change of energy in changing frame of reference

Let's imagine a car that can jump onto, or off a moving train. The train moves at 10m/s. The car, on a road next to the train, accelerates to the same 10m/s, jumps off a ramp and lands on the train, in a relative standstill in relation to the train.

To achieve that, a car of 1 ton needs to burn 50,000 Joules worth of gasoline. ($E=0.5mv^2 = 0.5 \cdot (1000kg) \cdot (10m/s)^2$). With 46MJ/kg of gasoline energy density that's about 1 gram of gasoline burnt and turned into the car's kinetic energy.

Now, we forget about the ground. We're on a massive train moving at a constant speed. The car accelerates in relation to the train. Again, to gain 10m/s relative to the train, from its relative 0, it needs 50,000J of energy, and burns another gram of gasoline.

It then reaches a ramp on one of the train cars, jumps off it, and now it moves at 20m/s over the road, all this at cost of 2 grams of gas.

Now let's try to reach 20m/s without the train. $E=0.5mv^2 = 0.5 \cdot (1000kg) \cdot (20m/s)^2 = 200,000J$. We need to burn 4 grams of gasoline to achieve this speed.

What happened? Where is the fallacy? What do I miss?

• For example, you can write its kinetic energy as $E = p^2 /2m$. Then if the mass is very high, $\Delta E \approx p \Delta p / m = v \Delta p$, which is independent of mass. May 9 '16 at 1:51