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Sachsen-Anhalt Abitur Physics study materials
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Sachsen-Anhalt Abitur Physics practice questions

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Question 1Mechanics

A car accelerates uniformly from rest and reaches 20 m/s after 8 s. What is its acceleration?

  1. 2.5 m/s² Correct answer.
  2. 0.4 m/s²
  3. 4.0 m/s²
  4. 20 m/s²

Explanation

For uniformly accelerated motion starting from rest, a = Δv/Δt = 20 m/s ÷ 8 s = 2.5 m/s². This is the constant rate at which the velocity increases over the 8 seconds.

Question 2Mechanics

An object is dropped from rest and falls freely for 3.0 s. Using s = ½gt² with g = 9.81 m/s², how far has it fallen?

  1. 44.1 m Correct answer.
  2. 88.3 m
  3. 29.4 m
  4. 14.7 m

Explanation

s = ½gt² = 0.5 × 9.81 × 3.0² = 0.5 × 9.81 × 9 ≈ 44.1 m. The factor ½ and the squared time are both essential features of uniformly accelerated motion starting from rest.

Question 3Mechanics

A ball is launched horizontally at v0 = 15 m/s from a height of h = 20 m (waagerechter Wurf). Using g = 9.81 m/s², what horizontal distance does it travel before landing?

  1. 30.3 m Correct answer.
  2. 26.2 m
  3. 21.4 m
  4. 42.8 m

Explanation

The fall time comes only from the vertical motion: t = √(2h/g) = √(2×20/9.81) ≈ 2.02 s. The horizontal range is then x = v0·t = 15 × 2.02 ≈ 30.3 m, since horizontal and vertical motion are independent (Superpositionsprinzip).

Question 4Mechanics

A net force of 45 N acts on a 15 kg object. Using Newton's second law F = ma, what is the object's acceleration?

  1. 3.0 m/s² Correct answer.
  2. 0.33 m/s²
  3. 675 m/s²
  4. 60 m/s²

Explanation

a = F/m = 45 N ÷ 15 kg = 3.0 m/s². Newton's second law directly relates net force, mass and acceleration for a constant force.

Question 5Mechanics

A 2 kg object moves at 6 m/s. Using E_kin = ½mv², what is its kinetic energy?

  1. 36 J Correct answer.
  2. 72 J
  3. 18 J
  4. 12 J

Explanation

E_kin = ½mv² = 0.5 × 2 kg × (6 m/s)² = 0.5 × 2 × 36 = 36 J. The velocity must be squared before multiplying by ½m.

Question 6Mechanics

Which of the following correctly defines mechanical work W done by a constant force on a moving object?

  1. W equals the component of the force in the direction of motion multiplied by the displacement, W = F·s·cosθ Correct answer.
  2. W equals the force multiplied by the time for which it acts
  3. W equals the force divided by the displacement
  4. W equals the force multiplied by the object's velocity

Explanation

Mechanical work is defined as W = F·s·cosθ, where θ is the angle between the force vector and the displacement. Only the force component along the direction of motion contributes to work; a force perpendicular to the displacement (θ=90°) does zero work.

Question 7Mechanics

A 2 kg cart moving at 4 m/s collides with a stationary 3 kg cart and sticks to it (perfectly inelastic collision). Using conservation of momentum, what is their common velocity after the collision?

  1. 1.6 m/s Correct answer.
  2. 2.0 m/s
  3. 2.4 m/s
  4. 8.0 m/s

Explanation

Momentum is conserved: p_before = m_A·v_A + m_B·v_B = (2×4) + (3×0) = 8 kg·m/s. After they stick together, v_common = p_total/(m_A+m_B) = 8/5 = 1.6 m/s.

Question 8Mechanics

A 0.5 kg ball hits a wall moving at 8 m/s and bounces straight back at 6 m/s. What is the magnitude of the impulse (Kraftstoß) the wall delivers to the ball?

  1. 7 N·s Correct answer.
  2. 1 N·s
  3. 3.5 N·s
  4. 14 N·s

Explanation

Taking the incoming direction as positive, the velocity change is Δv = (−6) − (8) = −14 m/s, since the ball reverses direction. The impulse is Δp = m·Δv = 0.5 × 14 = 7 kg·m/s = 7 N·s in magnitude.

Question 9Mechanics

In a v(t)-diagram, what feature indicates uniformly accelerated motion (gleichmäßig beschleunigte Bewegung)?

