Collision between two pucks, find the angle

[jahrollins]

Homework Statement

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.

Homework Equations

Conservation of linear momentum
m₁v_f1 + m₂v_f2 = m₁v_o1 + m₁v_o2

The Attempt at a Solution

So in the y:
m₁v_f1y + m₂v_f2y = m₁v_o1y + m₁v_o2y

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

In the x:
m₁v_f1x + m₂v_f2x = m₁v_o1x + m₁v_o2x

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.

 

[ideasrule]

That should be m₁v_f1 - m₂v_f2, since the two pucks head off in opposite directions.

The Attempt at a Solution

So in the y:
m₁v_f1y + m₂v_f2y = m₁v_o1y + m₁v_o2y

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

That should be 0.05(3sinΘa) - 0.1(2.5sinΘb), since the two pucks head off in opposite directions.

In the x:

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.

Solve the two equations you got for theta-a and theta-b. (Hint: use some trig identities.)

 

[jahrollins]

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