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 totally inelastically.If the first puck has a mass equal to half that of the second,what is the angle of their final velocity vector below the +x direction?
A) 9.5 degrees
B) 42 degrees
C) 34 degrees
D) 49 degrees
E) 37 degrees

Question

Quizplus · Accepted Answer

Initial momentum in the x-direction = (0.5)mv = (0.5)(m/2)(75) = 937.5 (mass cancels out)
Initial momentum in the y-direction = mv = (m)(-25) = -25m
Total initial momentum = 937.5i - 25j

Since the collision is inelastic, the two pucks stick together and move as one.

Final momentum is conserved in both the x and y directions.
Let the final velocity of the combined pucks be v_f at an angle θ from the +x direction.

Conservation of momentum in the x direction:
(0.5)m(75) = (m)v_f cosθ
v_f cosθ = (0.5)(75)
v_f cosθ = 37.5
cosθ = 37.5/v_f

Conservation of momentum in the y direction:
-m(25) = mv_f sinθ
v_f sinθ = 25
sinθ = 25/v_f

Squaring and adding the two equations:
cos^2θ + sin^2θ = 1
(37.5)^2/v_f^2 + 25^2/v_f^2 = 1
1384.375/v_f^2 = 1
v_f = sqrt(1384.375) = 37.18 m/s

Now, we can substitute this value of v_f in the equations for cosθ and sinθ to find θ:
cosθ = 37.5/v_f = 37.5/37.18 = 1.008
θ = cos^-1(1.008) = 34 degrees

Therefore, the angle of the final velocity vector below the +x direction is 34 degrees, which is closest to choice C).

A Sticky Hockey Puck Slides Along a Frictionless,horizontal Ice Surface