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Homework Help: Question about collisions in 2-dimensions

  1. Feb 21, 2006 #1
    Q.
    A 0.150 kg billiard ball (A) is rolling toward a stationary billiard ball (B) at 10.0 m/s. After the collision, ball A rolls off at 7.7 m/s at an andle of 40.0 degrees clockwise from its original direction. What is the speed and direction of Ball B after the collision?

    Known:
    M of A = .150 kg
    M of B = .150 kg
    Vi of B = 10.0 m/s
    Vf of A = 7.7 m/s[70degrees clockwise]

    Implied:
    Vi of A = 0 m/s
    Ff = 0N

    Unknown:
    Vi of B = ?

    I started out using the law of conservation of energy but I messed up somewhere early and got fully lost...I know you need to use this method to conserve the momentum of each object but I just can't remember how, can someone start please?
     
  2. jcsd
  3. Feb 21, 2006 #2
    This is a purely conservation of momentum problem. Pick your x and y axes, initially you have some ammount of momentum in x and y due to the motion of ball A and in the final situation you have the same ammount of momentum in x and y due to the combined motions of A and B.
     
  4. Feb 21, 2006 #3
    The conservation of momentum

    [tex]m\vec{v_{iA}}=m\vec{v_{fA}}+m\vec{v_{fB}}[/tex]
     
    Last edited: Feb 21, 2006
  5. Feb 21, 2006 #4
    Well, using conservation of momentum should do it. Remember, though, that you should treat the vector components separately, i.e. work it out for a chosen x-direction and y-direction. Easiest would be to have ball A roll in the x-direction.


    EDIT: Wow, the answers keep rolling in. :smile:
     
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