2025 OCR GCSE Twenty First Century Science - Physics B - J259 Practice Paper with Answers

wave speed = frequency × wavelength = \(5.0 \times 0.40 = 2.0\) m/s.

B (2.0 m/s) = 1 mark. DO NOT ALLOW more than one box ticked. AO2.1.

In a longitudinal wave the particles vibrate parallel to the direction of travel. Waves transfer energy, not matter.

C = 1 mark. DO NOT ALLOW more than one box ticked. AO1.1.

\(E_k = \frac{1}{2} m v^2 = 0.5 \times 2.0 \times 3.0^2 = 9.0\) J.

C (9.0 J) = 1 mark. Distractors: B from \(mv\), D from \(m v^2\). DO NOT ALLOW more than one box ticked. AO2.1.

\(V = I R = 0.50 \times 12 = 6.0\) V.

B (6.0 V) = 1 mark. Distractors: A from \(I/R\), D from \(R/I\). DO NOT ALLOW more than one box ticked. AO2.1.

speed = distance ÷ time = \(150 \div 20 = 7.5\) m/s.

B (7.5 m/s) = 1 mark. DO NOT ALLOW more than one box ticked. AO2.1.

\(a = F \div m = 1200 \div 800 = 1.5\) m/s².

B (1.5 m/s²) = 1 mark. Distractor A from \(m/F\). DO NOT ALLOW more than one box ticked. AO2.1.

An alpha particle is 2 protons and 2 neutrons, so the nucleus loses 4 from its mass number and 2 from its proton number.

A = 1 mark. Option C describes beta decay. DO NOT ALLOW more than one box ticked. AO1.1.

An iron core is magnetised by the coil and strengthens the field. Plastic has no effect, and reversing the battery only reverses the field direction.

B = 1 mark. DO NOT ALLOW more than one box ticked. AO1.1.

wave speed = frequency × wavelength = \(20 \times 0.15 = 3.0\) m/s.

B (3.0 m/s) = 1 mark (AO2.1). DO NOT ALLOW more than one box ticked. Distractors: A from 0.15 ÷ 20, C from adding, D from 20 ÷ 0.15.

\(E_k = \frac{1}{2} \times 4.0 \times 3.0^2 = \frac{1}{2} \times 4.0 \times 9.0 = 18\) J.

C (18 J) = 1 mark (AO2.1). Distractors: A from ½ × 4 × 3, B from 4 × 3, D from 4 × 9 (no ½).

charge = \(0.50 \times 40 = 20\) C.

B (20 C) = 1 mark (AO2.1). Distractors: A from 0.5 ÷ 40, C from adding, D from 40 ÷ 0.5.

acceleration = \(12 \div 8.0 = 1.5\) m/s\(^2\).

B (1.5 m/s²) = 1 mark (AO2.1). Distractors: A from 8 ÷ 12, C from 12 − 8, D from 12 × 8.

force = \(1500 \times 2.0 = 3000\) N.

C (3000 N) = 1 mark (AO2.1). Distractors: A from 1500 ÷ 2, B from subtracting, D from a power-of-ten slip.

Isotopes are atoms of the same element, so they have the same number of protons (and electrons when neutral), but different numbers of neutrons.

B = 1 mark (AO1.1). DO NOT ALLOW more than one box ticked.

The vibration is along the direction of travel, so the wave is longitudinal.

Using wave speed = frequency × wavelength: \( v = f \lambda = 5 \times 0.40 = 2.0 \) m/s.

Since \( \lambda = v / f \), doubling \( f \) at constant \( v \) halves the wavelength.

Tick one box in each row. 1 mark per row (AO1.1, AO2.1, AO2.1).

Row 1: longitudinal (1)
Row 2: 2.0 m/s (1)
Row 3: halves (1)

Guidance: DO NOT ALLOW a mark in any row where more than one box is ticked. Each row is marked independently, so there is no penalty across rows. IGNORE any working shown.

\( v = f \times \lambda = 4.0 \times 0.35 = 1.4 \) m/s

Check the answer line first. Correct answer 1.4 m/s = 2 marks.
Otherwise: substitution \( 4.0 \times 0.35 \) = 1 mark; correct value 1.4 with unit m/s = 1 mark.
ALLOW 1.40. DO NOT ALLOW a wrong unit for the final mark. AO2.1.

In a longitudinal wave the particles oscillate backwards and forwards along the same line as the direction of energy transfer. They return to their original positions and are not carried along with the wave.

Particles vibrate / oscillate / move back and forth (about a fixed position) = 1 mark.
Parallel to / in the same direction as the direction of wave travel (AW) = 1 mark.
DO NOT ALLOW particles travel along with the wave. IGNORE references to compressions only. AO1.1.

\( E_k = \tfrac{1}{2} m v^2 = 0.5 \times 1500 \times 12^2 = 0.5 \times 1500 \times 144 = 108\,000 \) J

Check the answer line first. 108 000 J (or 108 kJ) = 2 marks.
Otherwise: correct substitution \( 0.5 \times 1500 \times 12^2 \) = 1 mark; arithmetic = 1 mark.
If 12 is not squared (answer 9000) award 1 mark. ALLOW 1.08 × 10^5. Unit J required for full marks. AO2.1.

\( V = I \times R = 0.50 \times 12 = 6.0 \) V

Check the answer line first. 6.0 V = 2 marks.
Otherwise: substitution \( 0.50 \times 12 \) = 1 mark; answer 6 = 1 mark (unit V required for the second mark).
ALLOW 6. AO2.1.

