1st Year Physics — MCQs & Practice Questions

A base unit is one of the seven independent SI units, such as the metre. A derived unit is built from base units by multiplication or division, such as the newton, 1 N = 1 kg m s⁻².

The smallest measurement the instrument can read — the value of one division on its scale. A metre rule has a least count of 1 mm; a vernier calliper, 0.1 mm; a screw gauge, 0.01 mm.

Random errors scatter either side of the true value, so averaging makes them partly cancel. A single reading may happen to be one of the extreme ones.

1. Convert: 12.5 cm = 0.125 m, 0.1 cm = 0.001 mdivide by 100[1]
2. Percentage uncertainty = (Δl / l) × 100 = (0.1 / 12.5) × 100same units top and bottom, so the conversion cancels[1]
3. = 0.8%[1]

0.125 ± 0.001 m, or 0.8%

1. 0.51 mm is anomalous — it lies well away from the others[1]
2. It should be excluded from the mean, but reported rather than deletedthe reporting half is the mark most often missed[1]
3. Mean = (0.42 + 0.43 + 0.42 + 0.43) / 4four readings, not five[1]
4. = 0.425 ≈ 0.43 mm[1]

(a) 0.51 mm, excluded but reported (b) 0.43 mm

For every action there is an equal and opposite reaction. The two forces are of the same type and act on different bodies.

By Newton's first law the passenger continues moving at the same velocity because no resultant force acts on them; the bus decelerates beneath them, so they move forward relative to it.

The product of mass and velocity, p = mv. A vector quantity. SI unit: kg m s⁻¹.

1. States conservation of momentum: total before = total after[1]
2. Before: p = 2.0 × 3.0 + 4.0 × 0 = 6.0 kg m s⁻¹[1]
3. After: combined mass = 6.0 kg, so 6.0 = 6.0 × vthey stick together, so they share one velocity[1]
4. v = 1.0 m s⁻¹ in the original directiondirection expected for full marks[1]

1.0 m s⁻¹ in the direction of the original motion

1. Resultant force = 15 − 6.0 = 9.0 Nfriction opposes, so it subtracts[1]
2. Uses F = ma[1]
3. a = F/m = 9.0 / 3.0[1]
4. a = 3.0 m s⁻²unit required[1]

3.0 m s⁻²

A scalar has magnitude only, for example mass. A vector has both magnitude and direction, for example velocity.

Distance is the total path length, a scalar that only ever grows. Displacement is the straight line from start to finish, so returning to the starting point makes it zero.

The single force that has the same effect on the body as the two forces acting together.

1. Recognises the forces are perpendicular, so Pythagoras applies[1]
2. R = √(8.0² + 6.0²)[1]
3. R = 10.0 N[1]
4. Uses tan θ = 6.0 / 8.0[1]
5. θ = 36.9° north of easta direction with no reference line scores nothing[1]

10.0 N at 36.9° north of east

1. Replacing one vector by two components at right angles to each other[1]
2. Which together have the same effect as the original vector[1]
3. Uses W sin θ for the component along the slope[1]
4. = 250 × sin 20°[1]
5. = 85.5 N[1]
6. Uses W cos θ = 250 × cos 20°[1]
7. = 235 N[1]
8. The normal contact force from the surface of the slope[1]

(b) 85.5 N (c) 235 N

1. The current does not affect the crossing time: t = 30 / 1.2perpendicular components are independent[1]
2. t = 25 s[1]
3. Uses distance = 0.90 × 25[1]
4. = 22.5 m downstream[1]
5. Uses √(1.2² + 0.90²)[1]
6. = 1.5 m s⁻¹[1]

(a) 25 s (b) 22.5 m (c) 1.5 m s⁻¹

1. Selects v² = u² + 2asthe equation without t, since t is not given in part (a)[1]
2. 10.0² = 4.0² + 2 × a × 42 → 100 = 16 + 84a → a = 1.0 m s⁻²unit required for the mark[1]
3. Selects v = u + at (or s = ½(u+v)t)[1]
4. 10.0 = 4.0 + 1.0t → t = 6.0 s[1]

(a) 1.0 m s⁻² (b) 6.0 s

1. Uses s = ut + ½at² with u = 0"dropped from rest" is what tells you u = 0[1]
2. s = ½ × 9.81 × 3.2² = 50.2 ≈ 50 m[1]
3. Uses v = u + ator v² = u² + 2as[1]
4. v = 9.81 × 3.2 = 31.4 ≈ 31 m s⁻¹[1]
5. The actual speed would be lower / the stone would reach terminal velocitya statement about direction of change is enough[1]