  1. A straight line with a constant, nonzero slope Correct answer.
  2. A horizontal straight line
  3. A curve that bends increasingly steeply over time
  4. A straight line passing through the origin at 45°, regardless of the actual acceleration value

Explanation

In a v(t)-diagram, the slope represents acceleration. Uniformly accelerated motion has constant acceleration, which appears as a straight line whose (nonzero) slope equals that constant value - the line need not pass through the origin or have any particular angle.

Question 10Mechanics

A rocket in deep space, far from any gravitational field, ejects burnt fuel backward at high speed to accelerate forward. Which physical principle explains this Rückstoßprinzip?

  1. Newton's third law: the rocket exerts a force on the ejected fuel, and the fuel exerts an equal and opposite force on the rocket Correct answer.
  2. The ejected fuel pushes against the surrounding vacuum of space to propel the rocket forward
  3. The rocket's engine reduces the rocket's own gravitational pull, causing it to move forward
  4. Air resistance on the ejected fuel creates a forward thrust on the rocket

Explanation

Rocket propulsion is a direct application of Newton's third law and conservation of momentum: the engine pushes exhaust gas backward, and by the action–reaction principle the gas pushes the rocket forward with an equal and opposite force - no external medium is needed, which is exactly why rockets work in the vacuum of space.

Question 11Mechanics

A 0.2 kg object moves in a circle of radius 0.5 m at a constant speed of 4 m/s. Using F = mv²/r, what centripetal (radial) force acts on it?

  1. 6.4 N Correct answer.
  2. 3.2 N
  3. 1.6 N
  4. 1.28 N

Explanation

F = mv²/r = 0.2 × 4² / 0.5 = 0.2 × 16 / 0.5 = 3.2/0.5 = 6.4 N. Even at constant speed, an object moving in a circle constantly accelerates toward the centre, requiring this radial (centripetal) force.

Question 12Mechanics

Which of the following mechanical quantities is a vector (requiring both magnitude and direction to be fully specified)?

  1. Force Correct answer.
  2. Mass
  3. Kinetic energy
  4. Time

Explanation

Force is a vector quantity: pushing an object with the same magnitude of force in different directions produces different effects, so direction is essential information. Mass, kinetic energy and time are scalar quantities, fully described by a single numerical value (with units) and no direction.

Question 13Mechanics

A 4 kg block slides from rest down a frictionless incline and drops a vertical height of 3 m. Using energy conservation (mgh = ½mv²), what is its speed at the bottom?

  1. 7.67 m/s Correct answer.
  2. 5.42 m/s
  3. 58.9 m/s
  4. 29.4 m/s

Explanation

Energy conservation gives v = √(2gh) = √(2 × 9.81 × 3) = √58.86 ≈ 7.67 m/s. Notice the mass cancels out of mgh = ½mv², so the final speed does not depend on the block's mass.

Question 14Mechanics

In the simple Coulomb friction model used in the Fachlehrplan, the sliding (kinetic) friction force is modelled as F_R = μ·F_N, where F_N is the normal force. Which factor does this friction force NOT depend on?

  1. The apparent size of the contact area between the surfaces Correct answer.
  2. The normal force pressing the surfaces together
  3. The coefficient of friction μ, which depends on the two materials in contact
  4. Whether the surfaces are, for example, rubber on asphalt or steel on ice

Explanation

In the idealised model F_R = μ·F_N, the sliding friction force depends only on the normal force and the friction coefficient μ (itself a property of the two materials in contact) - not on how large the visible contact area is. This is a genuinely counter-intuitive but well-established result of the simplified friction model used at this level.

Question 15Mechanics

A car moving at 24 m/s brakes with a constant deceleration of magnitude 6 m/s². Using v² = v0² − 2·a·s, what is the stopping distance?

  1. 48 m Correct answer.
  2. 96 m
  3. 144 m
  4. 24 m

Explanation

Setting final velocity v = 0: 0 = v0² − 2as, so s = v0²/(2a) = 24²/(2×6) = 576/12 = 48 m. This s(t)-independent formula is useful whenever the stopping distance is needed without first finding the stopping time.

Question 16Electricity Magnetism

Two point charges, q1 = 2 μC and q2 = 3 μC, are separated by 0.5 m. Using Coulomb's law F = k·q1·q2/r² with k = 8.99×10⁹ N·m²/C², what is the force between them?

