Analyzing Two-Dimensional Collisions: Solving for Final Velocities

AI Thread Summary
The discussion focuses on solving a two-dimensional collision problem involving two pucks on an air-hockey table. Puck A, with a mass of 0.025 kg and an initial velocity of +5.5 m/s, collides with stationary puck B, which has a mass of 0.050 kg. The conservation of momentum is applied in both the x and y components to derive equations for the final velocities of both pucks. The initial momentum of puck A is calculated, and the conservation equations are set up but not fully solved. The participants emphasize the need to address the y-axis conservation equation to find a solution for the final velocities of both pucks.
chmilne
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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.
 
Last edited:
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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.
 
I'm not following you. Will you please elaborate and be more specific?
 
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