(a) 50 m (b) 31 m s⁻¹ (c) the speed would be less

1. Acceleration = gradient = (20 − 0) / 40gradient of a velocity–time graph is acceleration[1]
2. = 0.50 m s⁻²[1]
3. Recognises distance = area under the graphthis is the mark most often missed[1]
4. Triangle ½ × 40 × 20 = 400; rectangle 60 × 20 = 1200; triangle ½ × 30 × 20 = 300all three areas needed[1]
5. Total = 1900 m[1]
6. Average speed = 1900 / 130 = 14.6 ≈ 15 m s⁻¹total distance ÷ total time, not the mean of the velocities[1]

(a) 0.50 m s⁻² (b) 1900 m (c) 15 m s⁻¹

1. Uses vertical motion with u_y = 0: 45 = ½ × 10 × t²"thrown horizontally" means the initial vertical velocity is zero[1]
2. t² = 9.0, t = 3.0 s[1]
3. Uses s_x = u_x t with constant horizontal velocity[1]
4. s_x = 15 × 3.0 = 45 m[1]
5. Horizontal and vertical motion are independent / gravity acts only vertically, so the vertical motion is unaffected by the horizontal velocity[1]

(a) 3.0 s (b) 45 m (c) the two components are independent

1. Displacement is the straight-line distance from the starting point to the finishing point[1]
2. …together with its direction / it is a vector quantitythe direction is required for the second mark[1]
3. Any curved or non-straight path, e.g. a runner going round a bend, a car following a winding roada full circular lap, where displacement is zero, also earns this[1]

Distance is the total length of the path travelled, a scalar. Displacement is the straight line from start to finish together with its direction, a vector.

Yes. A ball thrown vertically upward is momentarily at rest at the top of its flight, but gravity still acts, so its acceleration is g downward.

1. Uses a = (v − u)/t = (0 − 25)/5.0[1]
2. a = −5.0 m s⁻², a deceleration of 5.0 m s⁻²the negative sign or the word deceleration, not both required[1]
3. Uses s = ½(u + v)t or v² = u² + 2as[1]
4. s = ½(25 + 0) × 5.0 = 62.5 m[1]

deceleration 5.0 m s⁻², distance 62.5 m

1. Uses p = mv[1]
2. p = 1200 × 15 = 18 000 kg m s⁻¹the stationary car contributes nothing[1]
3. Applies conservation: total after = 18 000 kg m s⁻¹[1]
4. Combined mass 2000 kg, so v = 18 000 / 2000 = 9.0 m s⁻¹[1]
5. Kinetic energy is not conserved[1]
6. Inelastic collisionobjects sticking together is always inelastic[1]

(a) 1.8 × 10⁴ kg m s⁻¹ (b) 9.0 m s⁻¹ (c) not conserved; inelastic

1. Total momentum before is zeronothing is moving[1]
2. 0 = 0.020 × 400 + 4.0 × v[1]
3. v = −8.0 / 4.0 = −2.0 m s⁻¹, i.e. 2.0 m s⁻¹ backwardsdirection required[1]
4. The rifle's momentum must equal the bullet's in size and be opposite in direction[1]
5. Since p = mv is fixed, a larger m gives a smaller v[1]

2.0 m s⁻¹ backwards

1. Takes the initial direction as positive, so the rebound velocity is −8.0 m s⁻¹the sign is the whole question[1]
2. Δp = m(v − u) = 0.15 × (−8.0 − 12) = −3.0 kg m s⁻¹a change of 20 m s⁻¹, not 4[1]
3. Uses F = Δp/Δt[1]
4. F = −3.0 / 0.050 = −60 N, i.e. 60 N away from the wall[1]

60 N, directed away from the wall

The total momentum of a system before an interaction equals the total momentum after it, provided no resultant external force acts on the system.

It increases the time over which the ball is brought to rest. Since F = Δp/Δt and the change in momentum is fixed, a longer time means a smaller force on the hands.

1. Uses Δp = m(v − u) with u = 0the ball starts at rest on the tee[1]
2. Δp = 0.045 × 60 = 2.7 kg m s⁻¹[1]
3. Converts the time: 0.50 ms = 5.0 × 10⁻⁴ s[1]
4. Uses F = Δp/Δt[1]
5. F = 2.7 / 5.0 × 10⁻⁴ = 5400 N[1]
6. Applies conservation of momentum to club and ball together[1]
7. Before: 0.30 × 70 = 21 kg m s⁻¹the ball contributes nothing[1]
8. After: 21 = 0.30 × v + 2.7[1]
9. v = 18.3 / 0.30 = 61 m s⁻¹[1]