\( a = \dfrac{\Delta v}{t} = \dfrac{9.0 - 0}{6.0} = 1.5 \) m/s²

Check the answer line first. 1.5 m/s² = 2 marks.
Otherwise: substitution \( 9.0 \div 6.0 \) = 1 mark; correct value with unit m/s² = 1 mark.
ALLOW m/s/s. DO NOT ALLOW m/s for the unit. AO2.1.

\( F = m \times a = 3.0 \times 2.5 = 7.5 \) N

Check the answer line first. 7.5 N = 2 marks.
Otherwise: substitution \( 3.0 \times 2.5 \) = 1 mark; correct value with unit N = 1 mark.
AO2.1.

Mass number = protons + neutrons = 6 + 8 = 14. Isotopes are atoms of the same element, so they have the same number of protons (6), but they have different numbers of neutrons (8 and 6).

Mass number 14 = 1 mark (part a).
Same number of protons AND different number of neutrons = 1 mark (part b).
Both ideas needed for the mark in part b. IGNORE electrons. DO NOT ALLOW different numbers of protons. AO1.1.

\( p_1V_1 = p_2V_2 \), so \( 100 \times 120 = p_2 \times 40 \). \( p_2 = 12\,000 \div 40 = 300 \) kPa

Check the answer line first. 300 kPa = 2 marks.
Otherwise: \( 100 \times 120 = p \times 40 \) (or equivalent rearrangement) = 1 mark; 300 = 1 mark.
ALLOW 300 000 Pa. Unit required for full marks. AO2.1.

Using the same spring means the only thing changing is the force. Reading the scale at eye level, with a pointer, reduces parallax error and improves precision.

Any one control variable, e.g. same spring / same ruler / same type of masses / not exceeding the elastic limit = 1 mark.
Any one way to improve precision, e.g. read scale at eye level (avoid parallax) / use a pointer or marker / use a ruler with smaller divisions = 1 mark.
DO NOT ALLOW the mass (it is the independent variable). DO NOT ALLOW 'repeat readings' for the precision mark. AO3.3a.

Waves transfer energy, not matter. The number of waves passing a point each second is the frequency.

Q1 | energy | 1 | AO1.1 | DO NOT ALLOW matter
Q1 | frequency | 1 | AO1.1 | IGNORE 'Hz' alone
Total 2

As the ball rises it slows down, so its kinetic store decreases. Its height increases, so its gravitational potential store increases.

Q2 | decreases | 1 | AO1.1 |
Q2 | increases | 1 | AO1.1 |
Total 2

Resistor A: \( R = V \div I = 6.0 \div 0.50 = 12\ \Omega \). Resistor B: \( V = I \times R = 0.30 \times 20 = 6.0\ \text{V} \).

Q3 | 12 (ohms) | 1 | AO2.1 |
Q3 | 6.0 (V) | 1 | AO2.1 | ALLOW 6
Total 2

Asha: \( 150 \div 25 = 6.0\ \text{m/s} \). Ben: time = distance ÷ speed = \( 90 \div 3.0 = 30\ \text{s} \).

Q4 | 6.0 (m/s) | 1 | AO2.1 | ALLOW 6
Q4 | 30 (s) | 1 | AO2.1 |
Total 2

Atoms of the same element have the same number of protons. Isotopes differ in the number of neutrons in the nucleus.

Q5 | protons | 1 | AO1.1 | ALLOW 'atomic number'
Q5 | neutrons | 1 | AO1.1 |
Total 2

Use wave speed = frequency × wavelength.
\( v = 8 \times 0.015 = 0.12 \) m/s.

Tick the first box (0.12 m/s) = 2 marks.
If a different box is ticked, 0 marks. Only one box should be ticked.
Mark 1 (AO2.1): correct selection of method (f × λ) – may be awarded only through the correct answer.
Mark 2 (AO2.1): correct value and unit.
DO NOT ALLOW more than one tick; ignore any working if the correct box is ticked.

Gravitational potential energy = mass × gravitational field strength × height.
\( E = 4 \times 10 \times 2.5 = 100 \) J.

Tick the fourth box (100 J) = 2 marks.
Otherwise 0 marks.
Mark 1 (AO2.1): selects/uses m × g × h.
Mark 2 (AO2.1): correct arithmetic.
DO NOT ALLOW more than one tick.

Use potential difference = current × resistance.
\( V = 0.30 \times 20 = 6 \) V.

Tick the second box (6 V) = 2 marks.
Otherwise 0 marks.
Mark 1 (AO2.1): uses V = I × R.
Mark 2 (AO2.1): correct arithmetic.
DO NOT ALLOW more than one tick. The distractors come from dividing (67 V), dividing the other way (0.015 V) and adding (20.3 V).

The distance between neighbouring crests is the wavelength, \(\lambda = 0.030\) m.
\(v = f \times \lambda\)
\(v = 8.0 \times 0.030 = 0.24\) m/s

Check the answer line first. Correct answer 0.24 (m/s) = 2 marks.

Otherwise:
1 mark: correct substitution \(8.0 \times 0.030\) (method).
1 mark: correct answer 0.24 with the unit m/s (accuracy).

ALLOW 0.24 without a unit for 1 mark only if working is shown. ALLOW 24 cm/s. DO NOT ALLOW an answer of 267 (division of frequency by wavelength) - this scores 0. ECF: allow 1 mark for a correct evaluation of a wrongly substituted product only if the equation used is speed = frequency x wavelength.

