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Collision between two pucks, find the angle

  1. Feb 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Two hockey pucks collide with each other. Puck 1 has a mass of 0.050 kg and an initial velocity of 5 m/s in the x direction. It collides with puck 2 which is initially at rest and has a mass of 0.10 kg. The collision isn't head-on. After the collision, Puck 1 flies off in the north-east direction with a velocity of 3 m/s in an angle θA above the x-axis. Puck 2 flies off in the southeast direction with a velocity of 2.5 m/s in an angle θB below the x-axis. Find θA and θB.


    2. Relevant equations
    Conservation of linear momentum
    m1vf1 + m2vf2 = m1vo1 + m1vo2


    3. The attempt at a solution
    So in the y:
    m1vf1y + m2vf2y = m1vo1y + m1vo2y

    0.05(3sinΘa) + 0.1(2.5sinΘb) = 0 + 0
    0.15sinΘa + 0.25sinΘb = 0

    In the x:
    m1vf1x + m2vf2x = m1vo1x + m1vo2x

    0.05(3cosΘa) + 0.1(2.5cosΘb) = 0.05(5) + 0
    0.15cosΘa + 0.25cosΘb = 0.25

    I'm lost as to where to go from here.
     
  2. jcsd
  3. Feb 14, 2010 #2

    ideasrule

    User Avatar
    Homework Helper

    That should be m1vf1 - m2vf2, since the two pucks head off in opposite directions.

    3. The attempt at a solution
    That should be 0.05(3sinΘa) - 0.1(2.5sinΘb), since the two pucks head off in opposite directions.
    Solve the two equations you got for theta-a and theta-b. (Hint: use some trig identities.)
     
  4. Feb 15, 2010 #3
    Alright so since sin2Θ + cos2Θ = 1 then...

    For y:
    0.15sinΘa - 0.25sinΘb = 0
    15sinΘa = 25sinΘb
    225sin2Θa = 625sin2Θb

    For x
    0.15cosΘa + 0.25cosΘb = 0.25
    15cosΘa = 25 - 25cosΘb
    225cos2Θa = (25 - 25cosΘb)2
    225cos2Θa = 625 - 1250cosΘb +625cos2Θb

    Add together:
    225 = 625 - 1250cosΘb + 625
    -1025/-1250 = cosΘb
    Θb = 34.915 = 35 degrees

    Put into the other equation (15sinΘa = 25sinΘb)
    15sinΘa = 25sin(34.915)
    Θa = 72.541 = 73 degrees
     
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