(a) 2.7 kg m s⁻¹ (b) 5400 N (c) 61 m s⁻¹

1. At the centre of gravity, which for a uniform beam is its midpoint, 1.0 m from the left end"uniform" is the word that tells you this[1]
2. Takes moments about the pivot; the beam's weight acts 1.0 − 0.50 = 0.50 m to the right of it[1]
3. Clockwise = anticlockwise: 40 × 0.50 = W × 0.50the load acts 0.50 m to the left of the pivot[1]
4. W = 40 N[1]

(a) at the midpoint, 1.0 m from the left end (b) W = 40 N

1. Passengers upstairs raise the centre of gravity of the bus[1]
2. A higher centre of gravity means a smaller tilt is needed before the vertical line through it falls outside the base / wheelbasethis is the key reasoning mark[1]
3. So the bus topples at a smaller angle / is less stable[1]

1. Recognises that only the perpendicular component turns the nutor equivalently uses the perpendicular distance[1]
2. Perpendicular component = 12 sin 30° = 6.0 N[1]
3. Moment = 6.0 × 0.25 = 1.5 N munit required[1]

1.5 N m

The turning effect of a force about a pivot, equal to force × perpendicular distance from the pivot to the line of action. SI unit: the newton metre, N m.

They are different quantities. Work is force acting along a displacement; a moment is force acting across a distance from a pivot. Sharing base units does not make them the same thing.

ΣF = 0 — no resultant force, so it does not accelerate. Σ moments = 0 — no resultant turning effect, so it does not rotate.

1. Left distance = 50 − 20 = 30 cm = 0.30 mdistance from the pivot, not from the end of the rule[1]
2. Anticlockwise moment = 3.0 × 0.30 = 0.90 N m[1]
3. Right distance = 70 − 50 = 20 cm = 0.20 m; for balance W₂ × 0.20 = 0.90principle of moments[1]
4. W₂ = 4.5 Nunit required[1]

4.5 N

Its direction of motion is changing continuously, so its velocity is changing. Velocity is a vector, and a changing velocity is an acceleration.

The tension in the string. If it breaks there is no longer a resultant force, so the stone flies off along the tangent in a straight line, not outwards along the radius.

No external torque acts, so angular momentum L = Iω is conserved. Pulling her arms in moves mass closer to the axis, reducing the moment of inertia, so the angular velocity must increase.

1. Uses ω = 2π/T[1]
2. ω = 2π / 5800 = 1.08 × 10⁻³ rad s⁻¹[1]
3. Uses v = rω[1]
4. v = 7.0 × 10⁶ × 1.08 × 10⁻³ = 7.6 × 10³ m s⁻¹[1]
5. a = ω²r = (1.08 × 10⁻³)² × 7.0 × 10⁶ = 8.2 m s⁻²close to g, as expected in low orbit[1]

ω = 1.08 × 10⁻³ rad s⁻¹, v = 7.6 km s⁻¹, a = 8.2 m s⁻²

1. Weight mg acting downwards[1]
2. Tension T also acting downwards, towards the centreat the top both point the same way[1]
3. At minimum speed the string just goes slack, so T = 0 and gravity alone provides the force[1]
4. mg = mv²/r, so v² = gr = 9.81 × 0.80the mass cancels[1]
5. v = 2.8 m s⁻¹[1]
6. At the bottom the tension acts up and the weight down: T − mg = mv²/r[1]
7. T = 0.25 × 6.0² / 0.80 + 0.25 × 9.81[1]
8. T = 11.25 + 2.45 = 13.7 N[1]

(b) 2.8 m s⁻¹ (c) 13.7 N

1. Uses KE = ½Iω²[1]
2. = ½ × 0.45 × 120² = 3240 J[1]
3. Uses L = Iω[1]
4. = 0.45 × 120 = 54 kg m² s⁻¹[1]
5. No external torque, so angular momentum is conserved: 54 = 0.60 ω[1]
6. ω = 90 rad s⁻¹kinetic energy has fallen — the collision is inelastic[1]

3240 J, 54 kg m² s⁻¹, 90 rad s⁻¹

Work = force × distance moved in the direction of the force, W = Fs cos θ. It is zero when the force is perpendicular to the motion, since cos 90° = 0.

Energy cannot be created or destroyed, only transferred from one store to another. The total energy of a closed system remains constant.

No. Energy is transferred to heat and sound in the ball and the floor at each bounce, so less remains as gravitational potential energy. The total is unchanged.

