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OPhO 2022, Теория — Задача 3. Rocket

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2022 Online Physics Olympiad: Invitational Contest — Theoretical Examination · 27 т.

Условие

OPhO organizers have a “propulsionless” rocket, which for simplicity can be assumed to be a 2-dimensional rectangular box of mass $2M$ and horizontal length $L$ . Assume that the horizontal sides of the box are massless while the vertical sides of the box each have mass $M$ . The rocket is initially at rest. We will now explore the mechanism for how this rocket move. Suppose we have $N$ particles of mass $m/N$ each on the left and right sides of the box. At time $t=0$ , we launch the $N$ particles on the left side of the box together to the right with velocity $\frac vN$ . In addition, in intervals of time $\frac Lv$ , starting at $t=0$ , we launch a particle from the right side of the box to the left side with velocity $v$ . Once a particle reaches the opposite side of the box, it is stopped. The particular mechanism to shoot and catch the particles can be ignored here. Assume that this mechanism can conserve energy. After time $t=\frac{NL}v$ , there will be $N$ particles on each side of the box, which is identical to the initial state.

Neglect relativistic effects in part (a) only.

a. (1 pt.) According to classical (Newtonian) mechanics, what happens to the rocket? Does it move?

b. (5 pts.) If $v\ll c$ , how far does the rocket move? Answer in lowest nonzero order in $v/c$ .

c. (10 pts.) How far does the center of mass of the rocket system move? Once again, answer in lowest nonzero order in $v/c$ . Justify your answer.

d. (6 pts.) Explain why this process cannot continue indefinitely. If it could continue forever, we would able to move the rocket indefinitely with no propulsion.

e. (5 pts.) Give an estimate for how long this process can continue. How far does the rocket move in this time?

Published organizer clarifications — separately quoted from the original clarification document, page 1:

There is a typo in the problem. It should say “After time t=NL/v, there will be N particles on either side of the box” instead of t=L/v.
Assume the box is a 2-dimensional rectangle with horizontal length L. There are 2N particles in total, N of which start on the left side of the box and N of which start on the right side. The particles are launched in the horizontal direction.

Решение

a.

At each moment, the particles from the left side of the rocket have momentum $N\frac mN\frac vN$ and the particles from the right side of the box have momentum $\frac{mv}N$ . As these are equal, by conservation of momentum for the entire system, the rocket must have 0 momentum, so it does not move.

b.

Let $\beta=v/c$ . The relativistic momentum of the particles from the left side of the rocket is

N\frac1{\sqrt{1-\frac{\beta^2}{N^2}}}\frac mN\frac{\beta c}N=\frac{\beta mc}{N\sqrt{1-\beta^2/N^2}}.

The momentum of the particle from the right side of the rocket is

\frac1{\sqrt{1-\beta^2}}\frac mN(\beta c)=\frac{\beta mc}{N\sqrt{1-\beta^2}}.

The difference in these quantities to lowest nonzero order in $\beta$ is

\frac{\beta mc}N\left(1+\frac{\beta^2}2-1-\frac{\beta^2}{2N^2}\right)=\frac{\beta^3mc(N^2-1)}{2N^3}.

This is the momentum of the rocket to the right. Thus, it is equal to $\gamma(2M+\frac{N-1}Nm)V$ if $V$ is the speed of the rocket. As the momentum is third order in $\beta$ , we can assume that $\gamma\approx1$ to find $V$ to lowest nonzero order. We obtain

V=\frac{\beta^3mc(N^2-1)}{2N^2(2MN+(N-1)m)}.

Now, the time is $t=\frac{NL}v=\frac{NL}{\beta c}$ , so the total distance traveled is

Vt=\frac{(N^2-1)m}{2N(2MN+(N-1)m)}\frac{v^2}{c^2}L.

Valid solutions that assume $N\gg1$ or $m\ll M$ are acceptable.

c.

The center of energy in a closed relativistic system with zero total momentum does not move, so the rocket’s center of mass at the end is exactly where it was in the beginning.

For a proof of this fact, consider a system of particles. We wish to show the claim that $\sum_i x_iE_i$ is constant. Whenever there is a collision between particles $i$ and $j$ , note that just before and after the collision, both particles are at the same position $x$ , so if $E_i'$ and $E_j'$ are the energies after the collision, then the change in $\sum_i x_iE_i$ during the collision is $xE_i'+xE_j'-xE_i-xE_j=x(E_i'+E_j'-E_i-E_j)=0$ . This is true as energy is conserved during the collision. We have the same conclusion when one particle decays for the same reason. Lastly, it suffices to show that $\sum_i x_iE_i$ is conserved for a set of particles moving freely with zero total momentum. In this case, $E_i$ is constant and equal to $\gamma_i m_i c^2$ , so the change in $\sum_i x_iE_i$ is $\sum_i(v_i t)\gamma_i m_i c^2=c^2t\sum_i\gamma_i m_i v_i=c^2t\sum_i p_i=0$ . Here, we have assumed that the particles move in one dimension, but it is simple to extend this to multiple dimensions. Therefore, we can conclude that the center of energy does not move.

Half credit will be given for the answer and a classical explanation. For full credit, a satisfactory relativistic explanation must be provided.

d.

The center of mass does not move although the rocket system moves to the right. This means that through this process, mass is transferred from the right side of the rocket to the left side. Every time this process is repeated, the same amount of mass is transferred from the right to the left side, and since the right side only starts with mass $M$ , the process cannot continue indefinitely. Unfortunately, we cannot construct a propulsionless rocket even with relativity.

e.

If a mass $\delta m$ is transferred from the left side to the right side in each step, then the center of mass of the rocket moves a distance $\frac{\delta m}{2M}L$ to the left from the geometric center of the rocket. In order for the center of mass of the rocket to not move, we must have

\frac{\delta m}{2M}L=\frac{(N^2-1)m}{2N(2M+(N-1)m)}\frac{v^2}{c^2}L,

\delta m=\frac{(N^2-1)mM}{N(2MN+(N-1)m)}\frac{v^2}{c^2}.

The number of times that this can continue is roughly

\frac M{\delta m}=\frac{N(2MN+(N-1)m)}{(N^2-1)m}\frac{c^2}{v^2}.

The total distance that can be traveled is roughly at most $L/2$ (if all of the mass from the right goes to the left).

Here, we implicitly assumed $m\ll M$ . Any solution with correct reasoning and a valid estimate is acceptable.

Organizer clarification document — other notes

Separately quoted from the organizer clarification document:

Other Notes

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Оригинал в Архива: theory.pdf · официални решения: theory_sol.pdf · допълнителни данни: Clarifications.pdf