  1. 0.216 N Correct answer.
  2. 0.108 N
  3. 2.4 × 10⁻¹¹ N
  4. 1.80 × 10⁵ N

Explanation

F = k·q1·q2/r² = 8.99×10⁹ × (2×10⁻⁶ × 3×10⁻⁶) / 0.5² = 8.99×10⁹ × 6×10⁻¹² / 0.25 ≈ 0.216 N. Coulomb's law has the same inverse-square structure as Newton's law of gravitation, but acts between electric charges rather than masses.

Question 17Electricity Magnetism

A parallel-plate capacitor has a potential difference of U = 300 V across a plate separation of d = 2.0 cm. Using E = U/d, what is the electric field strength between the plates?

  1. 15,000 V/m Correct answer.
  2. 150 V/m
  3. 6 V/m
  4. 6.67 × 10⁻⁵ V/m

Explanation

E = U/d = 300 V / 0.020 m = 15,000 V/m, since d = 2.0 cm = 0.020 m. The field strength between the plates of a parallel-plate capacitor is uniform (homogeneous) and equals the voltage divided by the plate separation in metres.

Question 18Electricity Magnetism

An electric field exerts a force on a charged particle placed within it. How does this force depend on the field strength E and the particle's charge q?

  1. The force is F = q·E - directly proportional to both the charge and the field strength Correct answer.
  2. The force is F = q/E - inversely proportional to the field strength
  3. The force depends only on the field strength E, not on the amount of charge q
  4. The force depends only on the charge q, not on the field strength E

Explanation

The defining relationship for electric field strength is E = F/q, rearranged to F = q·E. A stronger field, or a larger charge, both produce a proportionally larger force - this is exactly analogous to how gravitational field strength g relates to gravitational force via F = m·g.

Question 19Electricity Magnetism

Which combination of changes to a parallel-plate capacitor would increase its capacitance the most?

  1. Increasing the plate area and decreasing the plate separation, while inserting a dielectric with a higher permittivity Correct answer.
  2. Decreasing the plate area and increasing the plate separation
  3. Capacitance depends only on the applied voltage, not on the capacitor's geometry
  4. Removing any dielectric and leaving a vacuum always gives the highest possible capacitance for a given geometry

Explanation

For a parallel-plate capacitor, C = ε0·εr·A/d: capacitance increases with larger plate area A, decreases with larger plate separation d, and increases with a dielectric of higher relative permittivity εr (compared with vacuum, εr=1). Combining a larger area, a smaller separation, and a higher-εr dielectric together maximises capacitance.

Question 20Electricity Magnetism

A 10 μF capacitor is charged to a voltage of 50 V. Using E = ½CU², how much energy is stored in it?

  1. 12.5 mJ Correct answer.
  2. 25 mJ
  3. 0.25 mJ
  4. 6.25 mJ

Explanation

E = ½CU² = 0.5 × 10×10⁻⁶ F × 50² V² = 0.5 × 10×10⁻⁶ × 2500 = 0.0125 J = 12.5 mJ. This is the energy stored in the capacitor's electric field, released when the capacitor discharges.

Question 21Electricity Magnetism

An electron is accelerated from rest through a potential difference of 1000 V. Using E_kin = q·U, what kinetic energy (in joules) does it gain, given the elementary charge e = 1.602×10⁻¹⁹ C?

  1. 1.602 × 10⁻¹⁶ J Correct answer.
  2. 1.602 × 10⁻¹⁹ J
  3. 1.602 × 10⁻²² J
  4. 1.602 × 10⁻¹³ J

Explanation

E_kin = q·U = 1.602×10⁻¹⁹ C × 1000 V = 1.602×10⁻¹⁶ J. Conveniently, this same energy is exactly 1000 electron-volts (1 keV) by definition - one electron-volt is the energy gained by one elementary charge accelerated through a potential difference of one volt.

Question 22Electricity Magnetism

An electron (mass m = 9.109×10⁻³¹ kg, charge e = 1.602×10⁻¹⁹ C) starts from rest and is accelerated through a potential difference of U = 200 V. Using ½mv² = eU, what final speed does it reach?