\(E_k = \frac{1}{2} \times m \times v^2\)
\(E_k = 0.5 \times 1.5 \times 4.0^2 = 0.5 \times 1.5 \times 16 = 12\) J

Check the answer line first. Correct answer 12 (J) = 2 marks.

Otherwise:
1 mark: correct substitution into \(\frac{1}{2} m v^2\), e.g. \(0.5 \times 1.5 \times 4.0^2\) (method).
1 mark: correct answer 12 (accuracy).

ALLOW 12.0. IGNORE a missing unit if the answer line has the unit J printed. ALLOW 1 mark for 24 (half omitted) or 3.0 (speed not squared) where the correct equation is shown, as evidence of partial method. DO NOT ALLOW these for the accuracy mark.

Convert the time to seconds: 2.0 minutes = 120 s.
\(Q = I \times t\)
\(Q = 0.35 \times 120 = 42\) C

Check the answer line first. Correct answer 42 (C) = 2 marks.

Otherwise:
1 mark: converts time to 120 s and substitutes correctly, \(0.35 \times 120\) (method).
1 mark: correct answer 42 (accuracy).

ALLOW 1 mark for 0.7 (time left as 2.0 minutes, i.e. 0.35 x 2.0) as it shows use of Q = It without unit conversion. DO NOT ALLOW 0.35 / 120 or other division. IGNORE a missing unit if C is printed on the answer line.

Waves transfer energy from place to place. The water itself is not carried along with the wave.

Energy [1]. ALLOW 'information' in addition to energy. DO NOT ALLOW 'matter', 'water' or 'particles' as the only answer.

Wave speed = frequency × wavelength, so \( v = 4.0 \times 0.75 = 3.0 \) m/s.

Correct answer of 3 or 3.0 m/s [1]. Check the answer line first. ALLOW 3.00. Unit m/s required for the mark.

As the ball moves higher in the Earth's gravitational field, the energy in its gravitational store increases.

Gravitational (potential) [1]. ALLOW gravitational potential energy, GPE. DO NOT ALLOW kinetic or thermal.

An LDR has a resistance that falls as the light intensity increases.

Light-dependent resistor or LDR [1]. DO NOT ALLOW thermistor, photodiode or 'light resistor'.

Speed = distance ÷ time, so \( v = 600 \div 50 = 12 \) m/s.

Correct answer of 12 m/s [1]. Check the answer line first. Unit m/s required for the mark.

Isotopes have the same number of protons but different numbers of neutrons, so their nuclei differ in neutron number (and mass).

Different number of neutrons [1]. ALLOW different mass number or different nucleon number. DO NOT ALLOW different number of protons or electrons.

Frequency \( f = 12 \) Hz. Wavelength \( \lambda = 3.5 \) cm \( = 0.035 \) m.
\( v = f \times \lambda = 12 \times 0.035 = 0.42 \) m/s.

Check the answer line first. Correct answer of 0.42 (m/s) = 3 marks.
Otherwise:
1 mark: correct equation, wave speed = frequency × wavelength (AO1.2) OR conversion of 3.5 cm to 0.035 m.
1 mark: correct substitution, 12 × 0.035 (AO2.1) (ALLOW 12 × 3.5 = 42 for this mark).
1 mark: answer 0.42 with unit m/s (AO2.1).
ALLOW 42 cm/s for full marks. If 42 is given with m/s, award 2 marks maximum (unit conversion error). DO NOT ALLOW a final mark without a correct unit. ECF from an incorrect conversion.

\( E_p = m \times g \times h = 2.5 \times 10 \times 1.8 = 45 \) J.

Check the answer line first. Correct answer of 45 (J) = 3 marks.
Otherwise:
1 mark: correct equation, gravitational potential energy = mass × gravitational field strength × height (AO1.2).
1 mark: correct substitution, 2.5 × 10 × 1.8 (AO2.1).
1 mark: 45 (AO2.1).
ALLOW 0.045 kJ. ALLOW answers in J without the unit written if the unit is printed on the answer line. DO NOT ALLOW use of g = 9.8 unless it is also consistent with the question (gives 44.1, award 2 marks).

Time \( t = 2.0 \times 60 = 120 \) s.
\( Q = I \times t = 0.25 \times 120 = 30 \) C.

Check the answer line first. Correct answer of 30 (C) = 3 marks.
Otherwise:
1 mark: correct equation, charge = current × time (AO1.2).
1 mark: converts 2.0 minutes to 120 s (AO2.1).
1 mark: 0.25 × 120 = 30 (AO2.1).
If time is not converted, 0.25 × 2.0 = 0.5 scores 1 mark only (equation). ECF for a wrong conversion of time. IGNORE a unit written as coulombs, ALLOW 'C'.

Change in speed \( = 22 - 4.0 = 18 \) m/s.
\( a = \dfrac{\Delta v}{t} = \dfrac{18}{9.0} = 2.0 \) m/s².

Check the answer line first. Correct answer of 2.0 (m/s²) = 3 marks.
Otherwise:
1 mark: change in speed calculated as 18 m/s (AO2.1).
1 mark: correct equation, acceleration = change in speed ÷ time taken, with substitution 18 ÷ 9.0 (AO1.2 / AO2.1).
1 mark: 2.0 (AO2.1).
ALLOW 2 for the final mark. If 22 ÷ 9.0 = 2.4 is used (no subtraction), award 1 mark maximum for the equation. ECF from an incorrect change in speed.

