When I stop my car, how much energy is transferred to the brakes as heat, and how much is transferred to earth?

Question

Assume I'm travelling at 70mph (31.3m/s) in my car and suddenly slam on the brakes and come to a complete stop. I don't brake so hard that the tyres skid.

Assuming my car has a mass of 1000kg, that's a kinetic energy of 490KJ which was just dissipated.

Assuming my brake discs are made of steel and weigh 10kg in total, and given the specific heat capacity of steel is 502J/Kg/K, the brakes should heat up by about 98C.

I know that some energy is lost to air resistance and to friction in the bearings and tyres, but ignore that for the purposes of this question.

Even ignoring friction losses, I know that 98C is not the true figure, since some kinetic energy gets transferred to the earth.

How much kinetic energy gets transferred to the earth and how much to the brakes? Can we calculate that?

I think, obviously the earth must have been pushed forwards by some amount when my car slowed down, so can this be treated as a type of collision?

Answer 1

How much energy gets transferred to the earth as momentum?

Energy and momentum are two separate things. Both are conserved in a closed system. You can't convert one to the other.

Practically, all of the momentum of the car gets transferred to the Earth.

And practically all of the kinetic energy of the car gets converted to heat.

I know that some energy is lost to air resistance and to friction in the bearings and tyres,

Even this just means the energy gets turned into heat somewhere else instead of heat in the brakes.

Earth must have some kinetic energy from the collision no?

Let's say it's a big car, weighing 2000 kg.

The momentum transferred to the earth is

$$p = (30\ {\rm m/s})(2000 {\rm kg})= 60\times 10^3 {\rm kg \cdot m/s}$$

The mass of the Earth is about $6\times 10^{24} {\rm kg}$. So the velocity of the Earth changed by about

$$\Delta v = \frac{60\times 10^3}{6\times 10^{24}}{\rm m/s}=10^{-20}\ {\rm m/s}$$

and the kinetic energy imparted to the Earth was

$${\rm K.E.} = \frac{1}{2}(6\times 10^{24}\ {\rm kg})(10^{-20}\ {\rm m/s})^2=3\times 10^{-16}\ {\rm J}$$

Considering the initial K.E. of the car was about 900,000 J, this energy transfer to the Earth is negligible.

Answer 2

I think when you apply the brake , energy is lost in modes of friction as the tyre skids over the ground.So all the kinetic energy is suddenly changed into frictional and heat energy

Answer 3

Question like "How much energy is transferred [in a certain process]?" have a fundamental problem when they involve changes in the state of motion of various components of the system: the answer is observer dependent.

Consider the classic case of a 1D, elastic colission between in two equal masses with relative velocity $v_r$.

• In one frame (usually called the lab frame) one object start at rest and ends in motion while the other object experiences the opposite transition. Work done on either object: $\pm\frac{1}{2}mv_r^2$.
• In a different frame (center of momentum), both objects end with the same amount of kinetic energy the started with. Work done: $\pm\frac{1}{4}mv_r^2$ (each object goes from $\pm v_r/2$ to $\mp v_r/2$, right?).

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Now, in the case you ask about the difference is tiny becuase the mass ratio is so high, but to give an exact answer you should state the frame of reference you want to use, then calculate the total kinetic energy before and after the action and the difference gives you the thermal energy developed in the system. If we also assume the tire-ground interaction is elastic (imprecise, but not horrible in a no-skidding scenario) then that energy goes into the brake mechanism.