  1. 8.39 × 10⁶ m/s Correct answer.
  2. 5.93 × 10⁶ m/s
  3. 1.96 × 10⁵ m/s
  4. 1.68 × 10⁷ m/s

Explanation

Solving ½mv² = eU for v gives v = √(2eU/m) = √(2 × 1.602×10⁻¹⁹ × 200 / 9.109×10⁻³¹) = √(7.034×10¹³) ≈ 8.39×10⁶ m/s - about 2.8% of the speed of light, still safely within the non-relativistic regime.

Question 23Electricity Magnetism

In a Millikan-type experiment, an oil droplet of mass m = 1.63×10⁻¹⁴ kg is held stationary between two horizontal capacitor plates by a vertical electric field of E = 5.00×10⁵ V/m, so that the electric force balances gravity (qE = mg, with g = 9.81 m/s²). Approximately how many excess elementary charges (e = 1.602×10⁻¹⁹ C) does the droplet carry?

  1. 2 Correct answer.
  2. 1
  3. 3
  4. 4

Explanation

Setting the electric force equal to gravity: q = mg/E = (1.63×10⁻¹⁴ × 9.81)/(5.00×10⁵) ≈ 3.20×10⁻¹⁹ C. Dividing by the elementary charge, n = q/e ≈ 3.20×10⁻¹⁹/1.602×10⁻¹⁹ ≈ 2.0, so the droplet carries an excess charge of about 2 elementary charges - exactly the kind of quantised result (charge always comes in whole-number multiples of e) that Millikan's experiment was designed to reveal.

Question 24Electricity Magnetism

A charged test particle is moved along an equipotential surface within an electric field. How much work is done on the particle by the electric field during this motion?

  1. Zero, because there is no potential difference between any two points on an equipotential surface Correct answer.
  2. The work equals q times the field strength, regardless of the path taken
  3. The work depends on the length of the path travelled, even along an equipotential surface
  4. The work is always positive if the charge is positive

Explanation

Work done by an electric field moving a charge between two points is W = q·(V1−V2), the charge times the potential difference between start and end points. Every point on an equipotential surface, by definition, has the same potential, so V1−V2 = 0 and therefore W = 0, regardless of the (possibly curved) path taken along that surface.

Question 25Electricity Magnetism

Two point charges of equal magnitude but opposite sign produce electric fields at a point in space located exactly between them, on the line connecting them. According to the superposition principle, how do these two individual field contributions combine at that point?

  1. The two field vectors point in the same direction along that line and add together to give a stronger combined field Correct answer.
  2. The two fields always cancel exactly to zero at the midpoint, regardless of the charges' signs
  3. Only the field from the larger-magnitude charge is present at that point; the smaller field is entirely blocked
  4. Electric fields from different charges cannot be added together; only forces can be superposed

Explanation

The superposition principle states that the total electric field at any point is the vector sum of the fields produced by each individual charge. Midway between a positive and a negative charge of equal magnitude, the field from the positive charge points away from it, and the field from the negative charge points toward it - i.e. in the same direction - so the two contributions add constructively, producing a combined field stronger than either alone, pointing from positive to negative.

Question 26Modern Physics

Using E = h·f with Planck's constant h = 6.626×10⁻³⁴ J·s, what is the energy of a photon of visible light with frequency f = 6×10¹⁴ Hz?

  1. 3.98 × 10⁻¹⁹ J Correct answer.
  2. 1.99 × 10⁻¹⁹ J
  3. 2.48 J
  4. 9.06 × 10⁴⁷ J

Explanation

E = h·f = 6.626×10⁻³⁴ J·s × 6×10¹⁴ Hz ≈ 3.98×10⁻¹⁹ J (about 2.48 eV). Photon energy is directly proportional to frequency, with Planck's constant as the tiny proportionality factor - this is why individual visible-light photons carry such minuscule amounts of energy.

Question 27Modern Physics

What does the term 'wave-particle duality' describe in quantum physics?

  1. The fact that quantum objects such as photons and electrons exhibit both wave-like behaviour (e.g. interference) and particle-like behaviour (e.g. localized, discrete detection events), depending on the experimental setup Correct answer.
  2. The fact that light is purely a wave and electrons are purely particles, with no overlap between the two categories
  3. The fact that every physical object has exactly two velocities at once
  4. The fact that quantum objects switch permanently from being waves to being particles the first time they are ever observed

Explanation

Wave-particle duality describes how quantum objects show wave-like properties - such as interference and diffraction - in some experiments (e.g. the double-slit experiment), and particle-like properties - such as discrete, localized detection events - in others (e.g. the photoelectric effect). Neither the classical wave model nor the classical particle model alone fully captures the behaviour of quantum objects; both descriptions are needed depending on what is being measured.

