1st Balkan Physics Olympiad – 2019 BPO

July 14-18, Thessaloniki, Greece

Solution to Problem 1

1. In the simplest case the motion is along the vertical axis starting from the origin at the
bottom of the parabola with an arbitrary initial velocity v0. The motion is described by x =
0, y = v0t – gt2/2, vy = v0 – gt. The highest point of the trajectory is reached at time t = v0/g.
The period of the motion is T = 2v0/g.

2. Another simpler case is a trajectory that is symmetric to the parabola that bounds the
motion from below. This is illustrated with the following figure. In order to have periodic
motion the trajectory of the particle must reach the boundary at an angle of 90 o. The point
x0 is not arbitrary. It is specific for the given bounding parabola. If v0 is the speed at the
instant of collision with the boundary given by y = ax2 - b, a > 0, b > 0, the trajectory of the
particle is described by
x = - x0 + v0 cos α t, y = v0 sin α t – gt2/2, α = 45o.

Elimination of t from the above equations, with the additional requirement that the
trajectory has the symmetric form y = b – ax2, and taking into account that α = 45o, one finds
that the initial speed of the particle at the boundary should be v02 = 2gx0. Therefore the
period is T = 4x0/(v0 cos α) = 4√(x0/g).

3. The boundary may be represented as y = ax2. Let us take x0 > 0 and (± x0, ax02) as the end
points of the periodic trajectory. The tangents of the boundary at these end points, at the
right and left side, are y = ± 2ax0x – ax02. The orthogonal direction to the tangent on the left
is given by y = (x + x0)/(2ax0) + ax02. Motion of the point particle is described by

x = – x0 + v0 cos β t, y = y0 +v0 sin β t – gt2/2, tan β = 1/(2ax0).

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After elimination of t, the symmetry requirement implies the condition tan β = gx0/(v02 cos2
β) and the trajectory is given by y = y0 + g (x02 – x2)/(2v02 cos2β). The condition of
orthogonality at the boundary leads to 2ax0 gx0/(v02 cos2β) = 1. The speed at the boundary
reflection point is

v02 = 2agx02/cos2β = 2agx02 (1 + tan2 β) = 2agx02 (1 + 1/(4a2x02)) = g(1 + 4a2x02)/(2a) .

and the period T = 4x0/(v0 cos β) = 2√(2/ag) is independent of x0. This is in agreement with
the result of section 2, where tan α = 2ax0 = 1, and x0 = 1/2a, that is T = 2√(2/ag).

4. The trajectory in the following figure has a parabolic section in the middle and two
vertical sections at both ends. Indicated angles have to satisfy the relationship α + 2γ = π/2.

Such trajectory appears when the orthogonal line at the point of collision with the
boundary divides in two halves the angle between the vertical line and the tangent at the
end of the parabolic section of the trajectory. The period of this motion is given by T =
4x0/(v0 cos α) + 4v0/g and v02 = gx0/(sin α cos α) or

T = 4 [1 + √(1 + 1/tan2 α)] √[(x0/g) tan α] .

5. In this case the orbit, as shown in the figure, touches the boundary at three points. The
boundary and the right and left branches of the trajectory are given by y = ax2, y = bx2 ± (a –
b)x0x, with ±x0 the abscissa of the collision points with the boundary (a > 0, b < 0, y0 = ax02).
From the orthogonality condition at the boundary 2ax02(a + b) = 1, it follows b = 1/(2ax02) – a.

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The period is found from T = 4x0/(v0 cos α) where tan α = 1/(2ax0) and v0 is expressed from
y0 + 1/2a – (gx02/2v02)(1+ 1/4a2x02) = 0.

In analogy to the situation in section 4, the trajectory of section 5 can have vertical tails
under similar conditions. There are probably two additional periodic trajectories that are
not discussed.

Solution to Problem 2

a)

NA NB S
F’ F S’

ΒA ΒB ΒC

b)
The relevant equations are:
2𝑚𝑎𝐵 = 𝑚𝑔 − 𝑘𝑥
and
𝑚𝑎𝐴 = 𝑘𝑥.
Let us study the movement of B (and C) with respect to A: multiply the second eq. with 2
and subtract it from the first:
2𝑚(𝑎𝐵 − 𝑎𝐴 ) = 𝑚𝑔 − 3𝑘𝑥.
So, the problem is equivalent with suspending a body with the mass 2𝑚 by a spring with
the elastic constant 3𝑘. The elongation of this spring is
2𝑚𝑔
𝑥𝑚𝑎𝑥 = .
3𝑘
In conclusion, the maximum distance between A and B is
2𝑚𝑔
𝐿𝑚𝑎𝑥 = 𝐿0 + .
3𝑘

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(c)
NA NB S
fst S’

ΒA ΒB ΒC
Initially, d = l0. Since the bodies should remain at rest d = l0 ➔ F = 0

m.g -S = 0
S’-μ.m.g=0

m.g- μ.m.g = 0 ➔ μ = 1

(a) NA NB S
F’ F S’
fA fB
ΒA ΒC

Constant speed:
m.g-S =0
S’-F-fB=0
F’-fA=0
➔ m.g - fA - fB=0
m.g-2μ.m.g=0

➔ m.g = 2μ.m.g ➔ μ=1/2

F - fA = 0
k.x = μ.m.g

𝜇.𝑚.𝑔 𝑚.𝑔 𝑚.𝑔

𝑥= = ➔𝑙 = 𝑙0 +
𝑘 2𝑘 2𝑘

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Solution to Problem 3

Data:
m
l1 = 600 m , v1 = 1 ,
s
m
l2 = 800 m , v2 = 2
s

Expression for time t1 for the motion on the field as a function of x :

