Which will stop first a heavier car or a lighter car?

Question

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?

Answer 1

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.

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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.

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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.

Answer 2

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.

Answer 3

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.)