My son had an AP physics lab today where he rolled 2 carts toward each other and they "collided" highly elastically by using magnets with like poles directed toward the front of each cart. The students were instructed to find the momentum of the two cart system before and after the collision. They were also instructed to calculate the percent difference between the momentum before and after the collision. Percent difference is $\frac{|x-y|}{\frac{x+y}{2}} \times 100$

So in one example, the initial momentum of the 2 cart system was 0.01 kgms^-1 and the final momentum of the 2 cart system (after the collision) was -0.02 kgms^-1. This yields a percent difference of 600%. I am confused about 3 of things.

  1. Why is the percent difference useful in evaluating this experiment. The instructor is using percent difference to determine how well the experiment was performed. Had the initial momentum have been 1000 kgms^-1 and the final momentum had been -2000 kgms^-1 then the percent difference would still have been 600%.

  2. I also realized that using this formula, any percent difference between negative and positive values that average out to zero will be undefined. This also makes me question why percent difference is a good way to evaluate the experiment.

  3. The percent difference between ANY value and zero is always 200%.
    (5-0)/2.5=2 (10-0)/5=2 (1000-0)/500=2 It seems to me that having the momentum of a system change from 0kgms^-1 to 1000kgms^-1 would represent a much greater change than having the momentum change from 0kgms^-1 to 10kgms^-1, but they yield the same percent difference. This further leads me to question the value of percent difference in an experiment like this.

  4. If I took the example above where the initial momentum of the 2 cart system was 0.01 kgms^-1 and the final momentum of the 2 cart system (after the collision) was -0.02 kgms^-1 and which yielded a 600% difference, why is this a more significant difference than a change from say 1.04 kgms^-1 and 1.01 kgms^-1 which only yields a percent difference of 3%? This actually represents the exact same physical process occurring when viewed from a different inertial frame of reference.

  • 1
    $\begingroup$ I don't think percent difference is a particularly useful quantity in general, but it is especially useless when the quantity you are measuring is supposed to be zero according to theory. (This is all assuming that the two carts had equal (and opposite) momenta initially, so that the total momentum is zero. $\endgroup$
    – march
    Commented Jan 28, 2022 at 1:15
  • $\begingroup$ The formula you list is actually just one of several ways to do percent difference; see Wikipedia's article, for instance. Several of the formulas listed in that article can mitigate the issue you raise in Point #2. $\endgroup$
    – Kyle Kanos
    Commented Jan 28, 2022 at 1:44
  • $\begingroup$ I taught AP Physics at the HS level for 12 years, and I must note that your son's instructor certainly chose a strange equation to calculate percent difference for this lab. And I note that momentum conservation applies in all reference frames, meaning that in the center of mass reference frame, the total momentum both before and after the collision are supposed to be zero. 0/0 is an indeterminate form, which is supposedly what you are seeing in your results. $\endgroup$ Commented Jan 28, 2022 at 3:32
  • $\begingroup$ @DavidWhite, is there a percent difference equation that makes sense in this case. As I noted, using the formula above, the percent difference between 0 and 10 is 200%, the percent difference between 0 and 100 is 100%, the percent difference between 0 and 100,000 is 200% so how could it possible be a gauge of the degree to which momentum was conserved in an experiment, or rather "what would be" a good gauge? $\endgroup$ Commented Jan 28, 2022 at 16:08
  • $\begingroup$ @JosephHirsch, for this particular experiment, I have my doubts that a percent error calculation makes sense. One would like to see the case where that calculation remained the same in all reference frames, which doesn't appear to be the case for the proposed equation. $\endgroup$ Commented Jan 28, 2022 at 16:58

1 Answer 1


There are plenty of situations where the relative change of a measurand is a useful quantity. However, as you’ve correctly noticed, the relative change of something essentially equal to zero (to within the noise, at least) becomes meaningless. One would need to set up the experiment such that there is some non-negligible (total) momentum and look at the relative change in that. I.e. intentionally give one cart more momentum than the other.

Alternatively, select a different point of reference. Rather than dividing by the nominally zero total momentum, how about dividing by the average of the absolute momenta? This would give a useful error value relative to the scale of the experiment.


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