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NBPhO 2024, Теория и експеримент — Задача 1. Four charges

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Nordic-Baltic Physics Olympiad 2024 · 7 т.

Условие

— Päivo Simson, Jaan Kalda.

Four identical particles are initially in the corners of a square, as shown in the figure below. All particles have the same charge $q$ , mass $m$ , and the same magnitude of initial velocity $v_0$ . The directions of the initial velocities are indicated in the figure. You can assume $v \ll c$ and ignore gravity.

Four point particles labelled q, m at the corners of a square drawn with dashed sides of length L, the square standing on a vertex. Arrows show the initial velocities: the top particle moves horizontally to the left, the right particle moves vertically up (labelled v0 with a vector arrow), the left particle moves vertically down, the bottom particle moves horizontally to the right; each velocity makes a marked 45° angle with the adjacent side.

i) After a long time has passed, what is the magnitude of the final velocity $v_f$ of the particles with respect to the center of mass of the system? [2 т.]

ii) What is the angle between the initial velocity $\vec{v}_0$ and the final velocity $\vec{v}_f$ of a particle? [5 т.]

Решение

— Solution by Päivo Simson, grading schemes by Päivo Simson, Oleg Košik and Lasse Franti.

i) (2 т.)

Assuming $v \ll c$ and ignoring gravity, the initial total energy of the system is the sum of classical kinetic and potential energies

E_{total} = E_{kin} + E_{electric} =

E_{kin} + \frac{1}{2}\Sigma_{i\neq j}\, k\frac{q_iq_j}{r_{ij}} =

4\frac{mv_0^2}{2} + 4\frac{kq^2}{L} + 2\frac{kq^2}{\sqrt{2}L} =

= 4\left(\frac{mv_0^2}{2} + \frac{kq^2}{L}\frac{4+\sqrt{2}}{4}\right).

After the particles have moved infinitely far from each other, the potential energy becomes zero and the total energy is only kinetic:

E_{total} = 4\frac{mv_f^2}{2}.

Since the total energy does not change we have from the equality of these two expressions

v_f = \sqrt{v_0^2 + \frac{kq^2}{Lm}\frac{4+\sqrt{2}}{2}}.

Grading:

Idea of using energy conservation (0.2 pts)
Idea of symmetry and equality of quantities (0.2 pts)
Idea of total energy as a sum of kinetic and electrostatic (0.2 pts)
Formula for electrostatic energy including the two different distances (0.3 pts)
Factor $\frac{1}{2}$ from pairings (0.4 pts)
Final energy is purely kinetic (0.4 pts)
Correct final answer (0.3 pts)
If only dimensionless factors missing and final answer is reasonable (0.2 pts)

ii) (5 т.)

Let the required angle be $\phi$ .

The expression for the total energy reveals that the particles behave effectively independent of each other as if they are only subject to a central field generated by a single effective charge

Q_e = q\frac{2\sqrt{2}+1}{4} \approx 0.96q,

fixed at the center of mass of the system, and initially at a distance $L/\sqrt{2}$ from each particle. This observation allows us to reduce the initial four-body problem into four independent (and identical) two-body problems involving only the spatially fixed charge $Q_e$ and one moving charge $q$ .The trajectories are therefore hyperbolas with one focus located at the position of the effective charge. Since the force is repulsive, the effective charge is at the focus $F_1$ as shown in the following figure.

Hyperbolic trajectory drawn bold starting at a point on the x-axis with initial velocity v0 pointing up, passing through a point P; the two asymptotes cross at the origin, the other branch and the mirror branch are dashed. Foci F1 (charge Qe) on the left and F2 on the right; distances r1 and r2 from P to F1 and F2, angle β at F1, angles φ and α at the origin; braces mark L/√2, a and c along the x-axis; one asymptote is labelled asymptote.

Let’s now calculate the asymptote angle $\alpha$ , since the required angle $\phi$ is just $\pi/2 - \alpha$ .

A hyperbola is defined as a set of points $P$ , such that the absolute difference of the distances from $P$ to two fixed points $F_1$ and $F_2$ (the foci) is constant. Let the distances be $r_1$ and $r_2$ . From the figure it is easy to see that the constant is equal to $2a$ : if we take the point $P$ on the $x$ -axis, then $r_1 - r_2 = c + a - (c - a) = 2a$ .

To determine the asymptote angle $\alpha$ , let us look at the triangle $F_1PF_2$ . From the law of cosines, we have

r_2^2 = r_1^2 + 4c^2 - 4cr_1\cos\beta.

