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Homework Help: Momentum of ball on a pool table

  1. Jun 19, 2017 #1
    1. The problem statement, all variables and given/known data

    Picture looks like this:
    ----------------------- ○ (empty space) ●
    (Cue) (White ball) (Black Ball)

    Mass: Cue (595g), White Ball (170g), Black Ball (155g)

    Question: On the given pool table, the white ball bounces off of the black ball and moves in the direction opposite to its original direction. If the speed of the white ball immediately after the collision is 3.13m/s, then the speed of the black ball immediately after the collision is _______m/s. (Record your three-digit answer)

    2. Relevant equations
    Conservation of Momentum

    3. The attempt at a solution
    0=(-3.13m/s)×(170g) + (155g)×V
    532.1kgm/s = 155g×V
    V= 3.43m/s
    Apparently this is not the answer... What am I doing wrong??

    4. Keybook Solution

    MV (1) + MV (2) = MV'(1) + MV'(2)
    170g×(3.13m/s) + 0 = 170g×(-0.147m/s) + (155g)V '(2)
    V'(2) = 3.59m/s

    How do you get (-0.147m/s)???
  2. jcsd
  3. Jun 19, 2017 #2


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    The problem statement seems incomplete. What is the speed of the white ball before the collision?
  4. Jun 19, 2017 #3
    Are you sure it isn't supposed to be the speed immediately before the collision?
  5. Jun 19, 2017 #4


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    If the black ball is stationary and the white ball is more massive than the black then there is no way for an elastic collision to end with the white ball reversing course. [Conservation of momentum would require the black ball to move away rapidly. Its resulting kinetic energy would exceed that of the impacting white ball]

    If one assumes an elastic collision and a black ball more massive than the white then the problem becomes solvable.
  6. Jun 19, 2017 #5


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    That equation says the system has no momentum, but the white ball must have had some before the collision.
    As jjbriggs says, the masses must be wrong. Are you sure you quoted them correctly? Try swapping them around.
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