A hockey puck with mass 0.250 kg traveling along a line at 2.5 m/s strikes a stationary puck with the same mass.One puck exits the elastic collision in a direction that is 20 degrees away from the line at a speed of 2.35 m/s.What is the angle with respect to the line in which the second puck travels after the collision?

Question

Quizplus · Accepted Answer

Conservation of momentum in the x-direction:
0.25 kg * 2.5 m/s = 0.25 kg * v1x + 0.25 kg * v2x
where v1x is the x-velocity of the first puck after the collision and v2x is the x-velocity of the second puck after the collision.
Since the second puck is initially stationary in the x-direction, v2x = 0.
0.25 kg * 2.5 m/s = 0.25 kg * v1x
v1x = 2.5 m/s
Conservation of momentum in the y-direction:
0.25 kg * 0 m/s + 0 kg * 0 m/s = 0.25 kg * v1y + 0.25 kg * v2y
where v1y is the y-velocity of the first puck after the collision and v2y is the y-velocity of the second puck after the collision.
Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
(0.5 * 0.25 kg * (2.5 m/s)^2) + (0.5 * 0.25 kg * 0 m/s^2) = (0.5 * 0.25 kg * v1^2) + (0.5 * 0.25 kg * v2^2)
where v1 is the speed of the first puck after the collision and v2 is the speed of the second puck after the collision.
Simplifying and solving for v2:
v2 = sqrt((2.5^2 + 2*2.35^2) / 2) = 2.84 m/s
Using trigonometry, we can find the angles:
tan(20 degrees) = v1y / v1x, so v1y = 0.4636 m/s
tan(theta) = v2y / v1x, so theta = -70.53 degrees
Therefore, the second puck travels in a direction 70.53 degrees below the line.

A Hockey Puck with Mass 0 ∘\circ∘ B)-20 ∘\circ∘ C)-60 ∘\circ∘

A Hockey Puck with Mass 0 $\circ$
B)-20^$\circ$
C)-60^$\circ$