IB · Physics SL & HL · P2S

IB Physics Paper 2 · SL structured questions practice papers

SL · 50 marks · 1h 30m. Each generated paper follows the SL & HL Paper 2 · SL structured questions format, with the same number and style of questions as the real exam, and comes with an IB-style mark scheme.

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Sample Paper 2 · SL structured questions questions

Question 1Structured · 13 marksA* challenge

A planet of mass 6.4 × 1023 kg and radius R = 3.4 × 106 m has no atmosphere. G = 6.67 × 10−11 N m2 kg−2. Kepler’s third law also applies to an elliptical orbit if the radius is replaced by the semi-major axis (half the longest diameter of the ellipse).

(a)

A probe is launched vertically at 4.0 km s−1. Determine the maximum height above the surface that it reaches.

[3]
(b)

A second probe is launched horizontally from the surface. Its greatest distance from the centre of the planet is 3R. Determine its launch speed.

[5]
(c)

Determine the time the second probe takes to reach its greatest distance from the planet.

[3]
(d)

Show that the second probe has the same total energy as a probe in a circular orbit of radius 2R.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)≈ 6.0 × 10⁶ m
½v² − GM/R = −GM/r ✓
r = 9.4 × 106 m ✓
height ≈ 6.0 × 106 m ✓
3
(b)√(3GM/2R) ≈ 4.3 km s⁻¹
angular momentum conserved: v₁R = v₂(3R) ✓
energy: ½v₁² − GM/R = ½v₂² − GM/3R ✓
½v₁²(1 − 1/9) = 2GM/3R ✓
v₁ = √(3GM/2R) ✓
≈ 4.3 × 103 m s−1 ✓
5
(c)≈ 8.5 × 10³ s
semi-major axis = (R + 3R)/2 = 2R ✓
T = 2π√((2R)³/GM) ✓
half the period ≈ 8.5 × 103 s ✓
3
(d)both −GMm/4R
½(3GM/2R) − GM/R = −GM/4R per kg ✓
circular orbit at 2R: −GM/2(2R) = −GM/4R ✓
2
Question 2Structured · 9 marksA* challenge

A block of mass 0.25 kg on a frictionless horizontal surface is attached to a spring of spring constant 24.5 N m−1. It oscillates with simple harmonic motion of amplitude 0.05 m.

(a)

Show that the total energy ET can be written as ET = ½kx02.

[2]
(b)

Calculate the period of oscillation.

[2]
(c)

Calculate the speed of the block when its displacement is half the amplitude.

[2]
(d)

Determine the shortest time for the block to move from the equilibrium position to a displacement of half the amplitude.

[2]
(e)

Determine the fraction of the total energy that is kinetic at this displacement.

[1]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)E = ½mω²x₀² with ω² = k/m
ET = ½mω2x02 ✓
ω2 = k/m so ET = ½kx02 ✓
2
(b)0.635 s
T = 2π√(m/k) ✓
= 0.635 s ✓
2
(c)0.429 m s−1
v = ω√(x02 − x2) ✓
= 0.429 m s−1 ✓
2
(d)0.0529 s
x = x0sin ωt so ωt = π/6 ✓
t = T/12 = 0.0529 s ✓
2
(e)0.75
¾ ✓
1
Question 3Structured · 9 marks

Water falls through a vertical height of 150 m to the turbines of a hydroelectric power station at a rate of 20 m3 s−1. The density of water is 1000 kg m−3.

(a)

Calculate the rate at which the gravitational potential energy of the water decreases.

[2]
(b)

The overall efficiency is 80%. Determine the electrical power output.

[1]
(c)

Calculate the speed of the water as it reaches the turbines, assuming no energy losses in the pipes.

[2]
(d)

The station runs at full power for 5 hours each day. Calculate the electrical energy produced each day, in MJ.

[2]
(e)

State two energy transfers that reduce the efficiency.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)29.4 MW
mass per second = 20000 kg s−1 ✓
P = 20000 × 9.81 × 150 = 29.4 MW ✓
2
(b)23.5 MW
23.5 MW ✓
1
(c)54.2 m s−1
½v2 = gh ✓
v = 54.2 m s−1 ✓
2
(d)423792 MJ
E = Pt = 23.5 × 106 × 18000 ✓
423792 MJ ✓
2
(e)friction, heating
thermal energy from friction in the pipes/turbine ✓
resistive heating in generator coils/cables (or sound) ✓
2
Question 4Structured · 6 marks

A straight wire of length 0.25 m carries a current of 3.0 A in a uniform magnetic field of 0.40 T.

(a)

Calculate the force on the wire when it is perpendicular to the field.

[1]
(b)

Calculate the force when the wire is at 30° to the field.

[1]
(c)

Calculate the magnetic field strength 5.0 cm from a long straight wire carrying 3.0 A.

[2]
(d)

Explain why two parallel wires carrying currents in the same direction attract.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.30 N
F = BIL = 0.30 N ✓
1
(b)0.15 N
F = BIL sin30° = 0.15 N ✓
1
(c)1.2 × 10−5 T
B = μ0I/2πr ✓
= 1.2 × 10−5 T ✓
2
(d)each in other’s field
each wire is in the magnetic field of the other ✓
the force from F = BIL on each is towards the other wire ✓
2
Question 5Structured · 6 marks

An electric kettle of power 2.2 kW heats 1.5 kg of water from 20 °C to 100 °C. Specific heat capacity of water = 4200 J kg−1 K−1.

(a)

Calculate the minimum time taken.

[2]
(b)

The kettle is left on for a further 60 s. Calculate the mass of water that boils away. (L = 2.26 × 106 J kg−1)

[2]
(c)

Explain, in molecular terms, why the temperature stays constant while the water boils.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)230 s
Q = mcΔT = 504 000 J ✓
t = 229 s ✓
2
(b)0.058 kg
Q = 2200 × 60 = 132 000 J ✓
m = 0.058 kg ✓
2
(c)energy increases PE, not KE
energy supplied increases the potential energy of molecules (breaking bonds) ✓
mean kinetic energy, and so temperature, stays constant ✓
2
Question 6Structured · 10 marks

Coherent light of wavelength 532 nm is incident on two narrow slits 0.25 mm apart. Fringes are observed on a screen 2 m away.

(a)

Explain what is meant by coherent light and why it is needed.

[2]
(b)

Calculate the fringe spacing.

[2]
(c)

Explain, in terms of path difference, the formation of a dark fringe.

[2]
(d)

The apparatus is immersed in water of refractive index 1.33. Calculate the new fringe spacing.

[2]
(e)

One slit is covered with a filter that halves the intensity of the light passing through it. Describe the effect on the pattern.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)constant phase difference
constant phase difference (and same frequency) ✓
so that the interference pattern is stable ✓
2
(b)4.26 mm
s = λDd ✓
4.26 mm ✓
2
(c)half-wavelength path difference
the path difference is an odd number of half wavelengths ✓
waves arrive in antiphase and interfere destructively ✓
2
(d)3.20 mm
wavelength in water = 5321.33 = 400 nm ✓
spacing = 3.20 mm ✓
2
(e)dark fringes less dark
fringe positions unchanged ✓
minima are no longer completely dark (and maxima less bright), so contrast is reduced ✓
2

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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in