Question 28Modern Physics

In the photoelectric effect, light with photon energy 4.1 eV strikes a metal surface whose work function (Austrittsarbeit) is W_A = 2.3 eV. Using E_kin,max = E_photon − W_A, what is the maximum kinetic energy of the ejected photoelectrons?

  1. 1.8 eV Correct answer.
  2. 6.4 eV
  3. 9.43 eV
  4. 4.1 eV

Explanation

The Einstein photoelectric equation gives the maximum kinetic energy of ejected electrons as E_kin,max = E_photon − W_A = 4.1 eV − 2.3 eV = 1.8 eV. Only the energy in excess of the work function - the minimum energy needed to free an electron from the metal - becomes kinetic energy of the photoelectron.

Question 29Modern Physics

In a photoelectric-effect experiment, the maximum kinetic energy of ejected electrons is measured to be 1.8 eV. What stopping voltage U_stop (in volts) is needed to just prevent all photoelectrons from reaching the collector, using E_kin,max = e·U_stop?

  1. 1.8 V Correct answer.
  2. 3.6 V
  3. 0.9 V
  4. 1.12 × 10¹⁹ V

Explanation

Because kinetic energy was measured in electron-volts, and by definition 1 eV is the energy gained by charge e accelerated through 1 volt, the stopping voltage in volts equals the kinetic energy in eV directly: U_stop = E_kin,max/e = 1.8 eV/e = 1.8 V. This numerical shortcut works specifically because of how the electron-volt unit is defined.

Question 30Modern Physics

An electron (mass m = 9.109×10⁻³¹ kg) moves at v = 1×10⁶ m/s. Using the de Broglie relation λ = h/(mv) with h = 6.626×10⁻³⁴ J·s, what is its de Broglie wavelength?

  1. 7.27 × 10⁻¹⁰ m Correct answer.
  2. 1.37 × 10⁹ m
  3. 6.04 × 10⁻⁵⁸ m
  4. 6.63 × 10⁻⁴⁰ m

Explanation

λ = h/(mv) = 6.626×10⁻³⁴/(9.109×10⁻³¹ × 1×10⁶) ≈ 7.27×10⁻¹⁰ m ≈ 0.73 nm. This wavelength is comparable to atomic dimensions, which is why electron beams can be diffracted by crystal lattices - direct experimental evidence for the wave nature of matter.

Question 31Modern Physics

In a photoelectric-effect experiment, the stopping voltage is measured to rise from 0.5 V to 1.3 V as the light frequency increases from 5.5×10¹⁴ Hz to 7.5×10¹⁴ Hz. Since the Einstein equation predicts slope = ΔU_stop/Δf = h/e, what experimental value of Planck's constant h does this data give (using e = 1.602×10⁻¹⁹ C)?

  1. 6.41 × 10⁻³⁴ J·s Correct answer.
  2. 4.00 × 10⁻¹⁵ J·s
  3. 4.01 × 10⁻⁵ J·s
  4. 1.71 × 10⁻³⁴ J·s

Explanation

The slope of the stopping-voltage-vs-frequency data is ΔU_stop/Δf = (1.3−0.5)/(7.5×10¹⁴−5.5×10¹⁴) = 0.8/(2.0×10¹⁴) = 4.0×10⁻¹⁵ V·s. Since this slope equals h/e, h = slope × e = 4.0×10⁻¹⁵ × 1.602×10⁻¹⁹ ≈ 6.41×10⁻³⁴ J·s - remarkably close to the accepted value of 6.626×10⁻³⁴ J·s, illustrating how this classic experiment determines a fundamental constant.

Question 32Modern Physics

In the Compton effect, a high-energy photon scatters off a nearly free electron. What happens to the photon as a result of this collision?

  1. The photon loses some of its energy to the electron and emerges with a lower frequency (longer wavelength) than before the collision Correct answer.
  2. The photon's frequency and energy are completely unchanged; only its direction changes
  3. The photon gains energy from the electron and emerges with a higher frequency
  4. The electron absorbs the photon completely, and no photon exists after the collision

Explanation

The Compton effect treats the photon-electron interaction as an elastic collision, conserving both energy and momentum. Because some kinetic energy is transferred to the initially stationary electron, the scattered photon carries away less energy than before - and since E = hf, a lower photon energy means a lower frequency and longer wavelength. This angle-dependent wavelength increase was Compton's key experimental discovery, providing strong evidence for the particle nature of light.

