Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
Well think about this: if the deceleration is constant, what happens after the puck has slowed to a stop? – user2963 Dec 6 '11 at 23:11
up vote 1 down vote accepted

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.)

share|cite|improve this answer
Okay, I went for the 2nd model. It comes closer to what I want. Can you help me to calculate the total way the puck will move? I found that "fPredictedTime = Math.Log( this.fMiniumSpeed / fInitialSpeed, this.fFrictionFactor );", but how to calculate the total way? – Krumelur Dec 8 '11 at 20:30
Or maybe I will post a new question...that will be better I think. – Krumelur Dec 8 '11 at 20:34
@Krumelur: OK, if you want to use the strictly viscous model, the solution is very simple. Both velocity and position decay exponentially on the same time scale, which is $1/K$. Velocity decays toward zero, and position decays toward the final position, which you get by integrating $v$. – Mike Dunlavey Dec 8 '11 at 21:06

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.

share|cite|improve this answer
To the extent that the puck isn't touching the table, the normal approach for coefficient of friction won't work the same. It's hard to say what correlation it will follow for drag, or even if drag is a fully appropriate concept given the physics of the air under the puck. (i'm not the downvote btw) – Alan Rominger Dec 6 '11 at 23:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.