Avery big hockey puck with mass 2.5 kg traveling 20 degrees north of east at 10.0 m/s strikes a puck with a mass of 4.0 kg traveling north at 2 m/s. the 2.5 kg puck exits the collision in a direction that is 30 deg. north of east at a velocity of 8.0 m/s. what is the magnitude and direction of the 4.0 kg puck's velocity?

Velocity = 0.47 m/s

Direction = 65 degrees South of east.

(295 degrees counter clockwise from + X axis).

Explanation:

Conservation of momentum along the X and Y directions can be used to determine the velocity of the 4 kg puck.

Along the X direction momentum conservation is as follows:  

2.5 cos 20 + 4.0 cos 90 = 2.5 cos 30 + 4 v cos α

⇒  v cos α = 2.5 ( cos 20 - cos 30) ÷ 4 = 0.46 m/s

2.5 sin 20 + 4 sin 90 = 2.5 sin 30 + 4 v sin α

v sin α = (2.5 sin 20 - 2.5 sin 30) ÷ 4 =  -0.0987 m/s

v = √(v² cos² α + v² sin² α) = √ (0.46² + (-0.0987)² = 0.47 m/s

Direction = α = tan⁻¹ (-0.0987/0.46) = -65 degrees = 65 degrees clock wise from +X axis or 295 degrees counter clockwise from +X axis

hypothesising or postulating