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100+ Free OBF Practice Questions

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Sample OBF Practice Questions

Try these sample questions to test your OBF exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1A particle moves along the x-axis under a conservative force with potential energy U(x) = α x⁴ - β x², where α = 2.0 J/m⁴ and β = 8.0 J/m². The particle is released from rest with total mechanical energy E = 0 J. What are the positions of the stable equilibrium points and the maximum speed attained by the particle if its mass is m = 0.50 kg?
A.Stable equilibria at x = ±√2 m (≈ ±1.41 m) and maximum speed v_max = 4√2 m/s ≈ 5.66 m/s
B.Stable equilibria at x = 0 m and maximum speed v_max = 2.0 m/s
C.Stable equilibria at x = ±2.0 m and maximum speed v_max = 8.0 m/s
D.Stable equilibria at x = ±1.0 m and maximum speed v_max = 4.0 m/s
Explanation: Equilibrium points occur where dU/dx = 4αx³ - 2βx = 2x(2αx² - β) = 0, giving x = 0 (unstable, since d²U/dx² = -2β = -16 < 0) and x = ±√(β/(2α)) = ±√(8/4) = ±√2 m (stable, since d²U/dx² = 12αx² - 2β = 48 - 16 = +32 > 0). The potential minimum is U_min = U(±√2) = 2(2)² - 8(2) = 8 - 16 = -8.0 J. By conservation of energy E = K + U = 0, K_max = -U_min = 8.0 J. Thus, (1/2) m v_max² = 8.0 J ⇒ v_max = √(2 · 8.0 / 0.50) = √32 ≈ 5.66 m/s.
2A uniform solid cylinder of mass M and radius R rolls without slipping down an incline of angle θ. What is the linear acceleration of the cylinder's center of mass and the minimum coefficient of static friction μ_s required to prevent slipping?
A.a = (1/2) g sin θ and μ_s = (1/2) tan θ
B.a = (2/3) g sin θ and μ_s = (1/3) tan θ
C.a = (3/4) g sin θ and μ_s = (1/4) tan θ
D.a = (5/7) g sin θ and μ_s = (2/7) tan θ
Explanation: For a solid cylinder rolling down an incline, Newton's second law along the incline gives M g sin θ - f_s = M a. Rotational dynamics about the center of mass gives f_s R = I_cm α = (1/2 M R²)(a/R), so f_s = (1/2) M a. Substituting f_s yields M g sin θ - (1/2) M a = M a, which simplifies to a = (2/3) g sin θ. The friction force is f_s = (1/3) M g sin θ. To avoid slipping, f_s ≤ μ_s N = μ_s M g cos θ, which requires μ_s ≥ (1/3) tan θ.
3A small block of mass m rests on top of a frictionless sphere of radius R fixed to the ground. The block is given a tiny push so that it slides down the surface. At what vertical height h above the ground (measured from the base of the sphere) does the block lose contact with the sphere?
A.h = (4/3) R
B.h = (3/2) R
C.h = (5/3) R
D.h = (7/5) R
Explanation: Let θ be the angle measured from the vertical apex. Conservation of mechanical energy gives m g R (1 - cos θ) = (1/2) m v², so v² = 2 g R (1 - cos θ). Radial equation of motion is m g cos θ - N = m v² / R. Contact is lost when normal force N = 0, giving g cos θ = v² / R = 2 g (1 - cos θ). Solving yields 3 cos θ = 2, so cos θ = 2/3. The height above the center of the sphere is R cos θ = (2/3) R, and the total height above the base is R + (2/3) R = (5/3) R.
4A projectile is launched from the ground on a flat horizontal plane with initial speed v₀ at an angle α above the horizontal. Air resistance is negligible. If the projectile's maximum height equals its horizontal range (H = R), what is the launch angle α?
A.α = arctan(2) ≈ 63.4°
B.α = arctan(1) = 45.0°
C.α = arctan(3) ≈ 71.6°
D.α = arctan(4) ≈ 76.0°
Explanation: The maximum height is H = (v₀² sin² α) / (2g) and the horizontal range is R = (v₀² sin 2α) / g = (2 v₀² sin α cos α) / g. Setting H = R gives (v₀² sin² α) / (2g) = (2 v₀² sin α cos α) / g. Dividing both sides by (v₀² sin α) / g (since α ≠ 0) yields (sin α) / 2 = 2 cos α, which simplifies to tan α = 4, so α = arctan(4) ≈ 76.0°.
5A satellite of mass m is in a circular orbit of radius r around a planet of mass M. To transfer the satellite to a higher circular orbit of radius 2r using a standard Hohmann transfer orbit, what is the total speed change Δv_total = Δv₁ + Δv₂ required from the thrusters?
A.Δv_total = √(GM/r) [√(3/2) - 1 + 1/√2]
B.Δv_total = √(GM/r) [√(4/3) - 1 + 1/√2 - √(1/3)]
C.Δv_total = √(GM/r) [2 - √2]
D.Δv_total = √(GM/r) [√(2) - 1]