Resultant force \( = 2700 - 600 = 2100 \) N.
\( a = \dfrac{F}{m} = \dfrac{2100}{1500} = 1.4 \) m/s².

Check the answer line first. Correct answer of 1.4 (m/s²) = 3 marks.
Otherwise:
1 mark: resultant force calculated as 2100 N (AO2.1).
1 mark: correct equation, force = mass × acceleration, rearranged or substituted as 2100 ÷ 1500 (AO1.2 / AO2.1).
1 mark: 1.4 (AO2.1).
ALLOW 1.40. If 2700 ÷ 1500 = 1.8 is used (resistive force ignored), award 1 mark maximum. ECF from an incorrect resultant force.

For reflection at a plane mirror, the angle of reflection equals the angle of incidence. The angle of incidence is 35°, so the reflected ray is drawn from the point of incidence on the opposite side of the normal from the incident ray, at 35° to the normal. It is a straight line and travels away from the mirror.

Reflected ray drawn as a straight line from the point of incidence, on the opposite side of the normal to the incident ray, at 35° to the normal (AO2.1) [1]. ALLOW ±2° tolerance. ALLOW an arrow on the ray pointing away from the mirror. IGNORE the angle label if the ray is correct. DO NOT ALLOW a ray measured from the mirror surface (55° to the normal) or a ray that does not start at the point of incidence.

From Table 1, a force of 5.0 N gives an extension of 10.0 cm. Find 10.0 cm on the horizontal axis and 5.0 N on the vertical axis, and mark a cross where they meet.

Point plotted at (10.0 cm, 5.0 N) (AO2.1) [1]. ALLOW a cross or a dot, within half a small square of the correct position. DO NOT ALLOW a point with the axes reversed, for example (5.0 cm, 10.0 N). IGNORE a line of best fit drawn through the points.

Read across from the 6 s column in the table to the speed row: 1.8 m/s.

| Question | Expected response | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | 1.8 (m/s) | 1 | AO2.1 | Check the answer line first. ALLOW 1.80. DO NOT ALLOW 1.8 with a wrong unit, e.g. 1.8 km/h. |

Each small division is \( 1 \div 5 = 0.2 \) V. Three small divisions after 4 V is \( 4 + 3 \times 0.2 = 4.6 \) V.

| Question | Expected response | Marks | AO | Guidance |
|---|---|---|---|---|
| 2 | 4.6 (V) | 1 | AO2.1 | Check the answer line first. ALLOW 4.60. DO NOT ALLOW 4.3 or 4.06. |

Wavelength: \( 9.0 \div 5 = 1.8 \) cm \( = 0.018 \) m.
Frequency: \( f = 6.0 \) Hz.
Wave speed = frequency \( \times \) wavelength: \( v = f \lambda = 6.0 \times 0.018 = 0.108 \) m/s.

Check the answer line first. Correct answer of 0.108 (m/s) or 0.11 (m/s) with no working = 4 marks.

Otherwise award:
1. Wavelength found: \( 9.0 \div 5 = 1.8 \) cm (method mark). DO NOT ALLOW 9.0 cm used as the wavelength.
2. Conversion of cm to m: 1.8 cm = 0.018 m (method mark). ALLOW ECF for the wavelength from mark 1.
3. Substitution into \( v = f \lambda \): \( 6.0 \times 0.018 \) (method mark). ALLOW ECF.
4. Answer 0.108 (or 0.11) with unit m/s (accuracy mark). The unit is required for this mark. ALLOW ECF from marks 1 to 3.

Common error: not converting to metres gives 10.8 cm/s. This scores marks 1 and 3 only (2 marks) because the answer is not in m/s.

AO2.1 (2 marks), AO2.2 (2 marks).

Distance travelled in the first 4 s = speed × time = 3 × 4 = 12 m, so the first line goes from (0, 0) to (4, 12). The line is straight because the speed is constant. When the car is stationary its distance does not change, so the graph is horizontal at 12 m from 4 s to 6 s.

Mark 1: straight line drawn from the origin (0, 0) to (4 s, 12 m) with the end point within half a small square. [1]
Mark 2: horizontal straight line from (4 s, 12 m) to (6 s, 12 m). [1]
ALLOW the second mark by ECF if the first line ends at a wrong point but the horizontal line continues from that point to 6 s.
IGNORE thickness of the line if the position is clear.
DO NOT ALLOW a curved line or a line drawn without a ruler that is clearly not straight.
DO NOT ALLOW a line that goes down after 4 s.
AO2.1 / AO3.2a (M4a).

Distance from the first crest to the sixth crest = 13.0 − 3.0 = 10.0 cm.
There are 5 wavelengths between the first and sixth crests.
Wavelength = \( 10.0 \div 5 \) = 2.0 cm.

Check the answer line first. Correct answer of 2.0 cm = 2 marks.
If the answer is incorrect:
1 mark: finds the distance 10.0 cm OR divides a distance between crests by 5 (method mark).
1 mark: wavelength = 2.0 cm (accuracy mark).
ALLOW 2 (cm) without the decimal.
ALLOW 0.020 m if the unit is given.
DO NOT ALLOW 1.7 cm (dividing 10.0 by 6) for the accuracy mark; this scores the method mark only if 10.0 cm is shown.
IGNORE a missing unit if cm is printed on the answer line.

Change in speed = 4.5 − 1.5 = 3.0 m/s.
Time taken = 6 − 2 = 4 s.
Acceleration = \( 3.0 \div 4 \) = 0.75 m/s².

