Free Practice Questions for Slovak Physics Olympiad Cat A

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Slovak Mathematical Olympiad — Category A (Matematická olympiáda, kategória A)

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Key Facts: Slovak Physics Olympiad Cat A Exam

The 68th year of the Slovak Physics Olympiad takes place in the 2026/2027 school year.

NIVaM organizes the competition and JSMF is its expert guarantor.

Category A covers the whole secondary-school physics and mathematics curriculum and follows the IPhO syllabus.

The category A home round has 7 tasks including at least one experimental task.

The regional round has 4 theoretical tasks and lasts 4 hours.

Each task is worth 10 points and counts as successful from 5 points.

The national round invites about 40 contestants and combines 4 theoretical tasks with a 20-point experimental task.

The 2026/2027 national round of category A is scheduled for 8–11 April 2027.

The Slovak IPhO and EuPhO teams are selected from the national-round winners.

Tasks are set in Slovak, with Hungarian translations provided for Hungarian-speaking contestants.

Physics Olympiad category A (2026/2027) has a 7-task home round, a 4-hour regional round with 4 theoretical tasks (10 points each, success from 15 of 40 points with one solved task), and a national round with 4 theoretical tasks and a 20-point experiment on 8–11 April 2027.

Sample Slovak Physics Olympiad Cat A Practice Questions

Try these sample questions to review concepts for the Slovak Physics Olympiad Cat A exam. Each question includes a detailed explanation. Start the interactive quiz above for the full 102+ question experience with AI tutoring.

1A projectile is launched with an initial speed v_0 at an angle α above an inclined plane which itself makes an angle β with the horizontal (where α + β < 90°). Assuming a uniform gravitational acceleration g and neglecting air resistance, what is the range R of the projectile measured along the incline?

A.R = (2 v_0² sin α cos(α - β)) / (g cos² β)

B.R = (2 v_0² sin α cos(α + β)) / (g cos² β)

C.R = (v_0² sin(2α)) / (g cos β)

D.R = (2 v_0² sin(α + β) cos α) / (g sin² β)

Explanation: Setting up a tilted coordinate frame with the x-axis parallel to the incline and the y-axis perpendicular, the components of gravitational acceleration are g_x = -g sin β and g_y = -g cos β. The initial velocity components are v_{0x} = v_0 cos α and v_{0y} = v_0 sin α. The flight time until the projectile returns to the incline (y = 0) is t = (2 v_{0y}) / (g cos β) = (2 v_0 sin α) / (g cos β). Substituting t into the x-position equation x(t) = v_{0x} t - (1/2) g sin β t² yields R = (2 v_0² sin α) / (g cos² β) [cos α cos β - sin α sin β] = (2 v_0² sin α cos(α + β)) / (g cos² β).

2A small body of mass m is dropped from rest from a height h above the Earth's surface at latitude λ in the Northern Hemisphere. Taking Earth's angular velocity as ω and the local gravitational acceleration as g, what is the magnitude and direction of the horizontal deflection d of the body when it hits the ground (to first order in ω)?

A.d = (1/3) ω cos λ √(8 h³ / g) directed westward

B.d = (2/3) ω sin λ √(2 h³ / g) directed southward

C.d = (1/3) ω cos λ √(8 h³ / g) directed eastward

D.d = ω cos λ √(2 h³ / g) directed northward

Explanation: In the rotating reference frame of Earth, the Coriolis acceleration is a_cor = -2 (ω × v). As the body falls downward with velocity v_z ≈ -g t, the cross product between the Earth's angular velocity vector ω (pointing north at angle λ to the horizontal) and the downward velocity yields an eastward Coriolis acceleration a_x = 2 ω g t cos λ. Integrating twice with respect to time gives the eastward displacement x(t) = (1/3) ω g cos λ t³. Substituting the free-fall time t = √(2h/g) yields d = (1/3) ω g cos λ (2h/g)^{3/2} = (1/3) ω cos λ √(8 h³ / g) directed eastward.

3A single-stage sounding rocket has an initial total mass m_0 and burns fuel at a constant mass rate r = -dm/dt with constant exhaust speed u_{ex} relative to the rocket nozzle. The rocket is launched vertically from rest in a uniform gravitational field g. If the engine burns for a total duration t_b, what is the vertical velocity v(t_b) at engine burnout?