1. Uses E = mghthe useful energy is the gravitational potential energy gained[1]
2. E = 300 × 10 × 12 = 36 000 J[1]
3. Uses P = E/t[1]
4. P = 36 000 / 40 = 900 Wunit required[1]

900 W

1. Uses KE = ½mv²[1]
2. KE = ½ × 1200 × 20² = 240 000 Jsquare the velocity before multiplying[1]
3. Gain in KE = 240 000 J since it started from rest[1]
4. P = 240 000 / 8.0 = 30 000 W = 30 kW[1]

2.4 × 10⁵ J and 30 kW

1. Volume = 0.20 × 0.10 × 0.050 = 1.0 × 10⁻³ m³converting every length to metres first[1]
2. Density = 0.80 / 1.0 × 10⁻³ = 800 kg m⁻³[1]
3. Weight = mg = 0.80 × 10 = 8.0 Npressure needs force, and the force here is the weight[1]
4. Largest face area = 0.20 × 0.10 = 0.020 m²largest face gives the lowest pressure[1]
5. Pressure = 8.0 / 0.020 = 400 Pa[1]

(a) 800 kg m⁻³ (b) 400 Pa

1. Uses p = ρgh[1]
2. p = 1000 × 10 × 12 = 1.2 × 10⁵ Paunit required[1]
3. No effect / the pressure is unchanged[1]
4. Pressure in a liquid depends only on depth, density and g — not on the shape or width of the containerthe reasoning mark; the statement alone scores 1 of 2[1]

(a) 1.2 × 10⁵ Pa (b) no change — pressure depends only on depth

1. Extension = 12.0 − 8.0 = 4.0 cm = 0.040 mextension, not total length — the mark most often lost on this topic[1]
2. k = F/x = 5.0 / 0.040 = 125 N m⁻¹[1]
3. New extension = 8.0 / 125 = 0.064 m = 6.4 cm[1]
4. New length = 8.0 + 6.4 = 14.4 cmadding the natural length back on[1]

(a) 125 N m⁻¹ (b) 14.4 cm

Mass per unit volume, ρ = m/V. SI unit: kg m⁻³.

Broad feet spread the camel's weight over a larger area, so the pressure on the sand is smaller and it does not sink.

The extension of a spring is directly proportional to the load applied, provided the limit of proportionality is not exceeded.

1. Uses p = ρgh[1]
2. p = 800 × 10 × 1.5 = 12 000 Paunit required[1]
3. Rearranges p = F/A to F = pA[1]
4. F = 12 000 × 2.0 = 24 000 N[1]

p = 1.2 × 10⁴ Pa, F = 2.4 × 10⁴ N

Stress is force per unit cross-sectional area, σ = F/A, measured in pascals. Strain is extension divided by original length, a ratio with no units.

The maximum stress a material can take and still return to its original length when the load is removed. Beyond it the deformation is permanent.

The drag force grows with speed. As the ball accelerates the drag increases until drag plus upthrust equals the weight, the resultant force becomes zero, and the speed stops changing.

1. σ = F/A = 15 / 2.0 × 10⁻⁷[1]
2. σ = 7.5 × 10⁷ Pa[1]
3. ε = e/L = 1.2 × 10⁻³ / 1.8convert the mm first[1]
4. ε = 6.7 × 10⁻⁴no units[1]
5. E = σ/ε = 7.5 × 10⁷ / 6.7 × 10⁻⁴ = 1.1 × 10¹¹ Paabout right for copper[1]

σ = 7.5 × 10⁷ Pa, ε = 6.7 × 10⁻⁴, E = 1.1 × 10¹¹ Pa

1. Uses the equation of continuity A₁v₁ = A₂v₂[1]
2. 8.0 × 10⁻⁴ × 1.5 = 2.0 × 10⁻⁴ × v₂[1]
3. v₂ = 6.0 m s⁻¹area quartered, so speed quadrupled[1]
4. The pressure decreases[1]
5. By Bernoulli, where the speed is higher the pressure is lower[1]
6. The total energy per unit volume is constant, so a rise in kinetic energy must come at the expense of pressure energy[1]
7. The flow is steady and non-turbulent[1]
8. The fluid is incompressible and non-viscousany one more[1]

(a) 6.0 m s⁻¹ (b) pressure falls

1. Upthrust equals the weight of water displaced, ρgV[1]
2. = 1000 × 9.81 × 4.0 × 10⁻⁴[1]
3. = 3.9 N[1]
4. Weight in air = 1.1 × 9.81 = 10.8 N[1]
5. Apparent weight = 10.8 − 3.9 = 6.9 N[1]
6. It sinks — the upthrust is less than the weight, so its density exceeds that of water[1]

upthrust 3.9 N, apparent weight 6.9 N, sinks

1. Uses E = mcΔθ with Δθ = 70 °C88 − 18, not 88[1]
2. E = 1.5 × 4200 × 70 = 4.41 × 10⁵ J[1]
3. Uses t = E/P with P = 2000 Wconverting kW to W[1]
4. t = 441000 / 2000 = 220 saccept 220.5 s[1]
5. Energy is lost to the surroundings / the container is also heatedeither reason accepted[1]

