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If the friction from brakes, wind resistance and all such factors remain constat, which will stop first? A heavier car or a lighter car? How will the momentum of the car and graviational pull on a heavier object influence the stopping of the car?

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Note that the Physics 101 answer to this question (which is similar to your question in that it neglects many factors) may not agree with the real world answer because some of those things you are asking us to neglect are not negligible in real life. –  dmckee May 15 '11 at 19:47
    
Are you considering the friction from the road to be constant too? –  ysap May 15 '11 at 19:57

3 Answers 3

Physics Land:

The physics 101 answer to there kinds of questions goes like this:

Assume the cars will lock up their wheels so that I can apply a simple analysis of sliding friction. The frictional force $F_f$ is dependent only on the normal force $N$ between the car and the surface and on the coefficient of friction $\mu$ between the tires and the road.

$$ F_f = \mu N = \mu M g $$

where $M$ is the mass of the vehicle and $g$ is the acceleration due to gravity.

From this we can compute the acceleration due to friction as

$$ a_f = \frac{F_f}{M} = \mu g $$

and the stopping distance as

$$ d = \frac{v^2}{2(a_f)} = \frac{v^2}{2 \mu g} $$

where $v$ is the speed of the car at the moment the brakes are applied.

You will notice that this does not depend on the mass of the car. So both cars stop in the same distance.


Welcome to the real world:

  • The first expression above: $F_f = \mu N$ is an approximation that only applies when there are negligible deformations of either body. This is simply not true in the case of a car riding on inflated rubber tires in contact with a rough road surface. In the real world there are complicated interactions between the wheel and the tire and the road. The exact effects of this depend on the nature of the tire and the nature of the road. We are not in a position to say much.

  • The above analysis relies on both cars being able to lock up their wheels. In principle either car might fail to have the braking power to do so.

  • The above analysis relies on both cars skidding, but almost all modern cars have a sophisticated anti-lock braking system to prevent this and get a little more out of the brakes (skidding is slightly suboptimal, but better than a bad driver). Even without ABS sufficiently practiced and skilled drivers can sometimes outperform skidding stops by pumping the brakes and riding the edge of lock-up (this isn't easy, but we used to learn and practice it). The braking performance in a mixed skidding/rolling scenario is much harder to analyze.

So real world answer: beats the heck out of me.


Side note: the conventional wisdom has it that big rigs brake faster and better when fairly heavily loaded than either running empty or loaded to the gills.

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I don't think he was talking about sliding/skidding to a stop. –  JoeHobbit May 15 '11 at 21:02
    
@Joe: The analysis of a pure rolling stop is similar, but picks up a bunch of $\ge$ and $\le$ signs. –  dmckee May 15 '11 at 21:05
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Don't forget tires are non linear, and the more normal load applied the available traction friction reduces slightly. So if the two cars had the same tires, the smaller one can make better use of the tires. Of course, that is a 2nd or 3rd order effect compared to all the other factors. –  ja72 May 16 '11 at 18:00
    
@ja72: This is a detail I did not know. Thank you. Though I suppose I should have guess because high performance cars get big wheels... –  dmckee May 16 '11 at 18:12
    
@dmckee - a related reason why big wheels handle better is because the contact patch is wider (less crown across the tire profile) and thus they spread the pressure more evenly. Here is a nice read for you: racer.nl/reference/pacejka.htm –  ja72 May 16 '11 at 18:33

dmckee has answered what I believe to be the intent of the question; but Joe has correctly answered the absolute letter of the question.

Level ground seems to be implicitly assumed.

If all other things are EQUAL (e.g. coefficients of friction), except where they need to be PROPORTIONAL to be equal (e.g. wind resistance), then the cars will stop in the same distance.

If the cars both ride the the same coefficients of friction and have the same wind resistance (assumed to be a retarding force, not a tail wind) the lighter car will stop first.

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The lighter car will stop faster because the lighter car has less energy to dissipate than the heavier one, while the tires more equal limits of friction with the road.

That's because tires gain traction non-linearly with vertical load.

(Even if the braking systems can't reach the friction limit of the tires, the braking systems also have roughly equal capabilities to convert different amounts of energy to heat.)

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