Check the answer line first. Correct answer of 0.75 m/s² = 2 marks.
If the answer is incorrect:
1 mark: change in speed 3.0 m/s and time 4 s identified, or \( 3.0 \div 4 \) shown (method mark).
1 mark: 0.75 (accuracy mark).
ALLOW 0.8 (to 1 significant figure).
ALLOW 0.75 without a unit if m/s² is printed on the answer line.
ALLOW ECF for the accuracy mark if one value is misread from the table but the method is correct.
DO NOT ALLOW 1.5 m/s² (using 3.0 ÷ 2) or 0.5 (using 3.0 ÷ 6).

The distance between two neighbouring crests is the wavelength. The number of complete waves passing a point per second is the frequency, measured in hertz.

| Question | Expected response | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 (first blank) | wavelength | 1 | AO1.1 | DO NOT ALLOW any other word from the list |
| 1 (second blank) | frequency | 1 | AO1.1 | ALLOW Hz as an indication only if 'frequency' is not written; DO NOT ALLOW period |

Total: 2 marks. Award each mark independently.

Light is an electromagnetic wave, and the oscillations are at right angles to the direction of travel, so it is transverse. Sound is made of compressions and rarefactions along the direction of travel, so it is longitudinal. Water ripples move the surface up and down at right angles to the direction the wave travels, so they are transverse.

| Question | Expected response | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | Light: transverse (1) | 1 | AO1.1 | Award only if one box is ticked in the row. |
| 1 | Sound: longitudinal (1) | 1 | AO1.1 | DO NOT ALLOW if both boxes are ticked in the row. |
| 1 | Ripples on water: transverse (1) | 1 | AO1.1 | Total 3 marks. Each row is marked independently. |

As the ball rises it slows down, so its kinetic energy (\( \frac{1}{2}mv^2 \)) decreases. Its height increases, so its gravitational potential energy (\( mgh \)) increases. With no air resistance, no energy is dissipated, so the total stays the same. Energy is transferred from the kinetic store to the gravitational store.

| Question | Expected response | Marks | AO | Guidance |
|---|---|---|---|---|
| 2 | Kinetic energy: decreases (1) | 1 | AO1.1 | Award only if one box is ticked in the row. |
| 2 | Gravitational potential energy: increases (1) | 1 | AO1.1 | DO NOT ALLOW if more than one box is ticked in the row. |
| 2 | Total energy: stays the same (1) | 1 | AO2.1 | Total 3 marks. Each row is marked independently. |

wave speed = frequency × wavelength
\(v = 5.0 \times 0.030 = 0.15\) m/s

Tick in box 0.15 m/s = 2 marks.
If the wrong box is ticked, award 1 mark only if working shows correct substitution \(5.0 \times 0.030\).
DO NOT ALLOW more than one box ticked.

gravitational potential energy = mass × gravitational field strength × height
\(E = 4.0 \times 10 \times 2.5 = 100\) J

Tick in box 100 J = 2 marks.
If the wrong box is ticked, award 1 mark only if working shows \(4.0 \times 10 \times 2.5\).
DO NOT ALLOW more than one box ticked.

charge = current × time
\(Q = 0.40 \times 30 = 12\) C

Tick in box 12 C = 2 marks.
If the wrong box is ticked, award 1 mark only if working shows \(0.40 \times 30\).
DO NOT ALLOW more than one box ticked.

speed = distance ÷ time
\(v = 90 \div 15 = 6\) m/s

Tick in box 6 m/s = 2 marks.
If the wrong box is ticked, award 1 mark only if working shows \(90 \div 15\).
DO NOT ALLOW more than one box ticked.

Alpha particles have the lowest penetration and are stopped by paper. Gamma radiation is part of the electromagnetic spectrum.

Row 1: Alpha ticked = 1 mark.
Row 2: Gamma ticked = 1 mark.
DO NOT ALLOW more than one box ticked in a row.

The field of a solenoid is strengthened by increasing the current and by adding more turns of wire. An iron core (not plastic) also strengthens it.

Tick 'Increase the current in the wire' = 1 mark.
Tick 'Use more turns of wire' = 1 mark.
If more than two boxes are ticked, deduct 1 mark for each extra incorrect tick (minimum 0).

Use wave speed = frequency × wavelength.
\( v = 5.0 \times 0.040 = 0.20 \) m/s.

Q1 | Tick in box 0.20 m/s only | 1 mark | AO2.1 | DO NOT ALLOW more than one box ticked. The distractor 125 m/s comes from dividing frequency by wavelength; 0.008 m/s from dividing wavelength by frequency.

Use kinetic energy = ½ × mass × speed².
\( E_k = \tfrac{1}{2} \times 2.0 \times 3.0^2 = \tfrac{1}{2} \times 2.0 \times 9.0 = 9.0 \) J.

Q2 | Tick in box 9.0 J only | 1 mark | AO2.1 | DO NOT ALLOW more than one box ticked. The distractor 18 J comes from omitting the ½; 6.0 J from squaring incorrectly (using mass × speed).

The spring coils vibrate parallel to the direction in which the energy travels, so the wave is longitudinal.

Longitudinal (1 mark). ALLOW compression wave. DO NOT ALLOW transverse.

\( v = f \lambda = 5.0 \times 0.040 = 0.20 \) m/s

0.20 m/s (1 mark; answer and unit both needed). ALLOW 0.2 m/s. ALLOW 20 cm/s. DO NOT ALLOW the correct value with a wrong or missing unit.

