2
$\begingroup$

The question arose because of the following situation:

Two cars with the same mass are side by side on the same track with speed $v_1$ in the same direction. Suddenly, a third car, also with the same mass, comes in the opposite direction with speed $v_2$ and collides head-on with one of the cars. The total energy of the shock for an observer in the car that did not crash will be, relative to the cars, $E_{car}=E_1+E_2$, where $E_1=M(0)²/2$ and $E_2=M(v_1+v_2)²/2$, while that a stationary observer on earth goes $E_{earth}=E_1'+E_2'$, where $E_1'=Mv_1²/2$ and $E_2'=M(v_2)²/2$. Clearly the sum of energies are not the same: $E_{car}=M(v_1²+2v_1v_2+v_2²)/2 ≠ E_{earth}=M(v_1²+v_2²)/2$. But if it's the case that the mechanical energy is the same for any inertial frame, then these two equations must be the same, but the problem is that this creates a dependency between $v_1$ and $v_2$ that shouldn't exist, since each car has its own speed. Now the question is: if the mechanical energy depends on the inertial frame, then shouldn't the observer who sees the collision with more energy see a more serious accident between the cars?

$\endgroup$
1
  • $\begingroup$ Does this answer your question? $\endgroup$
    – user7896
    Jan 22 at 5:24

1 Answer 1

5
$\begingroup$

The kinetic energy does depend on the choice of inertial frame. We can see this easily by considering the kinetic energy of a single particle. We can make the kinetic energy of the particle take whatever value we want by boosting to a suitable inertial frame.

The fact that the kinetic energy is arbitrary does not mean that we can make our collision arbitrarily destructive. The thing that determines how much the cars are damaged in the collision is not the initial kinetic energy alone, but how much kinetic energy is lost in the collision. If we had a completely elastic collision, all of the initial kinetic energy of the cars would be converted to final kinetic energy, with no energy left over to deform or otherwise damage the cars.

Let's examine the collision in the two frames to see how much kinetic energy is lost. I'll assume the collision takes place along a line. The initial velocities of the cars in the Earth frame are $v_1$ and $-v_2$, and the final velocities in this frame are $u_1$ and $u_2$. I work in units with $m=2$ for simplicity. In the Earth frame, \begin{align} E_i' = v_1^2 + v_2^2\\ E_f' = u_1^2 + u_2^2 \end{align} and the energy lost in the collision is \begin{align} Q' = E_i' - E_f' = v_1^2 + v_2^2 - \left(u_1^2 + u_2^2\right) \end{align} In the car frame, \begin{align} E_i &= (v_1+v_2)^2 = v_1^2 + v_2^2 + 2v_1 v_2\\ E_f &= (u_1-v_1)^2 + (u_2-v_1)^2 = u_1^2 + u_2^2 + 2 v_1^2 - 2u_1v_1 - 2u_2 v_1 \end{align} and \begin{align} Q = E_i - E_f &= v_2^2 - v_1^2 + 2v_1 v_2 + 2(u_1 + u_2)v_1 - (u_1^2 + u_2^2) \end{align} Applying momentum conservation in the Earth frame, we have \begin{align} u_1 + u_2 = v_1 - v_2 \end{align} so \begin{align} Q &= v_2^2 - v_1^2 + 2v_1 v_2 + 2(v_1 - v_2)v_1 - (u_1^2 + u_2^2)\\ &= v_1^2 + v_2^2 - (u_1^2 + u_2^2)\\ &=Q' \end{align} We see that the kinetic energy lost in the collision (which is what will determine how much damage is sustained) is the same in both frames, even though the initial and final kinetic energies are frame-dependent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.