Question

Two identical pucks collide on an air hockey table. One puck was originally at rest.

(a) If the incoming puck has a speed of 6.00 m/s and scatters to an angle of 30.0

Answer 1

The momentum after collision must equal the momentum before. If you make X the direction of travel prior to collision then the moving puck has momentum in only the X direction. After collision there are both X and Y components. The X component after must equal the X component prior and the Y components after must be equal in magnitude and opposite in direction since the Y component before collision is 0.

u = speed of the original moving puck after collision (moving at 30 degrees to X)
v = speed of the other puck after collision (moving at 60 degrees to X)
m = mass of each puck

X direction.
Before = m*6
After = m*cos(30)*u + m*cos(60)*v

Equate before and after: m*6 = m*cos(30)*u + m*cos(60)*v
6 = u*sqrt(3)/2 + v/2

Y direction.
m*sin(30)*u = m*sin(60)*v
u/2 = v*sqrt(3)/2 ..... u = v*sqrt(3)

Combine the equations from before and after.
6 = v*sqrt(3)*sqrt(3)/2 + v/2
6 = v*3/2 + v/2 = 2*v
v = 3 and u = 3*sqrt(3)