  8. Oct 16, 2007 #7
    so, there can be no velocity or momentum in the y direction afterwards?
    so the right angle afterwards is irrelevant, and the change in momentum is just a complete transfer in the x direction?

    but say the ball still only has momentum in the x direction, but ball B hits ball A not straight on

    00
    00 00
    00

    wouldn't the ball be deflected in a way that has x and y momentums? such as in a game of pool?
     
  9. Oct 16, 2007 #8
    You are correct in saying it will have x- and y- components in its momentum. Remember I said be careful with your signs? If B has momentum Mvf in the positive direction, the A must have momentum -Mvf, i.e. in the negative direction. The total still adds up to zero! So the momentum in the y-direction IS conserved.
     
  10. Oct 16, 2007 #9
    ahh, how could I forget about the signs

    in the x direction

    mVai + Mvi = mVaf + Mvf

    0 + Mv = mVaf + 0
    Vaf = Mv/m

    in the y direction

    mVai + Mvi = -mVaf + Mvf
    0 + 0 = mVaf + Mv/2
    -mVaf = Mv/2
    Vaf = -Mv/2m

    Vbf = Mv/2m

    so for direction

    cosx = (Mv/m)/(-Mv/2m)
    cosx = -2

    something seems wrong still
     
  11. Oct 16, 2007 #10
    do i need to use sinx and coxs for the velocity components?
     
  12. Oct 16, 2007 #11
    anyone?
    i'm still lost on this problem

    thanks
     
  13. Oct 17, 2007 #12
    So x-component = (M/m)v
    ....y-component = -(M/2m)v

    What trig function involves x and y? Remeber that cos=adjacent/hypotenuse
    i.e. cos=x/r
    but we have x and y. You are not evaluating it with the correct trig function. Try to think of the trig function that involves x/y or y/x.
     
  14. Oct 17, 2007 #13
    tanx = (M/m)v/-(M/2m)v
    tanx = -2
    x = -63.4 degrees below horizontal
     
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