A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are μ_s = 0.40 and μ_k = 0.30, respectively. The puck's initial speed is 4.9 m/s. What speed does it have when it slides back down to its starting point?

Question

Quizplus · Accepted Answer

First, we need to find the maximum height that the puck will reach. Since the ramp is inclined at a 30° angle, we can use trigonometry to find the height:

sin(30°) = height / hypotenuse
height = sin(30°) x hypotenuse
height = sin(30°) x (distance up ramp)
height = 0.5 x (distance up ramp)

distance up ramp = (initial velocity)^2 x sin(2θ) / g
distance up ramp = (4.9 m/s)^2 x sin(2(30°)) / 9.81 m/s^2
distance up ramp = 0.62 m

height = 0.5 x 0.62 m
height = 0.31 m

Now, we can use energy conservation to find the final speed of the puck when it slides back down to its starting point. At the maximum height, all the initial kinetic energy has been converted into potential energy. When the puck slides back down, the potential energy is converted back into kinetic energy:

initial kinetic energy = final kinetic energy
0.5 x mass x (initial velocity)^2 = 0.5 x mass x (final velocity)^2

We can solve for the final velocity:

(final velocity)^2 = (initial velocity)^2 - 2gh
(final velocity)^2 = (4.9 m/s)^2 - 2(9.81 m/s^2)(0.31 m)
(final velocity)^2 = 2.70 m^2/s^2
final velocity = 1.6 m/s

However, we need to take into account the frictional force acting on the puck as it slides back down the ramp. Since we are given coefficients of static and kinetic friction, we need to determine if the puck will be sliding or rolling without slipping. If the puck is rolling without slipping, we can use a different equation to find the final velocity:

rolling without slipping: 
friction force = (radius of puck) x (mass of puck) x acceleration
friction force = 0.048 m x 0.2 kg x acceleration
friction force = 0.0096 kg m/s^2 x acceleration

initial torque = final torque
(initial angular velocity)(moment of inertia) = (final angular velocity)(moment of inertia)
(0.5 x mass x radius^2 x initial velocity) = (0.5 x mass x radius^2 x final velocity)
(initial velocity) = (final velocity) x sqrt(muS/5)

Set the potential energy at the puck's final position to be equal to its initial kinetic energy:

mgh = 0.5mv^2 + 0.5Iw^2 + fw(s)(-1)
0.5m(9.81 m/s^2)(0.62 m - 2*0.048 m) = 0.5m(v^2 + 5*v^2/4) + 0.0096*5*v^2/4
v = 2.8 m/s

Since the friction force is less than the maximum static friction force, the puck will be rolling without slipping. Using the equation given above, we can find the final velocity to be 2.8 m/s. Therefore, the answer is A.

A 200 G Hockey Puck Is Launched Up a Metal