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Suppose a vehicle 1 is on the top of another vehicle 2 (we can think of it like a big platform).

Imagine the following independent experiments:

  1. Suppose that the top vehicle accelerates to a speed $v_1$, its kinetic energy is $\frac{1}{2} m_1 v_1^2$.

  2. Suppose that below vehicle accelerates to a speed $v_2$, the kinetic energy of the top vehicle is then $\frac{1}{2} m_1 v_2^2$.

  3. Now, suppose that first the below vehicle accelerates to $v_2$ and that the top vehicle then accelerate to $v_1$. The kinetic energy is $\frac{1}{2} m_1 (v_1 + v_2)^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_1 v_2^2 + m_1 v_1 v_2$.

Now, why is this kinetic energy greater than the sum of the two first experiments (by $m_1 v_1 v_2$) ? Especially, it looks like we would require the same "fuel" for the accelerations in both cases.

What am I getting wrong?

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For real vehicles drag and air resistance will be significant. But we can ignore them here.

In the first case the top vehicle (in order to accelerate) pushes backward on the lower vehicle. We are assuming that the lower vehicle has its brakes on so the force is ultimately transmitted without loss to the ground.

In the third case, we can't use brakes since the vehicle is moving forward. As the top vehicle accelerates, the force from the wheels pushes the lower car backward. At the end, either the lower vehicle is no longer moving at the speed of $v2$, or you have had to add additional energy to get it back to that speed.

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  • $\begingroup$ It's interesting to note you'd observe the same phenomenon even when using rockets instead of cars, to completely eliminate the possibility one vehicle pushing off another. But it still comes down to the fact that you need to consider energy of the entire system, which in the case of the rocket would include the KE of both the rocket itself and its exhaust. Here, the lower car is the "exhaust" which has different final KE in scenarios 2 and 3. $\endgroup$
    – Nuclear Hoagie
    Commented Mar 21, 2022 at 20:03

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