Question 33Modern Physics

According to Heisenberg's uncertainty relation (in the form Δx·Δpx ≥ h used in the Fachlehrplan), what does this principle say about a quantum object such as an electron?

  1. It is fundamentally impossible to prepare a quantum state in which both the position and the momentum of the particle (along the same direction) are simultaneously known with arbitrary precision Correct answer.
  2. It only describes a practical limitation of current measurement technology, which will eventually be overcome with more precise instruments
  3. It states that position and momentum can both be measured to arbitrary precision individually, with no restriction on how precisely a state can simultaneously possess well-defined values of both
  4. It applies only to electrons and not to any other quantum particles such as photons or protons

Explanation

The Heisenberg uncertainty relation states that the product of the uncertainties in position and momentum (along the same axis) cannot be reduced below a fundamental limit, no matter how the measurement is performed: Δx·Δpx ≥ h. This is not a statement about the limits of measuring instruments - it reflects an intrinsic feature of quantum states themselves: no quantum state can simultaneously have both a perfectly sharp position and a perfectly sharp momentum along the same direction.

Question 34Modern Physics

In a double-slit experiment with single photons or electrons, an interference pattern builds up on the detector over many events. What happens to this interference pattern if a 'which-path' detector is added that determines which slit each particle passed through?

  1. The interference pattern disappears, because obtaining which-path information destroys the coherent superposition of the two paths, leaving only two overlapping single-slit distributions Correct answer.
  2. The interference pattern becomes sharper and more pronounced, since knowing the path improves the measurement precision
  3. Nothing changes; which-path information has no effect on the pattern as long as the detector does not physically block either slit
  4. The interference pattern only disappears if a human being consciously looks at the which-path data; an automated recording with no observer has no effect

Explanation

The Fachlehrplan explicitly names complementarity between Weginformation (which-path information) and Interferenzfähigkeit (capacity to interfere) as a core feature of quantum theory. As long as no which-path information exists anywhere, the two possible paths remain in coherent superposition and interfere, building up fringes. The moment which-path information becomes available in principle - even without a person reading it - the coherent superposition is destroyed (decoherence), and the interference pattern is replaced by the simple sum of two single-slit patterns.

Question 35Oscillations

A mass of 0.4 kg is attached to a spring with spring constant k = 25 N/m and set into oscillation. Using T = 2π√(m/k), what is the period of oscillation?

  1. 0.79 s Correct answer.
  2. 0.13 s
  3. 49.7 s
  4. 0.10 s

Explanation

T = 2π√(m/k) = 2π√(0.4/25) = 2π√0.016 ≈ 2π×0.1265 ≈ 0.79 s. The period of a spring-mass oscillator depends on the square root of mass over stiffness - a larger mass or a softer spring both increase the period.

Question 36Oscillations

A simple pendulum of length 1.0 m swings with small-angle oscillations near Earth's surface (g = 9.81 m/s²). Using T = 2π√(l/g), what is its period?

  1. 2.01 s Correct answer.
  2. 0.32 s
  3. 19.68 s
  4. 1.42 s

Explanation

T = 2π√(l/g) = 2π√(1.0/9.81) ≈ 2π×0.319 ≈ 2.01 s. For small-angle swings, the pendulum's period depends only on its length and the local gravitational field strength - not on the mass or the amplitude.

Question 37Oscillations

A pendulum completes one full oscillation every 0.50 s. What is its oscillation frequency?

  1. 2.0 Hz Correct answer.
  2. 0.5 Hz
  3. 4.0 Hz
  4. 0.25 Hz

Explanation

Frequency is the reciprocal of the period: f = 1/T = 1/0.50 s = 2.0 Hz - two complete oscillations occur every second.

Question 38Oscillations

A horizontal spring oscillator has spring constant k = 25 N/m and amplitude A = 0.05 m. Using E = ½kA², what is its total mechanical energy (the sum of kinetic and potential energy at any instant)?

  1. 31.3 mJ Correct answer.
  2. 62.5 mJ
  3. 625 mJ
  4. 15.6 mJ

Explanation

E = ½kA² = 0.5 × 25 N/m × (0.05 m)² = 0.5 × 25 × 0.0025 ≈ 0.0313 J = 31.3 mJ. This total energy is constant and continuously exchanges between kinetic energy (maximum at the equilibrium position) and elastic potential energy (maximum at the amplitude).