Explanation: Initial circular speed is v_c1 = √(GM/r) and final circular speed is v_c2 = √(GM/(2r)) = (1/√2)√(GM/r). The Hohmann transfer ellipse has semi-major axis a = (r + 2r)/2 = 1.5r = (3/2)r. By the vis-viva equation v² = GM(2/R - 1/a), at periapsis R = r, v_t1 = √[GM(2/r - 2/(3r))] = √(4/3)√(GM/r). Thus, Δv₁ = v_t1 - v_c1 = (√(4/3) - 1)√(GM/r). At apoapsis R = 2r, v_t2 = √[GM(2/(2r) - 2/(3r))] = √(1/3)√(GM/r). To circularize at 2r, Δv₂ = v_c2 - v_t2 = (1/√2 - √(1/3))√(GM/r). Summing both impulses gives Δv_total = √(GM/r)[√(4/3) - 1 + 1/√2 - √(1/3)].
6A uniform thin rod of length L and mass M is hinged at one end and free to rotate in a vertical plane. If the rod is released from rest in a horizontal position, what is the angular velocity ω and the magnitude of the horizontal force exerted by the hinge when the rod passes through the vertical position?
A.ω = √(3g/L) and F_horizontal = 0
B.ω = √(2g/L) and F_horizontal = (1/2) M g
C.ω = √(3g/L) and F_horizontal = (3/2) M g
D.ω = √(6g/L) and F_horizontal = 0
Explanation: Conservation of mechanical energy from horizontal to vertical gives M g (L/2) = (1/2) I ω², where I = (1/3) M L² about the pivot. Thus, M g (L/2) = (1/6) M L² ω², which gives ω = √(3g/L). At the vertical position, the rod's center of mass has only a vertical upward centripetal acceleration a_c = ω² (L/2) = (3g/L)(L/2) = (3/2)g and zero tangential acceleration (since torque due to gravity τ = 0). Since there is no horizontal acceleration of the center of mass, Newton's second law in the horizontal direction yields F_horizontal = 0.
7A cylindrical container of base area A is filled with an ideal incompressible fluid of density ρ to a height H. A small circular hole of area a (where a ≪ A) is opened in the flat base. How long does it take for the container to completely empty under gravity g?
A.t = (A / a) √(H / g)
B.t = (A / 2a) √(2H / g)
C.t = (2A / a) √(H / (2g))
D.t = (A / a) √(2H / g)
Explanation: By Torricelli's law, the efflux speed is v(y) = √(2gy), where y is the instantaneous fluid height. The rate of volume drainage is -A (dy/dt) = a v(y) = a √(2gy). Rearranging gives dt = - (A / a√(2g)) y^(-1/2) dy. Integrating from y = H to y = 0 yields t = (A / a√(2g)) ∫₀^H y^(-1/2) dy = (A / a√(2g)) [2√H] = (A / a) √(2H / g).
8Two identical blocks, each of mass m, are connected by a spring of stiffness k and placed on a frictionless horizontal floor. Initially, the spring is at its natural length. Block 1 is given an instantaneous impulse J directed along the line of the spring toward block 2. What is the maximum compression of the spring during the subsequent motion?
A.x_max = J / √(k m)
B.x_max = J / √(4 k m)
C.x_max = J / √(2 k m)
D.x_max = 2 J / √(k m)
Explanation: Initially, block 1 has speed v₀ = J/m and block 2 is at rest. Total linear momentum is P = J, so the center-of-mass velocity is v_cm = J / (2m). In the center-of-mass reference frame, the reduced mass is μ = m/2. The kinetic energy in the CM frame is K_rel = (1/2) μ v_rel² = (1/2) (m/2) (J/m)² = J² / (4m). At maximum compression, all relative kinetic energy is converted into elastic potential energy: (1/2) k x_max² = J² / (4m), which gives x_max² = J² / (2km), hence x_max = J / √(2km).
9A physical pendulum consists of a uniform circular hoop of mass M and radius R suspended from a knife-edge on its rim so that it oscillates in its own plane. What is the period T of small oscillations?
A.T = 2π √(2R / g)
B.T = 2π √(R / g)
C.T = 2π √(3R / (2g))
D.T = 2π √(R / (2g))
Explanation: For a physical pendulum, the period of small oscillations is T = 2π √(I_pivot / (M g d)), where d is the distance from the pivot to the center of mass (d = R). By the parallel-axis theorem, the moment of inertia about the rim pivot is I_pivot = I_cm + M R² = M R² + M R² = 2 M R². Substituting these values yields T = 2π √[(2 M R²) / (M g R)] = 2π √(2R / g).
10A bead of mass m slides without friction along a smooth wire bent into a parabola y = c x² in a uniform vertical gravitational field g (acting in the -y direction). For small oscillations near the vertex (x ≈ 0), what is the angular frequency ω₀?
A.ω₀ = √(g c)
B.ω₀ = √(2 g c)
C.ω₀ = √(4 g c)
D.ω₀ = √(g / (2c))
Explanation: The potential energy of the bead is U(x) = m g y = m g c x². For small oscillations near x = 0, the velocity along the wire is v = √(dx² + dy²) / dt = dx/dt √(1 + (2cx)²). Near x ≈ 0, (2cx)² ≪ 1, so kinetic energy is K ≈ (1/2) m (dx/dt)². Total energy is E = (1/2) m ẋ² + (1/2) (2mgc) x². Comparing with the standard harmonic oscillator E = (1/2) m ẋ² + (1/2) k_eff x² with k_eff = 2mgc, the angular frequency is ω₀ = √(k_eff / m) = √(2gc).

