Fields & Motion: A Level Physics questions and answers
10 free A Level Physics Fields & Motion questions written in the style of Cambridge Physics papers, each with its answer and mark scheme. 4 of them are A* challenge questions, built to catch the mistakes strong students still make.
This topic covers Motion in a circle, Gravitational fields, Oscillations, Electric fields, Capacitance, Magnetic fields, Alternating currents. Work through the questions below, then generate a full Fields & Motion paper with a fresh set every time.
A satellite of mass 500 kg is placed in a synchronous orbit around the Earth. The Earth has mass 5.97 × 1024 kg, radius 6.37 × 106 m and rotates once in 86400 s.
(a)
State two features of a synchronous (geostationary) orbit.
[2]
(b)
Show that the radius of the orbit is about 4.2 × 107 m.
[3]
(c)
Calculate the gravitational potential at the orbit.
[1]
(d)
Calculate the change in gravitational potential energy of the satellite when it is moved from the surface to the orbit.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
period = rotation period; above equator, same direction
period equal to the planet’s rotation period B1 orbit above the equator moving in the same direction of rotation B1
2
(b)
4.22 × 107
GMm/r2 = mω2r with ω = 2π/T C1 r3 = GMT2/4π2C1 r = 4.22 × 107 m A1
3
(c)
−9.43 × 106 J kg−1
φ = −GM/r = −9.43 × 106 J kg−1B1
1
(d)
2.65 × 1010 J
ΔE = GMm(1/R − 1/r) C1 = 2.65 × 1010 J A1
2
Question 2Structured · 8 marksA* challenge
A mass of 0.15 kg on a spring oscillates with simple harmonic motion of period 0.80 s and amplitude 3.0 cm.
(a)
Define simple harmonic motion.
[2]
(b)
Calculate the maximum speed of the mass.
[2]
(c)
Calculate the maximum acceleration.
[1]
(d)
Calculate the total energy of the oscillation.
[2]
(e)
Calculate the speed when the displacement is half the amplitude.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
a ∝ −x, towards fixed point
acceleration proportional to displacement B1 and directed towards a fixed point (opposite to displacement) B1
2
(b)
0.236 m s−1
ω = 2π/T = 7.85 rad s−1C1 v0 = ωx0 = 0.236 m s−1A1
2
(c)
1.85 m s−2
a = ω2x0 = 1.85 m s−2B1
1
(d)
4.16 × 10−3 J
E = ½mω2x02C1 = 4.16 × 10−3 J A1
2
(e)
0.204 m s−1
v = ω√(x02 − x2) = 0.204 m s−1B1
1
Question 3Structured · 8 marksA* challenge
A sinusoidal alternating voltage of peak value 17 V and frequency 60 Hz is applied across a 47 Ω resistor.
(a)
Calculate the root-mean-square voltage.
[1]
(b)
Calculate the mean power dissipated in the resistor.
[2]
(c)
Calculate the peak current and the angular frequency.
[2]
(d)
A diode is placed in series with the resistor. State the new mean power.
[1]
(e)
Explain how a capacitor connected across the resistor smooths the output.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
12.0 V
Vrms = V0/√2 = 12.0 V B1
1
(b)
3.07 W
P = Vrms2/R C1 = 3.07 W A1
2
(c)
0.362 A; 377 rad s−1
I0 = 0.362 A A1 ω = 2πf = 377 rad s−1A1
2
(d)
1.54 W
halved: 1.54 W B1
1
(e)
charges at peaks, discharges between
the capacitor charges when the output is near its peak B1 and discharges through the resistor between peaks, maintaining the pd B1
2
Question 4Structured · 10 marksA* challenge
The Earth has mass 5.98 × 1024 kg and radius 6.37 × 106 m. A satellite is in a geostationary orbit.
(a)
State two features of a geostationary orbit, other than its period.
[2]
(b)
Show that the radius of a geostationary orbit is about 4.2 × 107 m.
[3]
(c)
A satellite of mass 2000 kg is moved from a low orbit of radius 6.67 × 106 m to geostationary orbit. Calculate the change in gravitational potential energy.
[3]
(d)
Explain why the satellite’s kinetic energy is less in the geostationary orbit.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
equatorial orbit; moves west to east (same direction as Earth’s rotation)
above the equator B1 same direction of rotation as the Earth B1
2
(b)
r3 = GMT2 ÷ 4π2; r = 4.23 × 107 m
GMm/r2 = mω2r C1 ω = 2π ÷ 86 400 C1 r = 4.23 × 107 m A1
v2 = GM/r so speed decreases with radius; KE = GMm/2r is smaller
v2 = GM ÷ r B1 larger r, smaller speed and KE B1
2
Question 5Structured · 6 marks
A satellite orbits a planet of mass 6.4 × 1023 kg in a circular orbit of radius 4.4 × 106 m.