Question 39Oscillations

A system's natural (eigen) frequency is 5 Hz. According to the treatment of resonance in the Fachlehrplan, at what driving frequency will a lightly damped oscillator show the largest amplitude response?

  1. At a driving frequency close to 5 Hz, its own natural frequency Correct answer.
  2. At a driving frequency of exactly zero (a constant, unchanging force)
  3. At a driving frequency far above 5 Hz, since higher frequencies always transfer more energy
  4. Resonance amplitude is independent of the driving frequency and depends only on the driving force's magnitude

Explanation

Resonance occurs when the driving frequency matches (or is very close to) the system's own natural/eigen frequency, at which point energy is transferred most efficiently into the oscillation and, for light damping, the amplitude becomes very large. This is why, for example, a swing is pushed at its own natural rhythm to build up amplitude.

Question 40Oscillations

What is the essential difference between a free (freie) oscillation and a forced/driven (erzwungene) oscillation?

  1. A free oscillation occurs at the system's own natural frequency after a single initial disturbance, while a forced oscillation is continuously driven by an external periodic force and settles into oscillating at the driving frequency Correct answer.
  2. A free oscillation has no energy losses, while a forced oscillation always loses energy to friction
  3. A free oscillation only occurs in electrical circuits, while a forced oscillation only occurs in mechanical systems
  4. A free oscillation has a larger amplitude than a forced oscillation in every case

Explanation

In a free oscillation, a system is displaced once and then left alone; it oscillates at its own natural (eigen) frequency, gradually losing amplitude to damping. In a forced oscillation, a continuous external periodic force keeps driving the system; after an initial transient, it settles into oscillating at the driving frequency rather than necessarily its own natural frequency - except at resonance, where the two coincide.

Question 41Oscillations

Why does the amplitude of a real (freely swinging) pendulum gradually decrease over time rather than oscillating forever?

  1. Friction and air resistance continuously remove mechanical energy from the system, so it is damped (gedämpft) rather than ideally undamped Correct answer.
  2. The gravitational field weakens slightly each time the pendulum swings
  3. The pendulum's mass slowly decreases over time due to the oscillation
  4. Energy is not actually lost; the amplitude only appears to decrease because of an optical illusion

Explanation

An idealised, frictionless harmonic oscillator would oscillate forever with constant amplitude. Real pendulums experience air resistance and friction at the pivot, which continuously convert some mechanical energy into heat, so each successive swing has a slightly smaller amplitude - this is called damping (Dämpfung).

Question 42Waves Optics

A wave has a frequency of 50 Hz and a wavelength of 2 m. Using v = f·λ, what is its propagation speed?

  1. 100 m/s Correct answer.
  2. 25 m/s
  3. 0.04 m/s
  4. 52 m/s

Explanation

v = f·λ = 50 Hz × 2 m = 100 m/s. This wave equation links a wave's frequency, wavelength and propagation speed, and applies to any periodic wave, mechanical or electromagnetic.

Question 43Waves Optics

To measure the speed of sound, a student claps once toward a wall 686 m away and measures 4.0 s until the echo returns. What speed of sound does this give?

  1. 343 m/s Correct answer.
  2. 171.5 m/s
  3. 1372 m/s
  4. 686 m/s

Explanation

Sound travels to the wall and back - a total distance of 2×686 m = 1372 m - in 4.0 s. So v = distance/time = 1372/4.0 = 343 m/s, consistent with the accepted speed of sound in air at room temperature.

Question 44Waves Optics

A string of length 1.2 m is fixed at both ends and vibrates in its third harmonic, forming a standing wave pattern with 3 antinodes (Schwingungsbäuche). Using λ = 2L/n, what is the wavelength of this standing wave?

  1. 0.8 m Correct answer.
  2. 0.4 m
  3. 1.2 m
  4. 7.2 m

Explanation

For a string fixed at both ends vibrating in its nth harmonic, λ = 2L/n. With L = 1.2 m and n = 3: λ = 2×1.2/3 = 2.4/3 = 0.8 m.

Question 45Waves Optics

Sound waves in air are classified as longitudinal waves. What does this mean?

