# Collision Momentum

#### neoncrazy101

1. The problem statement, all variables and given/known data
The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.0180 kg and is moving along the x axis with a velocity of +7.71 m/s. It makes a collision with puck B, which has a mass of 0.0360 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the speed of (a) puck A and (b) puck B.
http://edugen.wileyplus.com/edugen/courses/crs2216/art/qb/qu/c07/ch07p_30.gif

2. Relevant equations
m1vf1x+m2vf2x = m1vo1+m2vo2

0 = Mf1V1sin(65) - M2Vf2sin(37)

M1 = .0180
M2 = .0360
Vo1 = 7.71
Vo0 = 0

3. The attempt at a solution

.018(v)+.036(v) = .018(7.71)
.054v = .13878
v = 2.57. (Terribly wrong)

.018(Vf1)sin(65) - .036(Vf2)sin(37)

My problem is I'm getting so confused as to what I am solving for. I mean, I know I'm solving for the final speeds of puck A and puck B but I'm confused as to how to get there. I'm just plan lost.

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#### tiny-tim

Homework Helper
hi neoncrazy101!
.018(v)+.036(v) = .018(7.71)
.054v = .13878
v = 2.57. (Terribly wrong)
but your two vs are different

also, momentum is conserved in both the x direction and the y direction, so you need an equation for each

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