Jurgen M said:
Car wheel don't act at Earth c.g. so car has tendency to push Earth back and rotate,translation+ rotation ?
If car is very very big he will push Earth just like little stone when comes at tyer?
The calculations are more complicated if we consider the rotation of the Earth, so let's simpify things by imagining a car of mass ##m## pushing against a large object of mass ##M##. The car has an engine which releases energy and creates relative motion between the mass ##m## and ##M##. The engineering details are not relevant to show where your mistake is.
First, we look at the initial rest frame. The car moves in one direction at ##v## and the large object in the other direction at speed ##V##. By conservation of momentum we have:$$mv = MV$$ Now, we calculate how much energy is needed:$$E = \frac 1 2 mv^2 + \frac 1 2 MV^2 = \frac 1 2 mv^2 + \frac 1 2 M(\frac{mv}{M})^2 = \frac 1 2 mv^2(1 + \frac m M)$$And if ##M## is much larger than ##m##, we see that ##\frac m M \approx 0## and we have: $$E \approx \frac 1 2 mv^2$$ We also see that the amount of energy needed increases as ##v^2##. E.g. if we want to car to travel at ##2v## relative to the large object, then we need ##4## times the energy.
Now, we try your trick of achieving a speed of ##v## and then moving to the rest frame of the car. What you say is that the car now has zero velocity again. But, what you forget is that in this frame of reference the large object now has speed ##v##. The KE of the system in the car's rest frame is: $$E_1 = \frac 1 2 M v^2$$ And the initial momentum is ##p_1 = -Mv## (if we take the direction of the car to be positive). I just want to stress that you cannot simply ignore this, as you have been doing in all your posts.
In any case, we now want to increase the relative speed of the car and object to ##2v##, which means approximately getting the car moving at speed ##v##. Again, by conservation of momentum we have $$mv - M(v + V) = -Mv$$ and again we have $$mv = MV$$ Now the total KE of the system is:
$$E = \frac 1 2 mv^2 + \frac 1 2 M(v+V)^2 = \frac 1 2 mv^2 + \frac 1 2 Mv^2(1 + \frac m M)^2$$
$$E = \frac 1 2 mv^2 + \frac 1 2 Mv^2(1 + 2\frac m M + \frac{m^2}{M^2}) = \frac 1 2 mv^2 + \frac 1 2 Mv^2 + mv^2 + \frac 1 2 mv^2(\frac m M)$$ Again, the last term is approx zero and we have the change in energy: $$E - E_1 = \frac 1 2 mv^2 + mv^2 = \frac 3 2 mv^2$$ And that is the same change in energy as we found in the original rest frame, where $$E' - E'_1 = \frac 1 2 m(2v)^2 - \frac 1 2 mv^2 = \frac 3 2 mv^2$$
It doesn't matter whether you change reference frames of not, the
change in kinetic energy is always the same. As it must be, as the energy came from the fuel consumption, which is independent of reference frame.