  1. The air particles oscillate back and forth along the same direction in which the wave travels Correct answer.
  2. The air particles oscillate perpendicular to the direction the wave travels
  3. The wave carries no energy, only a disturbance pattern
  4. The wave can only travel in a straight line and cannot be reflected

Explanation

In a longitudinal wave, the medium's particles vibrate parallel to (along) the direction of wave propagation, creating alternating regions of compression and rarefaction - this is exactly how sound propagates through air. This contrasts with transverse waves (like waves on a string or electromagnetic waves), where the oscillation is perpendicular to the direction of travel.

Question 46Waves Optics

According to Huygens' principle, how can the propagation of a wavefront be constructed?

  1. Every point on an existing wavefront acts as a source of a new secondary spherical (or circular) wavelet, and the new wavefront is the envelope tangent to all these wavelets Correct answer.
  2. The wavefront moves as a single rigid unit, and its shape can never change as it propagates
  3. Only the two edge points of a wavefront emit new wavelets; the interior points remain silent
  4. Huygens' principle only applies to light and cannot be used for mechanical waves such as water or sound waves

Explanation

Huygens' principle treats every point on a wavefront as a new point source emitting secondary wavelets in all directions; the new wavefront a short time later is the common tangent surface (envelope) of these wavelets. This construction explains phenomena such as diffraction, where waves bend around obstacles or through openings.

Question 47Waves Optics

Which frequency range is classified as ultrasound (Ultraschall)?

  1. Frequencies above about 20,000 Hz (20 kHz), above the upper limit of typical human hearing Correct answer.
  2. Frequencies below about 20 Hz, below the lower limit of typical human hearing
  3. Exactly the range 20 Hz to 20,000 Hz, the same as normal human hearing
  4. Only frequencies used in medical imaging, typically a single fixed frequency of 50 Hz

Explanation

The typical human audible range spans roughly 20 Hz to 20,000 Hz (20 kHz). Ultrasound refers to mechanical waves with frequencies above this upper limit - inaudible to humans but usable for applications such as medical imaging and sonar, since higher frequencies allow finer spatial resolution.

Question 48Waves Optics

At a certain point, two coherent sound sources produce waves of wavelength 0.5 m that have travelled a path difference of Δs = 0.75 m to reach that point. Is the interference at that point constructive or destructive?

  1. Destructive, because Δs/λ = 1.5 is a half-integer multiple of the wavelength Correct answer.
  2. Constructive, because Δs/λ = 1.5 rounds to an integer number of wavelengths
  3. Constructive, because any path difference smaller than 1 m always produces constructive interference
  4. No interference occurs unless the two sources have exactly the same amplitude

Explanation

Constructive interference occurs when the path difference is a whole number of wavelengths (Δs = k·λ), while destructive interference occurs when it is a half-integer number of wavelengths (Δs = (k+½)λ). Here Δs/λ = 0.75/0.5 = 1.5 = 1 + ½, a half-integer, so the two waves arrive out of phase by half a cycle and interfere destructively.

Question 49Electricity Magnetism

A charge of q = 1.6×10⁻¹⁹ C moves at v = 2×10⁶ m/s perpendicular to a magnetic field of B = 0.5 T. Using F = qvB, what force does it experience?

  1. 1.6 × 10⁻¹³ N Correct answer.
  2. 3.2 × 10⁻¹³ N
  3. 8 × 10⁻²⁰ N
  4. 2.56 × 10⁻³² N

Explanation

F = qvB = 1.6×10⁻¹⁹ C × 2×10⁶ m/s × 0.5 T = 1.6×10⁻¹³ N. This is the magnitude of the Lorentz force on a charge moving perpendicular to a magnetic field; its direction is given by the UVW (right-hand) rule, perpendicular to both v and B.

Question 50Electricity Magnetism

A long, straight wire carries a current of I = 10 A. Using B = μ0·I/(2πr) with μ0 = 4π×10⁻⁷ T·m/A, what is the magnetic flux density at a perpendicular distance of r = 5 cm from the wire?

  1. 4 × 10⁻⁵ T Correct answer.
  2. 8 × 10⁻⁵ T
  3. 4 × 10⁻⁷ T
  4. 31.8 T

Explanation

B = μ0·I/(2πr) = (4π×10⁻⁷ × 10)/(2π × 0.05) = 4×10⁻⁵ T = 40 μT - comparable in size to Earth's own magnetic field, which is why current-carrying wires can measurably deflect a nearby compass needle (the Oersted effect).