# Homework Help: Collision between two pucks, find the angle

1. Feb 14, 2010

### jahrollins

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. Feb 14, 2010

### ideasrule

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.)

3. Feb 15, 2010

### 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