A.v(t_b) = u_{ex} ln(m_0 / (m_0 - r t_b)) - g t_b

B.v(t_b) = u_{ex} (r t_b / m_0) - g t_b

C.v(t_b) = u_{ex} ln(m_0 / (m_0 - r t_b)) - (1/2) g t_b

D.v(t_b) = u_{ex} ln((m_0 - r t_b) / m_0) - g t_b

Explanation: From the variable-mass momentum balance m dv = -u_{ex} dm - m g dt, dividing by m gives dv = -u_{ex} (dm/m) - g dt. Integrating from t = 0 to t_b where m(t) = m_0 - r t yields v(t_b) = -u_{ex} ln(m(t_b)/m_0) - g t_b = u_{ex} ln(m_0 / (m_0 - r t_b)) - g t_b. This is the classical Tsiolkovsky rocket equation modified for vertical ascent under uniform gravity.

4What is the moment of inertia I_cm of a uniform thin planar equilateral triangular sheet of mass M and side length L about an axis perpendicular to the sheet passing through its center of mass?

A.I_cm = (1/6) M L²

B.I_cm = (1/8) M L²

C.I_cm = (1/18) M L²

D.I_cm = (1/12) M L²

Explanation: By scaling symmetry or decomposing the equilateral triangle into smaller self-similar triangles, the moment of inertia about the center of mass perpendicular to the plane satisfies I_cm = (1/12) M L². Alternatively, using the perpendicular axis theorem I_z = I_x + I_y and integrating along strips parallel to one base yields I_base = (1/6) M h² where h = (√3/2) L. Applying Steiner's parallel axis theorem with distance d = h/3 from base to center of mass gives I_{cm,x} = (1/6) M h² - M (h/3)² = (1/18) M h² = (1/24) M L². By three-fold rotational symmetry I_{cm,x} = I_{cm,y}, so I_z = 2 I_{cm,x} = 2 (1/24) M L² = (1/12) M L².

5A uniform solid sphere of mass M and radius R rolls without slipping from rest down an incline of angle θ and vertical height h. What is the translational speed v of the center of mass when it reaches the bottom of the incline?

A.v = √(2 g h)

B.v = √(10 g h / 7)

C.v = √(4 g h / 3)

D.v = √(6 g h / 5)

Explanation: Conservation of mechanical energy states that the potential energy drop M g h converts into translational and rotational kinetic energy: M g h = (1/2) M v² + (1/2) I ω². For a uniform solid sphere, I = (2/5) M R². Since it rolls without slipping, ω = v / R. Substituting gives M g h = (1/2) M v² + (1/2) (2/5 M R²) (v/R)² = (1/2 + 1/5) M v² = (7/10) M v². Solving for v yields v = √(10 g h / 7) ≈ 1.195 √(g h).

6A uniform thin rod of mass M and length L is pivoted freely about a frictionless horizontal axis at one end. The rod is held horizontally and released from rest. When the rod swings through the vertical position, what is the magnitude of the reaction force exerted by the pivot on the rod?

A.F_pivot = M g

B.F_pivot = 2 M g

C.F_pivot = (5/2) M g

D.F_pivot = 3 M g

Explanation: As the rod falls from the horizontal position to the vertical, its center of mass drops by h = L/2. By conservation of energy, M g (L/2) = (1/2) I_pivot ω². For a rod pivoted at its end, I_pivot = (1/3) M L². Thus M g L = (1/3) M L² ω², giving ω² = 3 g / L. At the bottom of the swing, the center of mass has upward centripetal acceleration a_c = ω² (L/2) = (3 g / L) (L/2) = (3/2) g. Applying Newton's second law vertically: F_pivot - M g = M a_c = (3/2) M g, which yields F_pivot = M g + (3/2) M g = (5/2) M g.

7A light horizontal circular wheel with moment of inertia I_0 is rotating freely at angular velocity ω_0 about a frictionless vertical axle. A small beetle of mass m is at the center of the wheel. The beetle begins walking radially outwards with constant speed u relative to the wheel along a spoke. What is the angular velocity ω(r) of the wheel when the beetle is at distance r from the axle?

A.ω(r) = (I_0 ω_0) / (I_0 + m r²)

B.ω(r) = ω_0 - (m u r) / I_0

C.ω(r) = ω_0 √(I_0 / (I_0 + m r²))

D.ω(r) = (I_0 ω_0) / (I_0 - m r²)

Explanation: Because no external torques act about the vertical axle, the total angular momentum of the system is strictly conserved. The total moment of inertia when the beetle is at radius r is I(r) = I_0 + m r². Equating initial and instantaneous angular momentum gives L = I_0 ω_0 = (I_0 + m r²) ω(r). Solving for ω(r) yields ω(r) = (I_0 ω_0) / (I_0 + m r²).

8What is the gravitational potential V at the center of the flat circular base of a uniform solid hemisphere of mass M and radius R (taking V = 0 at infinity)?