AD x 2 + l12
t1 = = (1)
v1 v1

Expression for time t2 for the motion on the road as a function of x :

DC l2 − x
t2 = = (2)
v2 v2

Expression for the total time t as a function of x :

1 1
t = t1 + t2 = x 2 + l12 + (l2 − x ) (3)
v1 v2

Solutions for the total time and position, t = t0 and x = x0 :

1 1
t0 = x0 2 + l12 + (l2 − x0 ) (4)
v1 v2
After some elementary algebraic manipulations in (4) we get
v12v22t02 + v12l22 + v12 x02 − 2v12v2t0l2 + 2v12v2t0 x0 − 2v12 x0l2 − v22 x02 − v22l12 = 0 (5)

“Shaking“ the solutions with small quantities t and x :

v12v2 2 (t0 + t )2 + v12l2 2 + v12 ( x0 + x)2 − 2v12v2 (t0 + t )l2 +
(6)
+2v12v2 (t0 + t )( x0 + x) − 2v12 ( x0 + x)l2 − v2 2 ( x0 + x) 2 − v2 2l12 = 0

(6)  v12 v2 2 (t0 2 + 2t0 t + t 2 ) + v12l2 2 + v12 ( x0 2 + 2 x0 x + x 2 ) −

−2v12 v2 (t0 + t )l2 + 2v12 v2 (t0 + t )( x0 + x) − 2v12 ( x0 + x)l2 − (7)
−v2 2 ( x0 2 + 2 x0 x + x 2 ) − v2 2l12 = 0

(5) → (7)  v12 v2 2 (2t0 t + t 2 ) + v12 (2 x0 x + x 2 ) −

−2v12 v2 tl2 + 2v12v2 (tx0 + t0 x + t x ) − 2v12 xl2 − (8)
−v2 (2 x0 x + x ) = 0
2 2

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 Neglecting all the terms containing very small products, t  t , t  x and x  x :
(8)  v12v22t0t + v12 x0x − v12v2tl2 + v12v2 (tx0 + t0x) − v12xl2 − v22 x0x = 0 (9)

Expression for t / x :
v2 2
l2 + (
− 1) x0 − v2t0
t v12
(9)  = (10)
x v2 2t0 − v2l2 − v2 x0

Equating x / t to zero and evaluation of x and t :

t
=0 (11)
x
v2 2
(10), (11)  l2 + ( − 1) x0 − v2t0 = 0 (12)
v12
l1
(4) → (12)  x0 = ... = (13)
v2 2
−1
v12

1  v2 
(13) → (4)  t0 = ... = l2 + l1 22 − 1  (14)
v2  v1 
(13)  x0 = ... = 346.8 m (15)
(14)  t0 = ... = 920 s (16)

Solution to Problem 4

As after the switching of the K1 the resistance of the circuit remains the same, the
potentials on the both ends of K1 are equal (voltage on K1 is zero). This situation is only
possible when the initial state of К2 is ”1”. Later, as the 𝑅ABγ > 𝑅ABδ , it can be concluded
that the last state of K1 is ”1”. So, using these conclusions we can fill the state of the
switches for all states of the circuit (see the table).
Now we can start calculations of the values of the resistors. As in the beginning both
switches are closed and the opening of the K1 does not change the circuit resistance, it
follows 𝑈𝑅1 = 𝑈𝑅2 and 𝑈𝑅3 = 𝑈𝑅4 . The relations of the currents are 𝐼𝑅1 = 𝐼𝑅3 and 𝐼𝑅2 = 𝐼𝑅4 .
𝑅2 𝑅
From these equations we can obtain that = 𝑅4 = 𝑥. To minimize the above expressions,
𝑅1 3

we can note 𝑅1 = 𝑅, 𝑅2 = 𝑥𝑅, let 𝑅3 = 𝑦𝑅, then 𝑅4 = 𝑥𝑦𝑅.

𝑅(1+𝑦)𝑅𝑥(1+𝑦) 𝑅𝑥(1+𝑦)
For circuit state “𝛽”: 𝑅(1+𝑦)+𝑅𝑥(1+𝑦) = 240 Ω . After simplification = 240 Ω (1).
1+𝑥
For circuit state “𝛾”: 𝑥(1 + 𝑦)𝑅 = 400 Ω . (2)
𝑥𝑅𝑅
For circuit state “𝛿”: 𝑥𝑅+𝑅 + 𝑥𝑦𝑅 = 280 Ω . (3)
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 400 Ω 400 2
Substituting (2) in (1), = 240 Ω, then 𝑥 = −1= .
1+𝑥 240 3
Subtracting (3) from (2), 𝑥(1 + 𝑦)𝑅 −
𝑅1 𝑅3 𝑥𝑅𝑅
+ 𝑥𝑦𝑅 = 120 Ω. After simplification,
𝑥𝑅+𝑅
K2 𝑥2𝑅
A B = 120 Ω. Substituting 𝑥 with the
𝑥+1
K1
obtained value, 𝑅 = 450 Ω. Now
substituting both 𝑥 and 𝑅 with their
𝑅2 𝑅4 2
values in (2), (1 + 𝑦)450 Ω = 400 Ω, we
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1
obtain 𝑦 = 3. So, the final values of the resistor are, 𝑅1 = 450 Ω, 𝑅2 = 300 Ω, 𝑅3 = 150 Ω,
and 𝑅4 = 100 Ω.

The answer is given in the table below.

State of State of State of 𝑅ABi , Resistor Value,
the circuit the K1 the K2 Ω 𝑅𝑗 Ω
𝛼 1 1 240 𝑅1 450
𝛽 0 1 240 𝑅2 300
𝛾 0 0 400 𝑅3 150
𝛿 1 0 280 𝑅4 100