Asked by Jeremiah
A hockey puck with mass 0.237 kg traveling along the blue line (a blue-colored straight line on the ice in a hockey rink) at 1.4 m/s strikes a stationary puck with the same mass. The first puck exits the collision in a direction that is 30° away from the blue line at a speed of 0.79 m/s (see the figure).
What is the direction and magnitude of the velocity of the second puck after the collision?
What is the direction and magnitude of the velocity of the second puck after the collision?
Answers
Answered by
drwls
Assume that total momentum is conserved, and solve for the speed and direction of the second puck.
Answered by
Erica
θ = 30°
x-direction:
vi + vf cos θ = x
1.4m/s - 0.79m/s(√3/2) = 0.716m/s
y-direction:
vf sin θ = y
0.79m/s * 0.5 = 0.395m/s
Speed =
√(x² + y²)
√[(0.716)² + (0.395²)] = 0.818 m/s
Direction:
Tan-¹ (y/x) = Tan-¹ (.395/.716) = 28.88° -> 28.9°
x-direction:
vi + vf cos θ = x
1.4m/s - 0.79m/s(√3/2) = 0.716m/s
y-direction:
vf sin θ = y
0.79m/s * 0.5 = 0.395m/s
Speed =
√(x² + y²)
√[(0.716)² + (0.395²)] = 0.818 m/s
Direction:
Tan-¹ (y/x) = Tan-¹ (.395/.716) = 28.88° -> 28.9°
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