The brick is raised against gravity, so its gravitational energy store increases.

Gravitational (potential) (1 mark). ALLOW GPE. IGNORE kinetic. DO NOT ALLOW elastic or thermal.

Frequency \( f = 8.0 \) Hz. Wavelength \( \lambda = 2.5 \text{ cm} = 0.025 \text{ m} \).
\( v = f \times \lambda = 8.0 \times 0.025 = 0.20 \) m/s.

Check the answer line first. Correct answer of 0.20 (m/s) = 3 marks.
Otherwise:
1. Correct equation, wave speed = frequency × wavelength (or \( v = f\lambda \)) [1] (AO1.2)
2. Substitution with the wavelength converted to metres, \( 8.0 \times 0.025 \) [1] (AO2.1)
3. Answer 0.20 with unit m/s [1] (AO2.1)
ALLOW 20 cm/s for full marks.
ALLOW 1 mark for 20 (m/s) or 200 with unit error from not converting cm to m (ECF for substitution mark: \( 8.0 \times 2.5 \)).
DO NOT ALLOW the final mark if the unit is missing or wrong.

\( E_k = \tfrac{1}{2} m v^2 = 0.5 \times 72 \times 6.5^2 \)
\( = 0.5 \times 72 \times 42.25 = 1521 \) J
To 3 significant figures: 1520 J.

Check the answer line first. Correct answer of 1520 (J) = 3 marks.
Otherwise:
1. Correct equation, kinetic energy = ½ × mass × speed² [1] (AO1.2)
2. Correct substitution, \( 0.5 \times 72 \times 6.5^2 \) [1] (AO2.1)
3. Evaluation 1521 or 1520 [1] (AO2.1)
ALLOW 1.52 × 10³ J.
ALLOW 1500 J (2 s.f.) or 1521 for full marks.
If the speed is not squared (\( 0.5 \times 72 \times 6.5 = 234 \)), award 1 mark for the equation only.

\( V = I \times R \), so \( R = \dfrac{V}{I} = \dfrac{9.0}{0.60} = 15 \) ohms.

Check the answer line first. Correct answer of 15 (ohms) = 3 marks.
Otherwise:
1. Correct equation or rearrangement, \( R = V \div I \) (or potential difference = current × resistance) [1] (AO1.2)
2. Correct substitution, \( 9.0 \div 0.60 \) [1] (AO2.1)
3. Answer 15 with unit ohms (or Ω) [1] (AO2.1)
ALLOW 1 mark for the equation \( V = IR \) followed by an incorrect rearrangement, with ECF on the arithmetic.
DO NOT ALLOW the final mark without a correct unit.

Distance between neighbouring crests = wavelength = 0.025 m.
\( v = f \times \lambda \)
\( v = 8.0 \times 0.025 = 0.20 \) m/s

Check the answer line first. Answer 0.20 (m/s) = 2 marks.
If the answer is incorrect, award 1 mark for correct substitution: \( 8.0 \times 0.025 \) (OR \( v = f \times \lambda \) with correct values).
Mark 1 (method): correct equation used or values correctly substituted. Mark 2 (accuracy): 0.20 m/s.
ALLOW 0.2 m/s. ALLOW 20 cm/s. ALLOW 1 mark for 0.2 with a wrong or missing unit if working shown (ECF for the numerical value).
DO NOT ALLOW 320 (from dividing 8.0 by 0.025) for the second mark; this scores 0 unless the substitution mark is earned by a correct substitution.
AO2.1 (2 marks).

(a) Distance for 5 waves = 36 cm = 0.36 m, so wavelength = \(0.36 \div 5 = 0.072\) m.
(b) wave speed = frequency × wavelength = \(6.0 \times 0.072 = 0.432\) m/s.

(a) Divides 36 by 5 (7.2 cm) [1]; converts to 0.072 m [1]. Correct answer of 0.072 m on the answer line gains both marks. ALLOW 0.07 m. DO NOT ALLOW 0.72 m for the second mark.
(b) Substitution \(6.0 \times 0.072\) (ECF from (a)) [1]; 0.432 (m/s) [1]. ALLOW 0.43 or 0.4. Correct answer gains both marks. IGNORE missing unit if given on the line.

(a) \(E = mgh = 12 \times 10 \times 1.5 = 180\) J.
(b) \(P = E \div t = 180 \div 3.0 = 60\) W.

(a) Substitution \(12 \times 10 \times 1.5\) [1]; 180 (J) [1]. Correct answer gains both marks.
(b) \(180 \div 3.0\) (ECF from (a)) [1]; 60 (W) [1]. Correct answer gains both marks. ALLOW J/s. DO NOT ALLOW wrong unit for the final mark.

(a) time = 2.0 × 60 = 120 s; \(Q = It = 0.50 \times 120 = 60\) C.
(b) \(V = IR = 0.50 \times 18 = 9.0\) V.

(a) Converts time to 120 s [1]; 60 (C) [1]. Correct answer gains both marks. Using 2.0 s gives 1.0 C: award 1 mark (equation used correctly).
(b) Substitution \(0.50 \times 18\) [1]; 9.0 (V) [1]. Correct answer gains both marks. ALLOW 9.

(a) \(a = (20 - 8.0) \div 6.0 = 2.0\) m/s².
(b) \(20^2 - 8.0^2 = 2 \times 2.0 \times s\), so \(336 = 4.0s\) and \(s = 84\) m.

