Analyzing Two-Dimensional Collisions: Solving for Final Velocities

[chmilne]

The Problem:

The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of 0.025 kg and is moving along the x-axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.050 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 final speed of (a) puck A and (b) puck B.

Work Done Thus Far:

I know that before the collision, puck A has momentum in its direction of motion, and all the energy (kinetic). After the collision the momentum of the two pucks must equal the original momentum of the first puck and the sums of their energies must equal the initial energy of puck A. So, here's what I've come up with.

P_{O1} = M_1V_{O1}
P_{O1} = .025kg * 5.5m/s
P_{O1} = .1375 kg*m/s

X-Component
M_{f1x}V_{f1x} + M_{f2x}V_{f2x} = M_{o1x}V_{o1x} + M_{o2x}V_{o2x}
.025kg(V_{f1x}) + .05kg(V_{f2x}) = .025kg(Vo1x) + .05kg(Vo2x)
.025kg(V_{f1x}) + .05kg(V_{f2x}) = .025kg(5.5m/s) + .05kg(0m/s)
.025kg(V_{f1x}) + .05kg(V_{f2x}) = .025kg(5.5m/s) + 0m/s
.025kg(V_{f1x}) + .05kg(V_{f2x}) = .1375kg*m/s
.025kg(V_{f1}cos65) + .05kg(V_{f2}cos37) = .1375kg*m/s

Y-Component
M_{f1y}V_{f1y} + M_{f2y}V_{f2y} = M_{o1y}V_{o1y} + M_{o2y}V_{o2y}
.025kg(V_{f1y}) + .05kg(V_{f2y}) = .025kg(0m/s) + .05kg(0m/s)
.025kg(V_{f1y}) + .05kg(V_{f2y}) = 0
.025kg(V_{f1y}) + .05kg(V_{f2y}) = 0
.025kg(V_{f1y}) + .05kg(V_{f2y}) = 0
.025kg(V_{f1}sin65) + .05kg(V_{f2}sin37) = 0



So,

.1375 kg*m/s = (.025kg * V_{f1}cos(65)) + (.05kg(V_{f2}cos(37))

I'm not sure what I missing to solve for the velocity of both the pucks.

 

[vsage]

You seem to have ignored your y-axis conservation equation. You have two equations and two unknowns, so a solution can be found fairly easily through substitution.

 

[chmilne]

I'm not following you. Will you please elaborate and be more specific?