Question about kinetic energy

Question

I have a question about kinetic energy:

Imagine two vehicles with velocities $v_1$ going at $30 \frac{\text{m}}{\text{s}}$ and $v_2$ at $50\frac{\text{m}}{\text{s}}$.

If they hit each other, will the kinetic energy be

$$ \frac{1}{2}m(v_1+v_2)^2 \text{ or } \frac{1}{2}m(v_1^2 +v_2^2) \text{ ?}$$

It should be the second formula, otherwise there would be more energy than necessary to reach this speed, but I don't understand why it is this way.

After all, if you had a vehicle hitting a wall at $80\frac{\text{m}}{\text{s}}$ it would have the same energy as 2 vehicles hitting each other at $50\frac{m}{s}$ and $30\frac{m}{s} $, right?

Can somebody explain this apparent discrepancy to me?

Answer 1

The total kinetic power of the system will be $\frac{1}{2}mv^2_1 +\frac{1}{2}mv^2_2$. The first equation that you mention is wrong, because this equation says that you have an object of mass $m$ with speed $v_1+v_2$. If you expand the squared term you will see that it is different.

Now, what do you mean hit each other? Do they have opposite velocities? In any case, there will be an energy exchange. You can find this by taking conservation of momentum:

$P_{before}=P_{after}$, where $P_{before}$ is the momentum of the system before the collision and $P_{after}$ is the momentum of the system after the collision. Beware though, that momentum is a vector and not a scalar quantity. If you assume motion on a line only, (only along one axis) then you can consider it as just a number along with its correct sign (denoting the direction of the velocity vector) and disregard any other directions (along axis y and z for example).

After you have found the momenta after the collision, you can then carry on to calculate the total energy of the system, after the collision.

Answer 2

Mythbusters ran an episode on this one. When you learn of relative velocity, you learn that from the perspective of one of the drivers, the other driver will appear to come at you at 80 m/s. So the thought is that it should be the same as hitting a wall at 80 m/s. The energy doesn't add up though.

Essentially, from the perspective of the driver, they are hit back by a force equal to the change in the momentum on each of them. A car going 30 m/s, when hit, doesn't change momentum to -50 m/s, for -80 m/s total change. It only changes momentum to reach the final system momentum, which is significantly less than 80 m/s This is the difference

The energy change is the same. Since each cars velocity only drops by the speed they were going, that's the amount of kinetic energy in the system converted to heat and other sources

$$KE = m1\frac {(v_1 - v_{final})^2}{2} + m2\frac {(v_2 - v_{final})^2}{2}$$

Even though the rotation of the earth still gives these cars a velocity even after a collision, that kinetic energy was not converted upon collision, so we can use the velocities relative to the earth.