About the OBF Exam

The Olimpíada Brasileira de Física (OBF) is Brazil's premier national physics competition organized annually since 1999 by the Sociedade Brasileira de Física (SBF). Open to elementary and high school students across four competition levels (Nível Júnior, Nível I, Nível II, Nível III), the OBF challenges students with deep conceptual physics, sophisticated algebraic and calculus-based modeling, and rigorous hands-on laboratory experimentation. The competition serves as the official qualifying pipeline for Brazil's delegations to the International Physics Olympiad (IPhO), Asian Physics Olympiad (APhO), and Ibero-American Physics Olympiad (OIbF), while providing direct admission opportunities to leading Brazilian universities through Olympic Quotas (Vagas Olímpicas). This practice question bank provides 100 fully worked, competition-grade multiple-choice physics problems adapted into English, complete with comprehensive algebraic solutions and distractor analyses.

Assessment

Phase 1: 20 objective multiple-choice questions taken online within a 4-hour window, across four school levels (Nível Júnior, I, II, III); Phase 2: in-person discursive theoretical physics exam at accredited application centers (CA/F2), scored on a 0–80 scale; Phase 3: in-person discursive theoretical exam plus an experimental laboratory exam for qualified finalists.

Time Limit

Phase 1: 4-hour online window; Phase 2: 4 hours; Phase 3: theoretical and experimental laboratory sessions

Passing Score

Merit-based qualification cutoff scores (Notas de Corte) per Phase and Level (Phase 1 scored 0–20, Phase 2 scored 0–80); National Gold, Silver, Bronze medals and Menção Honrosa

Exam Fee

Free for public school students; nominal registration fee for private educational institutions (Sociedade Brasileira de Física (SBF))

OBF Exam Content Outline

Not published

Overview of the Olimpíada Brasileira de Física (OBF)

Overview of the Olimpíada Brasileira de Física (OBF)

Not published

The Three Competition Phases & Examination Format

The Three Competition Phases & Examination Format

Not published

Syllabus Breakdown & Core Domain Weighting

Syllabus Breakdown & Core Domain Weighting

Not published

Experimental Requirements & Laboratory Physics

Experimental Requirements & Laboratory Physics

Not published

International Olympiads & University Admissions (Vagas Olímpicas)

International Olympiads & University Admissions (Vagas Olímpicas)

