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Homework Help: 2D Conservation of Momentum

  1. Mar 24, 2010 #1
    1. The problem statement, all variables and given/known data

    Object A has a mass of 2.0 kg and an initial velocity of 2.5 m/s. It strikes Object B, which is at rest and has a mass of 2.0 kg as well.

    After the collision, the objects travel in different directions, with Object B travelling at an angle of 44 degrees from its original position.

    What is the velocity of Object B after the collision and what is the displacement angle of Object A from the point of collision?

    2. Relevant equations

    Momentum before collision = momentum after collision

    Px = (m1 * vf1 * cosX) + (m2 * vf2 * cosY)

    Py = (m1 * vf1 * sinX) + (m2 * vf2 * sinY)

    KE before collision = KE after collision

    KEf = (.5)(m1)vf12 + (.5)(m2)vf22

    3. The attempt at a solution

    I first calculated the components of momentum and kinetic energy before impact.

    Px = 2 kg * 2.5 m/s = 5.0 kg*m/s
    Py = 0 kg*m/s
    KE = .5 * 2 kg * 2.5 m/s = 6.25 J

    Then, I setup equations relating the objects post-impact to the momentum and energy they should have.

    5 = 2*vf1*cosX + 2*vf2*cos44
    0 = 2*vf1*sinX + 2*vf2*sin44

    6.25 = vf12 + vf22

    I've tried using substitution to solve for one of the variables, but each time I end up getting arcsines within cosines equaling sines. And I really don't know how to solve from there.

    Am I at least on the right track? Should I solve for the angle first? Does it matter?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 24, 2010 #2

    kuruman

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    Gold Member

    This is good so far. Now suppose you took the first two (momentum conservation) equations and you solved for the components of the momentum of mass 1 in terms of the other quantities. In other words, put everything that has subscript 1 on one side and everything that has subscript 2 or is constant on the other side. What do you get?
     
  4. Mar 24, 2010 #3
    Awesome. Thanks!

    I super simplified the problem following your post:

    M = m1v1 + m2v2

    Mx = m1v1x + m2v2x = 5
    My = m1v1y - m2v2y = 0

    KE = .5m1v12 + .5m1v22

    KE = (.5*m1*v1x2 + v1y2) + (.5*m2*v2x2 + v2y2)

    Taking your advice, I basically got:

    v1y = v2y

    v1x = [tex]\frac{5}{m}[/tex] - v2x

    Plugged it all in and got my answers! Thanks again.
     
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