[repost] What distance will the puck move?

Question

The puck moving with velocity v = 10 m/s hits the floor
at angle 45 deg and continues to move sliding on the
floor.

http://alexandersemenov.tripod.com/puck/index.htm

Coefficient of friction μ = 0.3; g = 9.8 m/s².

Answer 1

At the moment the puck hits the floor, it will be traveling at it's x-component of velocity
Vx = 10m/s*cos(45)
Vx = 7.1 m/s

As the puck travels, it will lose speed from friction. Equating kinetic energy to the work done by friction, a puck will skid to a stop a distance "d" down the rink

1/2*m*v^2 ..........Kinetic Energy
m*g*μ*d.............Work done by friction
Equating kinetic energy and work, can cancel out the mass term on both sides.
1/2*v^2 = g*μ*d

Solving for d yields:
d = (1/2*v^2) / (μ * g)
d = (1/2*7.1^2) / (0.3*9.81) = 8.56 m <---- ANSWER
The puck will slide to a stop 8.56 metres from the landing point.

Initial speed = 7.1 m/s
Will slide to stop 8.56 metres away, decelerating at a rate of μ*g = -2.943 m/s^2

Answer 2

If we assume the KE of the puck is absorbed by the ice impact and by the sliding friction, then half the energy will be absorbed by the ice and the other half will be absorbed by the friction. This results by the 45 degree angle of the impact.

Therefore, (1/2 mv^2)/2 = Fd = k mg d; so that kgd = 1/4 v^2 and the distance of slide d = 1/4 v^2/(kg); where v = 10 cos(45), k = .3 your mu.

Lesson learned: The distance slid is independent of the mass of the puck. Also, the velocity along the ice depends on the angle of attack with the floor (ice). Finally, the obvious verified by the derived equation: lower k means a longer slide and higher velocity does as well. But the change in velocity has more impact on the slide distance because of the velocity square.