A sticky hockey puck slides along a frictionless, horizontal ice surface at 75 m/s in the +x direction. It collides with a second sticky puck originally sliding at 25 m/s in the -y direction. They collide perfectly inelastically. If the first puck has a mass twice that of the second, what is the angle of their final velocity vector below the +x direction?
A) 37 degrees
B) 9.5 degrees
C) 49 degrees
D) 42 degrees
E) 34 degrees

Question

Quizplus · Accepted Answer

The final velocity vector's angle can be found using the arctan of the ratio of the y-component to the x-component of the final velocity. Since the collision is perfectly inelastic, the pucks stick together, and we use conservation of momentum to find the components of the final velocity. The x-component of the final velocity (Vx) is due to the first puck only, and the y-component (Vy) is due to the second puck only. Given m1 = 2m and m2 = m, Vx = (2m*75)/(2m+m) = 50 m/s, and Vy = (m*25)/(2m+m) = 16.67 m/s. The angle θ = arctan(Vy/Vx) = arctan(16.67/50) ≈ 18.43 degrees. However, since the question asks for the angle below the +x direction and the calculation was based on standard trigonometric functions, it seems there was a mistake in the calculation process. Correctly calculating the angle should involve using the given velocities and masses to find the resultant velocity's direction accurately. Given the options and the nature of the mistake, the correct approach would involve recalculating the angle considering the conservation of momentum in both directions and the mass ratio correctly.

A Sticky Hockey Puck Slides Along a Frictionless, Horizontal Ice