- #1

Dakkers

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## Homework Statement

Two hockey pucks of equal mass undergo a collision on a hockey rink. One puck is initially at rest, while the other is moving with a speed of 5.4m/s. After the collision, the velocities of the pucks make angles of 33 degrees and 46 degrees relative to the original velocity of the moving puck. Determine the speed of each puck after the collision.

(Assuming the moving puck is object A and the stationary puck is object B)

Ma = Mb

Va1 = 5.4m/s

Vb1 = 0m/s

Deflection Angle of A = 33

Deflection Angle of B = 46

## Homework Equations

p(before) = p(after)

## The Attempt at a Solution

Okay, the first thing I did was broke it into components as it's in two dimensions. For x,

p(before) = p(after)

(Ma)(Va1x) + (Mb)(Vb1x) = (Ma)(Va2x) + (Mb)(Vb2x)

(5.4) + 0 = Va2x + Vb2x

5.4 = Va2x + Vb2x

for y,

p(before) = p(after)

(Ma)(Va1y) + (Mb)(Vb1y) = (Ma)(Va2y) + (Mb)(Vb2y)

0 + 0 = Va2y + Vb2y

Va2y = -Vb2y

And then I don't know where to go from there. I've thought about trig and using Pythagorean theorem but there's too many variables.