(a) Change in speed 12 [1]; 2.0 (m/s²) [1]. Correct answer gains both marks.
(b) Substitution \(400 - 64 = 2 \times 2.0 \times s\) (ECF from (a)) [1]; 84 (m) [1]. Correct answer gains both marks. ALLOW alternative method using average speed 14 m/s × 6.0 s = 84 m.

Wave speed is found from \(v = f \lambda\). Wavelength: use the photo or screen to measure the distance across 5 or 10 bright lines with the ruler, then divide by the number of gaps to get the mean wavelength (in metres). Frequency: count the number of waves passing a fixed point in a set time (such as 10 s) using the stopwatch, or read the motor setting, then divide by the time. Multiply frequency by wavelength. Repeat and take a mean to improve reliability.

Level 3 (5-6 marks): A complete method that covers measuring wavelength, finding frequency and using the equation. Steps are in a logical order, and the reason for measuring several waves is given. Scientific terms are used correctly.
Level 2 (3-4 marks): Describes how to find either wavelength or frequency in some detail, and mentions the other or the equation, but with gaps in detail or order.
Level 1 (1-2 marks): Gives one or two relevant points, such as using a ruler on the screen or stating wave speed = frequency × wavelength, with little or no method.
Level 0 (0 marks): No relevant content.

Indicative content:
- Use the photo or screen to measure across several wavelengths (for example 10) with the ruler, then divide by the number of waves to get one wavelength.
- Use the stopwatch to count the waves that pass a point in a measured time, then divide to get frequency in Hz. ALLOW using the dipper frequency from the motor.
- Use wave speed = frequency × wavelength (\(v = f\lambda\)).
- Repeat the measurements and calculate a mean.
- Measuring several wavelengths reduces the percentage error. ALLOW a lamp or a slow-motion or frozen image to make the waves easier to see.
- Convert units (cm to m) before calculating.
IGNORE safety points about water and electricity. DO NOT ALLOW measuring the speed with a stopwatch over the tank as the only method unless the distance is given. AO2.1 (2 marks), AO3.3a (2 marks), AO1.2 (2 marks).

Energy transfer: the gravitational potential energy store of the child decreases and the kinetic energy store increases. With no friction, all the lost gravitational energy becomes kinetic energy.

Gravitational potential energy lost: \(E = mgh = 30 \times 10 \times 2.5 = 750\) J.

Kinetic energy at the bottom = 750 J, so \(\frac{1}{2} m v^2 = 750\).
\(\frac{1}{2} \times 30 \times v^2 = 750\), so \(15 v^2 = 750\), \(v^2 = 50\).
\(v = \sqrt{50} = 7.07\) m/s, which is 7.1 m/s to 2 significant figures.

Check the answer line first. 7.1 (m/s) (ALLOW 7.07, 7) with a correct description gains at most Level 3 if the description is complete.

Level 3 (5-6 marks): Describes the transfer from gravitational potential to kinetic energy, with the idea that all the energy is transferred as there is no friction. The calculation is fully correct with correct equations, substitution and answer. Logical order.
Level 2 (3-4 marks): Describes the energy transfer and makes a correct start on the calculation (for example 750 J found), or gives a correct calculation with an incomplete description.
Level 1 (1-2 marks): One relevant point, such as the energy stores named or one equation correctly used.
Level 0 (0 marks): No relevant content.

Indicative content:
- Gravitational potential energy store decreases; kinetic energy store increases. ALLOW energy transferred by working or mechanically.
- No energy is dissipated, so loss in GPE = gain in KE.
- \(E = mgh = 30 \times 10 \times 2.5 = 750\) J.
- \(E_k = \frac{1}{2}mv^2\), so \(v^2 = 750 \div 15 = 50\).
- \(v = 7.1\) m/s.
ECF from an incorrect energy value. DO NOT ALLOW a unit error for the final mark (the unit is given). IGNORE mention of thermal energy unless it is claimed that this is significant. AO1.2 (2 marks), AO2.1 (2 marks), AO2.2 (2 marks).

wave speed = frequency × wavelength
\( v = 4.0 \times 0.15 = 0.60 \) m/s

Check the answer line first. Correct answer 0.60 (m/s) = 2 marks.
Otherwise: substitution of 4.0 × 0.15 (or correct equation with values) = 1 mark.
ALLOW 0.6. ALLOW 60 cm/s only if the unit is given. DO NOT ALLOW 0.60 with an incorrect unit for the second mark. AO2.1.

kinetic energy = ½ × mass × speed²
\( E_k = 0.5 \times 1.5 \times 6.0^2 = 0.5 \times 1.5 \times 36 = 27 \) J

Check the answer line first. Correct answer 27 (J) = 2 marks.
Otherwise: correct substitution 0.5 × 1.5 × 6.0² = 1 mark.
If speed is not squared (answer 4.5 J) = 1 mark for the correct equation and substitution only if the substitution is shown.
ALLOW 27.0. IGNORE missing unit if J is printed on the answer line. AO2.1.

potential difference = current × resistance
\( V = 0.30 \times 20 = 6.0 \) V

Check the answer line first. Correct answer 6.0 (V) = 2 marks.
Otherwise: substitution 0.30 × 20 = 1 mark.
ALLOW 6. DO NOT ALLOW 66.7 (division) for the arithmetic mark. AO2.1.

change in speed = 14 − 2.0 = 12 m/s
acceleration = change in speed ÷ time = \( 12 \div 4.0 = 3.0 \) m/s²