A.V = - (1/2) (G M / R)

B.V = - (G M / R)

C.V = - (4/3) (G M / R)

D.V = - (3/2) (G M / R)

Explanation: Divide the hemisphere into thin hemispherical shells of radius r and thickness dr centred on the centre of the flat base. Every point of such a shell is at distance r from that centre, and its volume is dV = 2π r² dr. With density ρ = M / ((2/3) π R³) = 3M / (2π R³), the potential at the centre is V = -G ∫ dm / r = -G ∫_0^R (ρ 2π r² dr) / r = -2π G ρ ∫_0^R r dr = -π G ρ R². Substituting ρ gives V = -π G (3M / (2π R³)) R² = -(3/2) (G M / R).

9A satellite moves in an elliptical Keplerian orbit of semi-major axis a and eccentricity e around Earth (mass M_E). What is the orbital speed v_p of the satellite at periapsis?

A.v_p = √((G M_E / a) · ((1 - e) / (1 + e)))

B.v_p = √((G M_E / a) · ((1 + e) / (1 - e)))

C.v_p = √((G M_E / a) · (1 + e²))

D.v_p = √((2 G M_E / a) · (1 + e))

Explanation: According to the vis-viva equation, the orbital speed at any radial distance r from the primary body is v² = G M_E (2/r - 1/a). At periapsis, the distance is r_p = a(1 - e). Substituting r_p into the vis-viva equation gives v_p² = G M_E (2 / (a(1 - e)) - 1/a) = (G M_E / a) ((2 - (1 - e)) / (1 - e)) = (G M_E / a) ((1 + e) / (1 - e)). Taking the square root yields v_p = √((G M_E / a) ((1 + e) / (1 - e))).

10A spacecraft in a circular parking orbit of radius r_1 around a central body of mass M performs a tangential burn to enter a Hohmann transfer ellipse that reaches a coplanar outer circular orbit of radius r_2 (where r_2 > r_1). What is the velocity increment Δv_1 required for this first burn?

A.Δv_1 = √(G M / r_1) · (√(2 r_1 / (r_1 + r_2)) - 1)

B.Δv_1 = √(G M / r_1) · (√(r_2 / r_1) - 1)

C.Δv_1 = √(G M / r_1) · (√(2 r_2 / (r_1 + r_2)) - 1)

D.Δv_1 = √(G M / r_2) · (1 - √(2 r_1 / (r_1 + r_2)))

Explanation: The initial circular orbit velocity is v_{c1} = √(G M / r_1). The semi-major axis of the Hohmann transfer ellipse is a = (r_1 + r_2) / 2. By the vis-viva equation, the velocity at periapsis (r = r_1) on the transfer orbit is v_{t1} = √(G M (2/r_1 - 1/a)) = √(G M (2/r_1 - 2/(r_1 + r_2))) = √(G M / r_1) √(2 r_2 / (r_1 + r_2)). The velocity increment needed is therefore Δv_1 = v_{t1} - v_{c1} = √(G M / r_1) (√(2 r_2 / (r_1 + r_2)) - 1).

About the Slovak Physics Olympiad Cat A Exam

Independent study practice for category A of the Slovak Physics Olympiad (Fyzikálna olympiáda), the national physics competition for students in their final year of secondary school. In the 68th year (2026/2027) the competition runs under the Physics Olympiad organizational rules (organizačný poriadok) and the ministry's directive No. 24/2026 on competitions, with NIVaM as organizer and JSMF as expert guarantor. Category A is based on the whole secondary-school physics and mathematics curriculum and the IPhO syllabus. Students solve 7 home-round tasks including an experimental one, 4 theoretical tasks in the 4-hour regional round, and in the national round 4 theoretical tasks and a 20-point experimental task. Each 10-point task counts as successful from 5 points. National-round winners form the pool for the IPhO and EuPhO teams. The 102 practice questions cover mechanics, thermodynamics, electromagnetism, optics, modern physics and experimental data analysis. This bank is an English-language, four-option multiple-choice study adaptation for checking ideas and results; it is not an official problem set and does not replace writing complete solutions or doing experiments.

Exam sponsor: NIVaM (Národný inštitút vzdelávania a mládeže) as organizer, with the Slovak Physics Olympiad Committee and JSMF as expert guarantor, under the Ministry of Education's directive on competitions. The requirements and fees below concern the certification or admission exam, separate from our free practice resources.

Assessment

Home round (domáca príprava; in 2026/2027 solutions are due by 11 January 2027) → regional round (krajské kolo; 10 February 2027; 4 theoretical tasks, 4 hours) → national round (celoštátne kolo; 8–11 April 2027; about 40 contestants; 4 theoretical tasks worth 10 points each and 1 experimental task worth 20 points). The Slovak IPhO and EuPhO teams are selected from the national-round winners.

