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100+ Free Catalonia PAU Physics / Física 2º Bachillerato (CIC Generalitat) Practice Questions
Catalonia PAU Physics 2026 (Proves d'Accés a la Universitat - Consell Interuniversitari de Catalunya) practice questions are available now; exam metadata is being verified.
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Sample Catalonia PAU Physics / Física 2º Bachillerato (CIC Generalitat) Practice Questions
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1Two point masses m1 = 2.0 × 10^4 kg and m2 = 5.0 × 10^4 kg are separated by a distance r = 50 m in empty space. What is the magnitude of the gravitational attraction force between them? (G = 6.674 × 10^-11 N·m²/kg²)
A.2.67 × 10^-5 N
B.1.34 × 10^-3 N
C.5.34 × 10^-5 N
D.6.67 × 10^-7 N
Explanation: Using Newton's law of universal gravitation F = G·m1·m2 / r², substituting the given values yields F = (6.674 × 10^-11) · (2.0 × 10^4) · (5.0 × 10^4) / (50)² = 2.67 × 10^-5 N. This mutual attractive force acts along the line connecting the two masses.
2Calculate the gravitational field strength (g) on the surface of a spherical planet with mass M = 4.0 × 10^24 kg and radius R = 5.0 × 10^6 m. (G = 6.674 × 10^-11 N·m²/kg²)
A.10.68 m/s²
B.5.34 m/s²
C.21.36 m/s²
D.5.34 × 10^7 m/s²
Explanation: The gravitational field strength at the planet surface is given by g = G·M / R². Plugging in values gives g = (6.674 × 10^-11) · (4.0 × 10^24) / (5.0 × 10^6)² = 10.68 m/s² directed radially inward.
3Determine the gravitational potential energy of a mass m = 250 kg positioned at a distance r = 1.0 × 10^7 m from the center of Earth (M_E = 5.97 × 10^24 kg). (G = 6.674 × 10^-11 N·m²/kg²)
A.-9.96 × 10^9 J
B.-9.96 × 10^16 J
C.+9.96 × 10^9 J
D.-4.98 × 10^9 J
Explanation: Gravitational potential energy in a central field is U = -G·M_E·m / r. Substituting the parameters yields U = -(6.674 × 10^-11) · (5.97 × 10^24) · 250 / (1.0 × 10^7) = -9.96 × 10^9 J, which is negative with reference at infinity.
4Planet A orbits a star at distance rA = 1.0 AU with an orbital period TA = 1.0 year. Planet B orbits the same star at rB = 9.0 AU. According to Kepler's third law, what is Planet B's orbital period TB?
A.27.0 years
B.81.0 years
C.9.0 years
D.3.0 years
Explanation: Kepler's third law states T² / r³ = constant. Therefore, (TB / TA)² = (rB / rA)³, which gives TB = TA · (9.0)^(3/2) = 1.0 · 27.0 = 27.0 years.
5Calculate the speed of a satellite moving in a circular orbit at radius r = 1.60 × 10^7 m around Earth (M_E = 5.97 × 10^24 kg). (G = 6.674 × 10^-11 N·m²/kg²)
A.4.99 km/s
B.7.91 km/s
C.3.53 km/s
D.24.9 km/s
Explanation: Equating gravitational force to centripetal force yields v = √(G·M_E / r). Substituting values gives v = √[(6.674 × 10^-11 · 5.97 × 10^24) / (1.60 × 10^7)] = √(2.49 × 10^7) = 4990 m/s = 4.99 km/s.
6What is the escape velocity from the surface of the Moon (M_M = 7.35 × 10^22 kg, R_M = 1.74 × 10^6 m)? (G = 6.674 × 10^-11 N·m²/kg²)
A.2.38 km/s
B.1.68 km/s
C.11.2 km/s
D.5.66 km/s
Explanation: Escape velocity is given by v_esc = √(2·G·M_M / R_M). Inserting Moon parameters gives v_esc = √[2 · (6.674 × 10^-11) · (7.35 × 10^22) / (1.74 × 10^6)] = √(5.638 × 10^6) = 2375 m/s = 2.38 km/s.
7If the acceleration due to gravity on Earth's surface is g0 = 9.81 m/s², what is the gravitational field strength at an altitude h = 2 R_E above the surface?
A.1.09 m/s²
B.3.27 m/s²
C.2.45 m/s²
D.4.91 m/s²
Explanation: At altitude h = 2 R_E, the distance from Earth's center is r = R_E + h = 3 R_E. Since g ∝ 1/r², g(h) = g0 / 3² = 9.81 / 9 = 1.09 m/s².
8Calculate the gravitational potential V created by Earth (M_E = 5.97 × 10^24 kg) at a distance r = 2.00 × 10^7 m from its center. (G = 6.674 × 10^-11 N·m²/kg²)
A.-1.99 × 10^7 J/kg
B.-9.96 × 10^13 J/kg
C.+1.99 × 10^7 J/kg
D.-3.98 × 10^7 J/kg
Explanation: Gravitational potential is defined as V = -G·M_E / r. Plugging in values yields V = -(6.674 × 10^-11 · 5.97 × 10^24) / (2.00 × 10^7) = -1.99 × 10^7 J/kg.
9What is the orbital period T of a satellite in a circular orbit of radius r = 1.00 × 10^7 m around Earth (M_E = 5.97 × 10^24 kg)? (G = 6.674 × 10^-11 N·m²/kg²)
A.9954 s (~2.77 h)
B.3140 s (~0.87 h)
C.15800 s (~4.39 h)
D.86400 s (24 h)
Explanation: The orbital period is T = √[4π²·r³ / (G·M_E)]. Substituting values gives T = √[39.478 · 1.00 × 10^21 / (3.984 × 10^14)] = √(9.909 × 10^7) = 9954 s ≈ 2.77 hours.
10Calculate the total mechanical energy E of a satellite of mass m = 800 kg in a circular orbit of radius r = 1.28 × 10^7 m around Earth (M_E = 5.97 × 10^24 kg). (G = 6.674 × 10^-11 N·m²/kg²)
A.-1.245 × 10^10 J
B.-2.490 × 10^10 J
C.+1.245 × 10^10 J
D.-6.225 × 10^9 J
Explanation: For a circular orbit, mechanical energy is E = -G·M_E·m / (2r). Substituting values yields E = -(6.674 × 10^-11 · 5.97 × 10^24 · 800) / (2 · 1.28 × 10^7) = -1.245 × 10^10 J.
About the Catalonia PAU Physics / Física 2º Bachillerato (CIC Generalitat) Practice Questions
Verified exam format metadata for Catalonia PAU Physics 2026 (Proves d'Accés a la Universitat - Consell Interuniversitari de Catalunya) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.