(a)
Show that T2 = 4π2r3GM.
[2]
(b)
Calculate the orbital period.
[2]
(c)
State two features of a geostationary orbit.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
gravitational force = centripetal force
GMm/r2 = mrω2M1 ω = 2π/T leads to result A1
2
(b)
8880 s
substitution C1 8880 s A1
2
(c)
period 24 h; equatorial; west to east
Any 2 of: period 24 hours; above the equator; moves west to east (same direction as Earth rotates) B2
2
Question 6Structured · 10 marks
A proton moving at 6.0 × 105 m s−1 enters a uniform magnetic field of flux density 0.50 T at right angles to the field.
(a)
Define the tesla.
[2]
(b)
Explain why the proton moves in a circle.
[2]
(c)
Calculate the radius of the path. (mass of proton = 1.67 × 10−27 kg)
[3]
(d)
Show that the time for one orbit does not depend on the speed of the proton, and calculate it.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
B = F/(IL); 1 N on 1 m of wire carrying 1 A perpendicular
unit of magnetic flux density B1 a field in which a straight wire of length 1 m carrying 1 A at right angles to the field experiences a force of 1 N B1
2
(b)
force always perpendicular to velocity; provides centripetal force, speed constant
the magnetic force is always perpendicular to the velocity B1 so it does no work and provides the centripetal force (constant speed) B1
2
(c)
0.0125 m
Bqv = mv2/r C1 r = mv/(Bq) C1 0.0125 m A1
3
(d)
T = 2πm/(Bq) = 1.31 × 10−7 s
T = 2πr/v C1 = 2πm/(Bq), independent of v A1 1.31 × 10−7 s A1
3
Question 7Structured · 5 marks
Two parallel plates are 4 cm apart with a potential difference of 1000 V between them.
(a)
Calculate the electric field strength between the plates.
[1]
(b)
Calculate the force on an electron between the plates and its acceleration.
[2]
(c)
Describe the path of an electron that enters the field moving parallel to the plates.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
25000 V m−1
25000 V m−1B1
1
(b)
4.00 × 10−15 N; 4.39 × 1015 m s−2
F = eE: 4.00 × 10−15 N A1 a = F/m: 4.39 × 1015 m s−2A1
2
(c)
parabola towards positive plate
parabolic path B1 curving towards the positive plate B1
2
Question 8Structured · 4 marks
A satellite orbits the Earth at a height of 2400 km. The Earth has radius 6.4 × 106 m and mass 6.0 × 1024 kg. Calculate the orbital period in hours. (G = 6.67 × 10−11 N m2 kg−2)
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
2.28 hours
GMm/r2 = mω2r C1 r = 8.80 × 106 m C1 T = 2π√(r3/GM) C1 2.28 A1
4
Question 9Structured · 10 marks
A 470 μF capacitor is charged through a 2.2 kΩ resistor from a 6.0 V supply, starting uncharged.
(a)
Sketch graphs of current and of capacitor p.d. against time during charging.
Fig. 9.1
[2]
(b)
Calculate the initial current.
[1]
(c)
Calculate the p.d. across the capacitor after one time constant.
[2]
(d)
Calculate the time for the p.d. to reach 90% of the supply p.d.
[3]
(e)
Explain why the current decreases as the capacitor charges.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
current decays exponentially from V/R; p.d. rises to V exponentially
current: exponential decay from maximum B1 p.d.: exponential rise approaching 6.0 V B1
2
(b)
2.73 mA
I = V/R = 2.73 mA B1
1
(c)
3.79 V
V = V₀(1 − e−1) C1 3.79 V A1
2
(d)
2.38 s
0.90 = 1 − e−t/RCC1 t = RC ln 10 C1 2.38 s A1
3
(e)
p.d. across capacitor opposes supply; p.d. across resistor falls
the capacitor p.d. increases and opposes the supply B1 so the p.d. across the resistor, and hence the current, decreases B1
2
Question 10Structured · 5 marks
A ball of mass 0.2 kg on a string moves in a horizontal circle of radius 0.6 m on a smooth table, making 1.5 revolutions per second.
(a)
Calculate the angular speed of the ball.
[1]
(b)
Calculate the tension in the string.
[2]
(c)
Explain why the ball accelerates even though its speed is constant.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
9.42 rad s−1
9.42 rad s−1B1
1
(b)
10.7 N
F = mrω2C1 10.7 N A1
2
(c)
direction changes; centripetal force
velocity changes direction continuously B1 resultant force towards centre gives centripetal acceleration B1
In the 9702 syllabus, Fields & Motion covers Motion in a circle, Gravitational fields, Oscillations, Electric fields, Capacitance, Magnetic fields, Alternating currents.
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in