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2026 Statistics
Key Facts: Extremadura PAU Physics Exam
UEx 2026
Organized by Universidad de Extremadura and Junta de Extremadura
Comisión Organizadora PAU Extremadura
90 Mins
Official examination duration
UEx PAU Guidelines
Min 4.0
Minimum mark required in Access Phase to combine with Bachillerato GPA
PAU Extremadura Regulations
5 Core Blocks
Gravitational Field, Electromagnetism, Waves & Optics, Relativity, Quantum & Nuclear Physics
LOMLOE 2º Bachillerato Physics Syllabus
100
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Sample Extremadura PAU Physics Practice Questions
Try these sample questions to test your Extremadura PAU Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.
1Two point masses of m1 = 500 kg and m2 = 800 kg are separated by a distance of r = 2.0 m. Using Newton's law of universal gravitation with G = 6.674 x 10^-11 N*m^2/kg^2, what is the magnitude of the gravitational force between them?
A.6.67 x 10^-6 N
B.1.33 x 10^-5 N
C.3.34 x 10^-6 N
D.2.67 x 10^-5 N
Explanation: According to Newton's law of universal gravitation, F = G * m1 * m2 / r^2. Substituting the values: F = (6.674 x 10^-11) * (500) * (800) / (2.0)^2 = (6.674 x 10^-11) * 400,000 / 4 = (6.674 x 10^-11) * 100,000 = 6.674 x 10^-6 N.
2Planet A orbits a star at a mean orbital radius rA, with an orbital period of TA = 8.0 years. Planet B orbits the same star at a mean orbital radius rB = 4 * rA. According to Kepler's Third Law (T^2 / r^3 = constant), what is the orbital period of Planet B?
A.32 years
B.64 years
C.16 years
D.128 years
Explanation: Kepler's Third Law states that T^2 / r^3 is constant. Therefore, (TB / TA)^2 = (rB / rA)^3 = (4)^3 = 64. Taking the square root gives TB / TA = 8. Thus, TB = 8 * TA = 8 * 8.0 years = 64 years.
3What is the gravitational field strength g at an altitude h = RE above the surface of the Earth, where RE is the Earth's radius and g0 = 9.80 m/s^2 is the surface gravitational acceleration?
A.4.90 m/s^2
B.1.23 m/s^2
C.2.45 m/s^2
D.9.80 m/s^2
Explanation: The distance from Earth's center to the altitude h = RE is r = RE + h = 2 * RE. Gravitational field strength varies inversely with the square of distance: g(r) = G * ME / r^2 = G * ME / (2 * RE)^2 = (1/4) * (G * ME / RE^2) = g0 / 4 = 9.80 / 4 = 2.45 m/s^2.
4Which statement correctly describes the sign and physical interpretation of gravitational potential energy Ep = -G * M * m / r for a two-body bound system?
A.Ep is always negative because energy must be added to the system to separate the masses to infinity where Ep = 0.
B.Ep is always positive because gravity is an attractive force.
C.Ep is zero at the surface of the primary body and becomes negative at infinity.
D.Ep is positive for bound orbits and negative for unbound hyperbolic trajectories.
Explanation: Gravitational potential energy is conventionally set to zero at infinity (r -> infinity). Because gravity is attractive, work must be done against the gravitational field to move a mass from distance r to infinity. Hence, Ep(r) < 0 for any finite distance r.
5A satellite of mass m = 1000 kg orbits Earth in a circular path at distance r = 4 * RE from Earth's center. Given Earth's surface gravity g0 = 9.80 m/s^2 and RE = 6.37 x 10^6 m, what is the satellite's circular orbital speed?