(a) 4.4 × 10⁵ J (b) 220 s (c) heat lost to surroundings

1. Temperature rises from −10 °C to 0 °C[1]
2. Horizontal section at 0 °C while the ice melts[1]
3. Temperature rises from 0 °C to 100 °C[1]
4. Longer horizontal section at 100 °C while the water boilsthe boiling plateau must be longer than the melting one[1]
5. During the flat sections the energy supplied breaks bonds between particles rather than raising kinetic energythe explanation mark[1]

1. Heating makes particles vibrate more and push further apart, so each metal expands[1]
2. The two metals expand by different amounts for the same temperature risethis is the essential point[1]
3. The strip bends toward the metal that expands less[1]
4. Used in a thermostat / fire alarm / oven switchany valid use[1]

Temperature is the average kinetic energy of the particles, measured in °C or K. Heat is the total thermal energy of all the particles, measured in joules, and depends on how many there are.

The two metals expand by different amounts for the same temperature rise, so the strip curves toward the metal that expands less.

The energy supplied is used to break the bonds holding the particles in the lattice, not to increase their kinetic energy. Since temperature measures kinetic energy, it does not change.

1. Recognises two stages: melting, then warminga single-stage answer cannot score more than two[1]
2. Melting: E₁ = mL = 0.20 × 3.34 × 10⁵ = 6.68 × 10⁴ J[1]
3. Warming: E₂ = mcΔθ = 0.20 × 4200 × 20[1]
4. E₂ = 1.68 × 10⁴ J[1]
5. Total = 6.68 × 10⁴ + 1.68 × 10⁴ = 8.36 × 10⁴ Jadding the two stages[1]

8.36 × 10⁴ J

For a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume. Both the mass and the temperature must be constant.

The particles gain kinetic energy and move faster, so they strike the walls harder and more frequently. With the same wall area, the force per unit area increases.

The gas does work pushing back the surroundings, so W is negative. With Q = 0, ΔU is negative, the internal energy falls and the temperature drops.

1. Convert: 17 °C = 290 K, 137 °C = 410 K[1]
2. Constant pressure, so V₁/T₁ = V₂/T₂Charles's law[1]
3. 0.015 / 290 = V₂ / 410[1]
4. V₂ = 0.015 × 410 / 290[1]
5. V₂ = 0.021 m³a 41% rise in kelvin gives a 41% rise in volume[1]

0.021 m³

1. Uses 1 − T_c/T_h with kelvin[1]
2. 1 − 300/500 = 0.40, so 40%[1]
3. Maximum work = 0.40 × 6000 = 2400 J[1]
4. Heat rejected = 6000 − 2400[1]
5. = 3600 Jthis cannot be avoided — the second law requires it[1]
6. Actual efficiency = 1800/6000 = 0.30, or 30%[1]
7. Lower than the maximum because the real engine is not reversible[1]
8. Friction, turbulence and heat lost to the surroundings all reduce the work obtained[1]

(a) 40% (b) 2400 J of work, 3600 J rejected (c) 30%

1. The increase in internal energy equals the heat supplied plus the work done on the system[1]
2. ΔU = Q + W[1]
3. W = +250 J (work done ON the gas), Q = −100 J (heat lost)the signs are the difficult part[1]
4. ΔU = −100 + 250 = +150 J[1]
5. The temperature rises[1]
6. Internal energy has increased, and for a gas that means greater mean kinetic energy[1]

ΔU = +150 J, so the temperature rises

Wavelength is the distance between two neighbouring points that are in phase, for example crest to crest. Frequency is the number of complete waves passing a fixed point each second.

In a transverse wave the vibration is perpendicular to the direction of travel, for example light. In a longitudinal wave it is parallel to the direction of travel, for example sound.

The frequency stays the same. The speed decreases and, since v = fλ, the wavelength decreases in proportion.