Solving for $\cos\beta$ and using $r_2 = r_1 - 2a$ we get

\cos\beta = \frac{r_1^2 - r_2^2 + 4c^2}{4cr_1} = \frac{a}{c} + \frac{c^2 - a^2}{cr_1}.

As $r_1 \to \infty$ , then $\beta \to \alpha$ , and we get the expression for the cosine of the asymptote angle:

\cos\alpha = \frac{a}{c}.

Now using $c = L/\sqrt{2} - a$ and knowing that

E = \frac{kqQ_e}{2a} = \frac{mv_0^2}{2} + \frac{\sqrt{2}kqQ_e}{L}

(the vis-viva equation for the hyperbola), from which

\frac{1}{a} = \frac{mv_0^2}{kqQ_e} + \frac{2\sqrt{2}}{L},

we finally get

\sin\phi = \cos\alpha = \frac{a}{c} = \frac{1}{\frac{L}{\sqrt{2}a} - 1} =

= \frac{1}{1 + \frac{Lmv_0^2}{\sqrt{2}kqQ_e}} = \frac{1}{1 + \frac{4Lmv_0^2}{kq^2(4+\sqrt{2})}}.

Grading:

Effective repulsive central force $F = \frac{A}{r^2}$ pointing from the COM (explicit expression or statement is required) (1 pts)
Realising hyperbolic motion (0.5 pts)
Central charge is located at the correct focus (0.5 pts)
Expression for the angle $\phi$ or $\alpha$ in terms of geometrical parameters e.g. $\cos\alpha = a/c$ , $\tan\alpha = b/a$ (1 pts)
Vis-viva equation OR angular momentum conservation (1 pts)
Deriving final answer (1 pts)

Subpoints for angular momentum conservation:

Mentioned that angular momentum is conserved (0.3 pts)
Writes initial angular momentum around COM using $L$ and $v_0$ (0.3 pts)
Writes final angular momentum around COM using $v_f$ (0.4 pts)

Variation 1.

If the student doesn’t know the vis-viva equation for the hyperbola

E = \frac{kqQ_e}{2a},

which differs from the corresponding equation for the ellipse only by the sign of the total energy, it can be derived as follows.

From the asymptote angle formula

\cos\alpha = \frac{a}{c}

the geometric relationships shown in the following figure follow immediately.

Geometry of the hyperbola near its vertex: focus F1 (charge Qe) on the left, F2 on the right, the asymptote labelled asymptote through the centre at angle α; two identical right triangles with sides a, b, c and angle α; braces mark c, a and L/√2 along the axis; the bold hyperbola branch starts at the vertex with a vertical segment b.

We see that the two right triangles on the figure are identical.

From the conservation of angular momentum (with respect to the point $F_1$ ) we have

m(c + a)v_0 = mbv_f \implies v_f = v_0\frac{c + a}{b},

and from the conservation of energy using $b^2 = c^2 - a^2$

\frac{kqQ_e}{a + c} = \frac{mv_f^2}{2} - \frac{mv_0^2}{2} =

= \frac{mv_0^2}{2}\left(\frac{(c + a)^2}{c^2 - a^2} - 1\right) = \frac{mv_0^2}{2}\frac{2a}{c - a},

from which

\frac{mv_0^2}{2} = \frac{c - a}{2a}\cdot\frac{kqQ_e}{c + a}.

Inserting this into the total energy expression we finally get

E = kqQ_e\left(\frac{1}{2a}\cdot\frac{c - a}{c + a} + \frac{1}{c + a}\right) = \frac{kqQ_e}{2a}.

Variation 2.

Using the algebraic equation for the hyperbola

\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,

and knowing that $b^2 = c^2 - a^2$ , we get

\tan\alpha = \lim_{x\to\infty}\frac{y}{x} = \lim_{x\to\infty} b\sqrt{\frac{1}{a^2} - \frac{1}{x^2}} = \frac{b}{a}.

Variation 3.

The vis-viva equation step can be done using conservation of angular momentum around COM, as all the forces are radial.

\mathcal{L} = mv_0\frac{L}{\sqrt{2}} = mv_fb

since we already know $a + c = \frac{L}{\sqrt{2}}$ , $c^2 = a^2 + b^2$ , we can solve for the hyperbolas parameters.

Оригинал в Архива: NBPhO_2024_problems_v1.0.pdf · официални решения: NBPhO_2024_solutions_v0.98.pdf