How to Pass the OBF Exam

What You Need to Know

  • Passing score: Merit-based qualification cutoff scores (Notas de Corte) per Phase and Level (Phase 1 scored 0–20, Phase 2 scored 0–80); National Gold, Silver, Bronze medals and Menção Honrosa
  • Assessment: Phase 1: 20 objective multiple-choice questions taken online within a 4-hour window, across four school levels (Nível Júnior, I, II, III); Phase 2: in-person discursive theoretical physics exam at accredited application centers (CA/F2), scored on a 0–80 scale; Phase 3: in-person discursive theoretical exam plus an experimental laboratory exam for qualified finalists.
  • Time limit: Phase 1: 4-hour online window; Phase 2: 4 hours; Phase 3: theoretical and experimental laboratory sessions
  • Exam fee: Free for public school students; nominal registration fee for private educational institutions

Keys to Passing

  • Work through all 100 available questions
  • Review every answer and explanation
  • Track weak areas and revisit them
  • Use our AI tutor for tough concepts

OBF Study Tips from Top Performers

1Master the Fundamentals of Free-Body Diagrams: In mechanics, always draw clear vector diagrams with coordinate systems before setting up Newton's second law (ΣF = ma) or torque equilibrium (Στ = 0).
2Prioritize Conservation Laws: For complex interactions, examine whether mechanical energy, linear momentum, or angular momentum is conserved before attempting direct kinematic integration.
3Work with Symbolic Algebra First: Always solve physics problems symbolically in terms of variables before plugging in numerical values; this enables dimensional checks and prevents premature rounding errors.
4Learn Experimental Uncertainty Propagation: Phase 3 requires mastery of Gaussian error propagation, reading vernier scales, and determining physical constants from graphical slopes.
5Practice Timed Discursive Derivations: For Phases 2 and 3, train yourself to write clear, structured, step-by-step mathematical justifications that examiners can award full partial credit for.
6Master Modern Physics and Relativity Fundamentals: Understand photon quantization (E = hf), stopping potential calculations, and relativistic energy-momentum triangles to secure high marks in the advanced sections.

Frequently Asked Questions

What is the difference between OBF and regular high school physics exams in Brazil?

While school exams and ENEM focus largely on qualitative understanding and direct formula application, OBF demands deep theoretical intuition, multi-step algebraic derivations, complex vector geometry, calculus-ready modeling, non-standard physical setups, and rigorous experimental data and uncertainty analysis.

Who is eligible to participate in OBF?

Any regularly enrolled student in a registered Brazilian public or private elementary or secondary school is eligible, competing in the level matching their school year under the 2026 regulation: Nível Júnior (6º and 7º anos EF), Nível I (8º and 9º anos EF), Nível II (1ª and 2ª séries EM), or Nível III (3ª série EM).

Are calculators permitted in OBF examinations?

Under SBF regulations, simple scientific calculators are generally permitted in the 2nd and 3rd Phases (discursive theoretical and experimental exams), while programmable, graphing, or internet-connected smart devices are strictly prohibited. In the 1st Phase objective test, calculators are generally not permitted unless specified by current annual guidelines.

How are the medals and awards distributed in OBF?

Medals (Gold, Silver, Bronze) and Menções Honrosas are awarded nationally per competition level based on cumulative performance in the 2nd and 3rd Phases. Additionally, top-performing public school students and regional state champions receive special recognition and trophies.

How does OBF select students for the International Physics Olympiad (IPhO)?

Top medalists from Níveis 2 and 3 in the national OBF final are invited to the selective training program (Processo de Seleção Internacional da SBF). Qualifying students undergo intensive theoretical and laboratory training rounds before the final 5-member IPhO, 8-member APhO, and 5-member OIbF national teams are named.

What are Vagas Olímpicas and how do they relate to OBF?

Vagas Olímpicas are special undergraduate admission spots allocated by major Brazilian universities (such as UNICAMP, USP, and UNESP) for students with national academic olympiad medals. OBF Gold, Silver, and Bronze medalists can secure direct admission into competitive programs like Physics, Computer Engineering, Aerospace Engineering, and Applied Mathematics without sitting for standard vestibular exams.

Why is this practice bank presented in English?

This practice bank is engineered as a bilingual training adaptation. It prepares Brazilian students for international olympiads (IPhO, APhO) which are administered exclusively in English, while providing international physics students with access to Brazil's acclaimed national physics olympiad problem-solving traditions.