1. Uses λ = length ÷ number of waves[1]
2. λ = 3.0 / 15 = 0.20 m[1]
3. Uses v = fλ[1]
4. v = 12 × 0.20 = 2.4 m s⁻¹[1]

λ = 0.20 m, v = 2.4 m s⁻¹

1. The waves spread out after passing through the gap[1]
2. This is diffraction[1]
3. The wavelength and frequency are unchanged; only the shape of the wavefront changes[1]
4. The waves spread out more[1]
5. Because the gap width is closer to the wavelengthmaximum spreading when gap ≈ λ[1]
6. Converts 25 mm = 0.025 m[1]
7. Uses v = fλ[1]
8. v = 8.0 × 0.025 = 0.20 m s⁻¹[1]

(c) 0.20 m s⁻¹

1. The waves change direction at the boundary[1]
2. They bend towards the normal[1]
3. The waves travel more slowly in the shallow water[1]
4. The end of each wavefront entering the shallow region slows first[1]
5. So the wavefront pivots, changing the direction of travel[1]
6. The direction does not change[1]
7. Because the whole wavefront slows at the same instant, so there is nothing to pivot aboutthe wavelength still shortens[1]

Sources with the same frequency and a constant phase difference. Without that, the phase relationship changes randomly and no stable pattern of fringes is seen.

Light is a transverse wave, so its vibrations are perpendicular to the direction of travel and one plane can be selected. Sound is longitudinal, vibrating along the direction of travel, so there is no plane to select.

Diffraction is greatest when the gap is comparable to the wavelength. Sound wavelengths are around a metre, similar to a doorway, while visible light is about 5 × 10⁻⁷ m — far too small to spread noticeably.

1. d = 1 / (300 × 10³) = 3.33 × 10⁻⁶ m300 per mm is 3 × 10⁵ per metre[1]
2. Uses d sin θ = nλ with n = 1[1]
3. sin θ = 590 × 10⁻⁹ / 3.33 × 10⁻⁶ = 0.177[1]
4. θ = 10.2°[1]
5. Highest order: n ≤ d/λ = 5.6, so n = 5sin θ cannot exceed 1[1]

θ = 10.2° for the first order; 5 orders visible

1. The two paths to the centre are equal in length[1]
2. So the path difference is zero, the waves arrive in phase and interfere constructively[1]
3. Uses x = λD/a[1]
4. x = (5.9 × 10⁻⁷ × 2.4) / 4.0 × 10⁻⁴[1]
5. x = 3.5 × 10⁻³ m, about 3.5 mm[1]
6. The fringe spacing doubles[1]
7. Because x is inversely proportional to a[1]
8. The fringes become more widely separated and easier to measure[1]

(b) 3.5 mm (c) spacing doubles

1. A laser beam is split down two perpendicular arms and recombined[1]
2. The arms are set so the returning beams cancel by destructive interference[1]
3. A tiny change in one arm's length upsets the cancellation, so light appears at the detectora fraction of a wavelength is enough[1]
4. Events such as black hole mergers emit almost no light[1]
5. But they do radiate gravitational waves, which also pass through intervening dust and gas unimpeded[1]

Electrons are transferred from the cloth onto the rod. The rod gains electrons and becomes negative; the cloth is left with a positive charge.

A region in which a charge experiences a force. The direction of the field is the direction of the force on a small positive test charge.

The rod induces a separation of charge in the paper, drawing the opposite charge to the near side. That near charge is closer to the rod, so its attraction outweighs the repulsion of the far side.

1. Converts 25 mm = 0.025 m[1]
2. Uses E = V/d = 5000 / 0.025[1]
3. E = 2.0 × 10⁵ V m⁻¹[1]
4. F = QE = 3.2 × 10⁻¹⁹ × 2.0 × 10⁵ = 6.4 × 10⁻¹⁴ N[1]

E = 2.0 × 10⁵ V m⁻¹, F = 6.4 × 10⁻¹⁴ N

1. Friction between the tanker and the air, and between the fuel and the tank walls, transfers electrons[1]
2. The tanker is insulated by its rubber tyres, so the charge cannot escape and accumulates[1]
3. The charge raises the potential of the tanker[1]
4. Eventually a spark jumps to a nearby earthed object[1]
5. The spark could ignite the fuel vapour and cause an explosion[1]
6. The wire provides a conducting path to earth[1]
7. Electrons flow along it until the tanker is at the same potential as the earth[1]
8. So charge leaks away steadily instead of building to a spark[1]

1. The lines are parallel and equally spaced, showing a uniform field[1]
2. They run from the positive plate to the negative plate[1]
3. The force is upward, towards the positive plate[1]
4. Because the drop is negative, so the force is opposite to the field direction[1]
5. Increase the potential difference across the plates[1]
6. Move the plates closer together, or increase the charge on the dropany one[1]

(b) upward, towards the positive plate

The rate of flow of electric charge, I = Q/t. Its unit is the ampere (A), where one ampere is one coulomb per second.

The current through a metallic conductor is directly proportional to the potential difference across it, provided the temperature stays constant.