A.7.90 km/s
B.5.59 km/s
C.3.95 km/s
D.11.2 km/s
Explanation: For a circular orbit, gravitational force provides centripetal acceleration: G * ME * m / r^2 = m * vorb^2 / r => vorb = sqrt(G * ME / r). Since g0 = G * ME / RE^2, we have G * ME = g0 * RE^2. Thus vorb = sqrt(g0 * RE^2 / (4 * RE)) = sqrt(g0 * RE / 4) = 0.5 * sqrt(g0 * RE) = 0.5 * sqrt(9.80 * 6.37 x 10^6) = 0.5 * 7901 m/s = 3950 m/s = 3.95 km/s.
6What is the escape velocity v_esc from the surface of Earth (RE = 6.37 x 10^6 m, g0 = 9.80 m/s^2)?
A.7.90 km/s
B.11.2 km/s
C.15.8 km/s
D.9.80 km/s
Explanation: Escape velocity is obtained by setting total mechanical energy E = E_k + E_p = 0: (1/2) * m * v_esc^2 - G * ME * m / RE = 0 => v_esc = sqrt(2 * G * ME / RE) = sqrt(2 * g0 * RE). Substituting RE = 6.37 x 10^6 m and g0 = 9.80 m/s^2: v_esc = sqrt(2 * 9.80 * 6.37 x 10^6) = sqrt(1.2485 x 10^8) = 11174 m/s approx 11.2 km/s.
7According to Kepler's Second Law (Law of Equal Areas), how does the speed of a planet at perihelion (closest distance rp) compare to its speed at aphelion (farthest distance ra)?
A.Speed at perihelion is higher, satisfying vp * rp = va * ra due to angular momentum conservation.
B.Speed at aphelion is higher because the planet is farther from the Sun.
C.Speed is equal at perihelion and aphelion because mechanical energy is constant.
D.Speed at perihelion is lower because gravitational potential energy is maximum.
Explanation: Kepler's Second Law is a direct consequence of the conservation of angular momentum (L = m * r * v * sin(theta) = constant). At perihelion and aphelion, velocity is perpendicular to the position vector, so L = m * rp * vp = m * ra * va => vp * rp = va * ra. Since rp < ra, vp > va.
8A comet in an elliptical orbit around the Sun has a perihelion distance rp = 0.50 AU with speed vp = 54 km/s. What is its orbital speed va at aphelion, located at ra = 4.5 AU?
A.6.0 km/s
B.12 km/s
C.27 km/s
D.3.0 km/s
Explanation: Using conservation of angular momentum at perihelion and aphelion: vp * rp = va * ra => va = vp * (rp / ra) = 54 km/s * (0.50 / 4.5) = 54 / 9 = 6.0 km/s.
9How much work must be done by an external agent to lift a m = 200 kg satellite from Earth's surface (r1 = RE) to an altitude h = RE (r2 = 2 * RE)? (Use RE = 6.37 x 10^6 m, g0 = 9.80 m/s^2).
A.1.25 x 10^10 J
B.6.24 x 10^9 J
C.3.12 x 10^9 J
D.2.49 x 10^10 J
Explanation: Work done W = Ep(r2) - Ep(r1) = -G * ME * m / (2 * RE) - (-G * ME * m / RE) = G * ME * m / (2 * RE) = (1/2) * m * g0 * RE. Substituting m = 200 kg, g0 = 9.80 m/s^2, RE = 6.37 x 10^6 m: W = 0.5 * 200 * 9.80 * 6.37 x 10^6 = 6.2426 x 10^9 J approx 6.24 x 10^9 J.
10Two point masses M1 = 16 * M and M2 = M are fixed at a distance d = 10 m apart. At what distance x from M1 along the line connecting them does the net gravitational field intensity equal zero?
A.8.0 m
B.6.0 m
C.5.0 m
D.9.0 m
Explanation: The gravitational fields due to M1 and M2 point in opposite directions between the masses. Setting g1 = g2: G * M1 / x^2 = G * M2 / (d - x)^2 => 16 * M / x^2 = M / (10 - x)^2. Taking square roots: 4 / x = 1 / (10 - x) => 4 * (10 - x) = x => 40 - 4x = x => 5x = 40 => x = 8.0 m.
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