- #76

bob012345

Gold Member

- 488

- 89

Thanks, I did the math, though probably not as formally as you would. While I agree what you and jbriggs are saying, the example I picked does not convey the dilemma fully. I'll try and think of another.Work through the math - the recoil doesn't change the velocity by much, but it changes it by enough. I really cannot stress enough how important it is that you actually work through the math - until and unless you've worked through the math you won't really understand how the physics works here.

It will be clearer if you work through the math yourself. The change in the plane's speed and hence the momentum change from the recoil is the same in both frames. The energy change is not the same and cannot be. Let's say that the plane is moving at speed ##u## before the gun is fired, and the recoil changes its speed by an amount ##\Delta{v}##. The kinetic energy before was ##E_0=m(u^2)/2##, the kinetic energy afterwards is $$E_1=m(u-\Delta{v})^2/2=m(u^2-2u\Delta{v}+\Delta{v}^2)/2\approx{m(u^2-2u\Delta{v}})/2=E_0-2u\Delta{v}$$

(The approximation is allowed because ##\Delta{v}## is small compared with ##u## - if you work through the math you will see that for reasonable assumptions about the mass of the plane and of the bullet ##\Delta{v}^2## will be a one part in a billion rounding error).

So the change in kinetic energy is ##E_1-E_0=-2u\Delta{v}##, and that's going to be different for different values of ##u## - it is frame-dependent.

Note also that the frame in which the plane is at rest is the only one in which the change in the plane's kinetic energy doesn't matter. It doesn't matter because in that frame ##u## is zero so ##2u\Delta{v}## is also zero.

Did I mention that you should work through the math yourself? No? I'm sorry, that was an oversight. I certainly meant to say it.