The metal ions vibrate more strongly, so the moving electrons collide with them more often. Each collision impedes the flow, so the resistance rises.

1. Converts the area: 0.50 mm² = 0.50 × 10⁻⁶ m²the step most often dropped[1]
2. Uses ρ = RA/L[1]
3. ρ = (0.068 × 0.50 × 10⁻⁶) / 2.0[1]
4. ρ = 1.7 × 10⁻⁸ Ω mcopper[1]
5. Doubling the length doubles the resistance: 0.136 Ω[1]

ρ = 1.7 × 10⁻⁸ Ω m; R = 0.14 Ω

1. The graph passes through the origin[1]
2. It is a straight line at low potential difference[1]
3. It then curves towards the V axis, so the gradient falls[1]
4. A larger current heats the filament[1]
5. The ions vibrate more and the electrons collide with them more often[1]
6. So the resistance increases and the current no longer rises in proportion[1]
7. R = V/I = 6.0 / 0.50 = 12 Ω[1]
8. P = VI = 6.0 × 0.50 = 3.0 W[1]

(c) 12 Ω and 3.0 W

1. Total resistance = 20 + 25 = 45 Ω[1]
2. Uses I = V/R[1]
3. I = 9.0 / 45 = 0.20 A[1]
4. Uses V = IR for the thermistor[1]
5. V = 0.20 × 25 = 5.0 V[1]
6. The potential difference across the thermistor decreases[1]
7. Its resistance falls as it warms, so it takes a smaller share of the supply voltage[1]

(a) 0.20 A (b) 5.0 V (c) it falls

1. Uses V_s/V_p = N_s/N_p[1]
2. V_s = 230 × 5000/200[1]
3. V_s = 5750 V[1]
4. Step-upmore turns on the secondary[1]
5. Direct current produces a constant magnetic field in the core[1]
6. A changing field is required to induce an e.m.f., so nothing is induced in the secondarythe word "changing" is the mark[1]

(a) 5750 V (b) step-up (c) d.c. gives a constant field, and induction needs a changing one

1. The needle deflects to one side while the magnet is moving[1]
2. Move the magnet faster[1]
3. Use a stronger magnet, or a coil with more turnsany second valid change[1]
4. The needle returns to zero — no deflection[1]
5. There is no change in the magnetic field through the coil, and induction requires a change[1]

1. For a given power, a higher voltage means a smaller currentfrom P = VI[1]
2. Power lost as heat in the cables is I²R, so a smaller current wastes much less energythe I² is the key point[1]
3. A step-up transformer raises the voltage at the power station[1]
4. A step-down transformer lowers it again for safe use in homes[1]

Increase the current through the coil, or use a stronger magnetic field. (Also acceptable: increase the number of turns on the coil.)

Soft iron is easily magnetised and demagnetised, so it follows the rapidly alternating field of the primary and carries the changing flux efficiently to the secondary.

1. Uses I = P/V = 20 × 10⁶ / 25 × 10³[1]
2. I = 800 A[1]
3. Uses P = I²R = 800² × 4.0[1]
4. = 2.56 × 10⁶ W = 2.56 MWnearly 13% of the output[1]
5. New current = 20 × 10⁶ / 400 × 10³ = 50 A[1]
6. New loss = 50² × 4.0[1]
7. = 1.0 × 10⁴ W = 10 kW[1]
8. A reduction by a factor of 256, because the current fell by 16 and the loss depends on I²the point of the question[1]
9. Turns ratio N_s : N_p = 400 : 25 = 16 : 1[1]

(a) 800 A (b) 2.56 MW (c) 50 A and 10 kW (d) 16 : 1

The laws of physics are the same in all inertial frames. The speed of light in a vacuum is the same for all observers, regardless of the motion of the source or the observer.

They travel at close to the speed of light, so from our frame their internal clocks run slow and they survive far longer than their rest-frame half-life. Equivalently, in the muon's frame the atmosphere is length contracted, so there is less distance to cover.

Its relativistic mass is γm₀, and γ tends to infinity as v approaches c. An infinite amount of energy would therefore be needed to reach light speed.

