If a puck on ice deccelarates, is its speed "v = v0 - t * a" or is it "v = v0 -k * v0"?

Question

I've been wondering about a puck sliding on ice or a puck in an air hockey table game. If a puck is hit and we start watching it as soon as it has reached its maximum speed and starts becoming slower, what will this deccelaration look like?

I assume we take a snapshot of the puck in intervals.

a = Deccelaration
t = time passed since last snapshot
vCurrent = the current speed to be calculated when taking the snapshot
vPrevious = the speed it had at the last snapshot

• Is the deccelaration constant, so that vCurrent = vPrevious - a * t?
• Or is vCurrent = vPrevious - k * vPrevious, where k is some factor in the range of 0.01 to 0.1?

The first assumption seems to make sense to me but I wrote a simulation of it and it just does not look right. The second one looks better, but is it real?

• If the latter one is the true, what would that "k" be? Does it have a name?
• What "kind" of deccelaration is the 2nd one?
• For the first assumption I can easily calculatethe way and time (v² = 2*a*s) it takes until the puck comes to a rest, but how do I have to calculate way and time for the second case?

Answer 1

Your first model is the usual way of expressing sliding friction, where

$a = - g F (v/|v|)$

where $v$ and $a$ are vectors, $F$ is the coefficient of friction, and $g$ is the acceleration of gravity, as long as $v$ is non-zero. When $v$ is zero, $a$ is zero.

Your second model looks like viscous friction, where a fluid is between the sliding surfaces, as in

$a = - K v$

so that as it slows down, the deceleration decreases as well. ($K$ is the viscosity, or proportional to it.)

You can certainly make a model that's a weighted sum of these.

Then of course you write your differential equations:

$dv/dt = a$
$dx/dt = v$

and integrate them any way you like. (What you described looked like the Euler algorithm, the simplest one, which might be quite adequate for your purpose. If you want more accuracy, you might check out a simple low-order Runge-Kutta.)

Answer 2

Strictly speaking, I think it's both.

friction vs the surface should be (relatively) constant, so you'd get: v = v0 - a*t

But friction vs the air is fluid dynamics, and basically it's proportional to the speed, so you get the 2nd equation.

I guess it would depend which effect is dominant. For nicely-shapped, very heavy, very slow obects, then air friction would be a small factor. For fast, lightweight objects, air friction can be a major factor.

Air hockey, I assume the table design is intended to significantly reduce surface friction, so it's likely that air friction is a heavy factor, which I think explains your experimental result.