Time Limit

Regional round 4 hours; national round 5 hours (theory) plus at least 3 hours (experiment)

Passing Score

A task is successful with at least 5 of 10 points; regional-round success requires at least one successful task and at least 15 of 40 points; national-round success threshold is half of the average of the three best scores.

Exam / Certification Fees

Free; the competition is organized by NIVaM for students registered by their schools.

Exam sponsor website

Fees, eligibility, and exam policies can change. Confirm them with the exam sponsor before applying or paying.

Official sources

Our practice resources: topics covered

We aim to reflect publicly available exam outlines and topic information in our study resources. Coverage, format, and difficulty may differ from the actual exam, and we cannot guarantee that every detail is accurate or current. Confirm exam requirements, fees, and policies with the official exam sponsor.

No official weighting; 32 of 102 practice questions

Pokročilá mechanika (Advanced Mechanics)

Kinematics and dynamics, rigid bodies, oscillations, gravitation and orbits, fluids, non-inertial frames, and experimental methods such as graph linearization and uncertainty estimates.

No official weighting; 20 of 102 practice questions

Termodynamika a štatistická fyzika (Thermodynamics and Statistical Physics)

Ideal and real gases, the laws of thermodynamics, cycles and efficiency, entropy, kinetic theory, heat transfer and phase changes.

No official weighting; 25 of 102 practice questions

Elektromagnetizmus a obvody (Electromagnetism and Circuits)

Electrostatics, capacitors, DC and AC circuits, magnetic fields, electromagnetic induction, motion of charges in fields, and energy flow.

No official weighting; 13 of 102 practice questions

Vlnová a geometrická optika (Wave and Geometric Optics)

Refraction and lenses, optical instruments, interference, diffraction, gratings and polarization.

No official weighting; 12 of 102 practice questions

Kvantová, relativistická a jadrová fyzika (Quantum, Relativistic and Nuclear Physics)

Photons and the photoelectric and Compton effects, atomic spectra, the uncertainty principle, special relativity, radioactive decay and nuclear reactions.

Preparing for the Slovak Physics Olympiad Cat A Exam

What You Need to Know

Passing score: A task is successful with at least 5 of 10 points; regional-round success requires at least one successful task and at least 15 of 40 points; national-round success threshold is half of the average of the three best scores.
Assessment: Home round (domáca príprava; in 2026/2027 solutions are due by 11 January 2027) → regional round (krajské kolo; 10 February 2027; 4 theoretical tasks, 4 hours) → national round (celoštátne kolo; 8–11 April 2027; about 40 contestants; 4 theoretical tasks worth 10 points each and 1 experimental task worth 20 points). The Slovak IPhO and EuPhO teams are selected from the national-round winners.
Time limit: Regional round 4 hours; national round 5 hours (theory) plus at least 3 hours (experiment)
Exam / certification fees: Free; the competition is organized by NIVaM for students registered by their schools. Official sources

Using Our Practice Resources

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Slovak Physics Olympiad Cat A: Suggested Study Strategy

1Solve past Slovak Physics Olympiad category A problems in full, writing the general solution first and substituting numbers only at the end, as the official instructions require.

2Practise experimental tasks: plan measurements, linearize graphs, choose sensible scales and estimate uncertainties, since the home and national rounds include experiments.

3Work through the IPhO syllabus systematically, paying special attention to rotational dynamics, thermodynamic cycles, electromagnetic induction and wave optics.

4Train for the 4-hour regional round by solving four unseen theoretical problems under time pressure.

5Check every result with dimensional analysis and limiting cases before moving on.

Frequently Asked Questions

Who can compete in category A of the Physics Olympiad?

Category A is for students in the final year of secondary school: the 4th year of four-year gymnázia, the 8th year of eight-year gymnázia, or the maturita year of other secondary schools. Every student may compete in their own category or a higher one, and the school registers participants in EduPage.

What are the rounds of category A?

In 2026/2027 the home round (domáca príprava) has 7 tasks including at least one experimental task, with solutions due by 11 January 2027. Successful solvers advance to the regional round on 10 February 2027 (4 theoretical tasks in 4 hours). About 40 of the best regional solvers are invited to the national round on 8–11 April 2027.

How is the Physics Olympiad scored?

Each task is worth 10 points and is considered successfully solved with at least 5 points. In the home round, a successful solver submits all tasks and solves at least five successfully, including the experimental work. In the regional round, success requires at least one successful task and at least 15 points out of 40. The national round has 4 theoretical tasks (40 points) and an experimental task (20 points); its success threshold is half of the average of the three best scores.

Is this practice bank an official Physics Olympiad problem set?

No. This is independent practice by OpenExamPrep: an English-language, four-option multiple-choice study adaptation covering topics from the IPhO-level syllabus used in category A. It is not an official problem set or translation, and the real competition requires complete written solutions and experimental work.