1. γ = 1/√(1 − 0.60²) = 1/√0.64[1]
2. γ = 1/0.80 = 1.25[1]
3. Uses L = L₀/γdivide, because the moving object is shorter[1]
4. L = 2.0/1.25 = 1.6 m[1]
5. The width is unchanged at 0.10 m — contraction acts only along the direction of motion[1]

γ = 1.25, length 1.6 m, width still 0.10 m

1. γ = 1/√(1 − 0.64)[1]
2. γ = 1.67[1]
3. From Earth, t = distance/speed = 8.0/0.80light-years over fractions of c gives years[1]
4. t = 10 years[1]
5. Crew time = t/γ = 10/1.67the crew measure the proper time[1]
6. = 6.0 years[1]
7. In the crew's frame the distance is length contracted[1]
8. It is only 8.0/1.67 = 4.8 light-years, and at 0.80c that takes 6.0 years — the two frames agree[1]

γ = 1.67; 10 years from Earth; 6.0 years for the crew

1. Energy per second = 3.8 × 10²⁶ Ja watt is a joule per second[1]
2. Rearranges E = mc² to m = E/c²[1]
3. m = 3.8 × 10²⁶ / 9.0 × 10¹⁶ = 4.2 × 10⁹ kgabout four million tonnes every second[1]
4. It is not destroyed — mass and energy are equivalent[1]
5. In fusion the products have slightly less mass than the reactants, and that mass defect is radiated as energy[1]

4.2 × 10⁹ kg per second

1. 92 protonsthe proton number[1]
2. 235 − 92 = 143 neutrons[1]
3. 92 electrons, since the atom is neutral[1]
4. Same number of protons (92); different number of neutrons (146 instead of 143)[1]
5. Chemical behaviour depends on the electrons[1]
6. Both have 92 electrons arranged identically, so they react in the same way[1]

(a) 92 p, 143 n, 92 e (b) same Z, different A (c) identical electron arrangement

1. Most alpha particles passed straight through[1]
2. So the atom is mostly empty space[1]
3. Some were deflected through large angles[1]
4. So there is a concentrated positive charge repelling them[1]
5. A very few were reflected almost straight back[1]
6. So the nucleus is very small and contains most of the atom's mass[1]

1. In that model the positive charge is spread thinly throughout the atom[1]
2. So the repulsive force on an alpha particle anywhere would be small[1]
3. It could not produce the large-angle deflections or backscattering that were observed[1]

The nucleon number A is the total number of protons and neutrons in the nucleus. The proton number Z is the number of protons alone.

They have the same number of protons and therefore the same number of electrons in the same arrangement. Chemical behaviour is decided by the electrons, not by the number of neutrons.

1. Almost all the alpha particles passed straight through[1]
2. A very small fraction were deflected through large angles, some straight back[1]
3. Passing straight through shows the atom is mostly empty space[1]
4. Large deflections show a concentrated region of positive charge[1]
5. Which repels the positive alpha particle[1]
6. Backward scattering shows that region is also very massive, and very small[1]
7. So each alpha particle meets at most one nucleus[1]
8. A thicker foil would cause multiple scattering and the result could not be interpreted[1]

1. 18 hours is 3 half-lives18 ÷ 6[1]
2. Halves three times: 800 → 400 → 200 → 100[1]
3. 100 counts per minute[1]
4. Each half-life removes only half of what remains, so some always remains[1]
5. Temperature or pressure[1]
6. Chemical state or physical form of the sampleany second valid factor[1]

(a) 100 counts/min (b) halving never reaches zero (c) temperature, pressure, chemical state

1. Alpha: A = 218 − 4 = 214[1]
2. Z = 84 − 2 = 82lead[1]
3. Beta: A unchanged at 214[1]
4. Z = 82 + 1 = 83bismuth[1]
5. A neutron changes into a proton and an electron, and the electron is emitted[1]

(a) ²¹⁴₈₂Pb (b) ²¹⁴₈₃ (c) a neutron becomes a proton plus an emitted electron

1. Gamma[1]
2. It is penetrating enough to leave the body and be detected, and least ionising so it does least damageboth halves wanted[1]
3. Short, so the activity falls quickly and the patient is not exposed for long[1]
4. But not so short that it decays away before the scan can be completedthe balance is the point of the question[1]

Handle it with long tongs to increase the distance from the body, and return it to its lead-lined container immediately after use. (Also acceptable: never point it at anyone; minimise exposure time.)

The low level of ionising radiation always present in the environment. Sources include radon gas from rocks, cosmic rays, medical X-rays and food.

1. The time taken for the number of undecayed nuclei in the sample to halve[1]
2. Equivalently, the time for the count rate to fall to half its value; the process is random so this is an average[1]
3. 32 days is 4 half-lives[1]
4. Uses 640 ÷ 2⁴[1]
5. = 40 counts per minute[1]
6. The extra 15 counts per minute is background radiation[1]
7. It is measured with the source removed and subtracted from every reading[1]
8. The proton number increases by 1[1]
9. The nucleon number is unchangeda neutron becomes a proton plus an electron[1]

(b) 40 counts per minute