Check the answer line first. Correct answer 3.0 m/s² = 2 marks (1 mark for value, 1 mark for unit).
Otherwise: change in speed 12 m/s, or 12 ÷ 4.0 = 1 mark.
ALLOW 3 m/s². ALLOW ECF for an incorrect change in speed (for example 14 ÷ 4.0 = 3.5) with the correct unit for 1 mark. DO NOT ALLOW m/s for the unit. AO2.1.

force = mass × acceleration, so acceleration = force ÷ mass
\( a = 3000 \div 1200 = 2.5 \) m/s²

Check the answer line first. Correct answer 2.5 (m/s²) = 2 marks.
Otherwise: correct rearrangement a = F ÷ m, or substitution 3000 ÷ 1200 = 1 mark.
ALLOW 2.50. DO NOT ALLOW 400 000 (multiplication). AO2.1.

The alpha particle removes 2 protons: 88 − 2 = 86 protons.
It removes 4 nucleons in total (2 protons + 2 neutrons): 226 − 4 = 222.

Number of protons is 86 = 1 mark.
Mass number is 222 = 1 mark.
ALLOW the answers in either order if clearly labelled. IGNORE the name radon. AO2.1.

The magnetic field of a solenoid is strengthened by increasing the current through it or by increasing the number of turns. An iron core strengthens the field further, so a better core also helps.

Any two from:
- increase the current OR add more cells OR use a higher potential difference (1)
- more turns of wire OR coil the wire more closely (1)
- use an iron core that is larger OR use a core made of a better magnetic material (1)
ALLOW a bigger battery. DO NOT ALLOW use a thicker nail unless linked to a larger iron core. IGNORE use a stronger magnet. AO1.1.

The wavelength is the distance between neighbouring crests.
\( 7.5 - 3.0 = 4.5 \) cm (check: \( 12.0 - 7.5 = 4.5 \) cm).

| Question | Expected response | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | 4.5 (cm) | 1 | AO2.1 | Check the answer line first. ALLOW 4.5 with unit cm. ALLOW 0.045 m if unit given. DO NOT ALLOW 9.0 (distance between crests 1 and 3) or 1.5 (4.5 ÷ 3). IGNORE missing unit if the answer line already shows cm. |

Each small division is 0.02 A. The pointer is 3 small divisions above 0.6 A.
\( 0.6 + 3 \times 0.02 = 0.66 \) A.

| Question | Expected response | Marks | AO | Guidance |
|---|---|---|---|---|
| 2 | 0.66 (A) | 1 | AO2.1 | Check the answer line first. ALLOW 0.66 with unit A. DO NOT ALLOW 0.63 (counting 0.01 A divisions) or 0.68. IGNORE missing unit if A is printed on the answer line. |

Wavelength is unchanged, so one full wave still spans 4 cm along the x-axis. Amplitude is doubled from 1 cm to 2 cm, so crests are at +2 cm and troughs at -2 cm. The wave crosses the axis at x = 0, 2 and 4 cm.

1 mark: a smooth wave drawn with wavelength 4 cm (one complete wave between x = 0 and x = 4 cm) AND amplitude 2 cm, starting at the origin and rising to a crest. ALLOW small tolerance of half a small square on crest and trough heights. DO NOT ALLOW straight-line (zigzag) waves or a wave with the wrong wavelength. AO2.1.

Distance in the first 10 s: \( 6 \times 10 = 60 \) m. So the first line goes from (0, 0) to (10, 60). While she is stopped the distance does not change, so the line is horizontal at 60 m from 10 s to 15 s.

1 mark: straight line from the origin to (10 s, 60 m) AND a horizontal straight line from (10 s, 60 m) to (15 s, 60 m). Both parts are needed for the mark. ALLOW tolerance of half a small square. DO NOT ALLOW a curved line. AO2.1, M4a.

Wasted energy = 200 − 20 = 180 J. The input of 200 J is 20 squares wide, so 10 J is one square. The wasted energy is \( 180 \div 10 = 18 \) squares wide.

1 mark: wasted energy arrow drawn leaving the main arrow, 18 squares wide (ALLOW 17.5 to 18.5 squares) AND labelled as thermal energy (OR heat OR wasted energy). ALLOW arrow drawn up or down. DO NOT ALLOW an arrow of the wrong width. AO2.1, M4a.

(a) Wave speed = frequency × wavelength, so \( v = f\lambda = 8.0 \times 0.015 = 0.12 \) m/s.
(b) Waves transfer energy (not matter) from the source to the edge of the tank.

Q(a) [2 marks, AO2.1]: Check the answer line first. Answer 0.12 (m/s) = 2 marks. Otherwise: correct substitution \( 8.0 \times 0.015 \) = 1 mark. ALLOW 1 mark for 0.12 with a wrong or missing unit (ECF on arithmetic only). ALLOW 1.2 × 10^-1. DO NOT ALLOW 0.12 with unit m/s² or similar for the unit mark.
Q(b) [1 mark, AO1.1]: energy. ALLOW wave energy OR kinetic energy. DO NOT ALLOW water OR matter OR particles OR force. IGNORE information, vibrations.

(a) Water ripples move up and down while the wave travels forward, so they are transverse. The crest-to-crest distance is the wavelength.

(b) Wave speed = frequency × wavelength. Frequency = 8.0 Hz. Wavelength = 2.5 cm = 0.025 m.
\( v = f \lambda = 8.0 \times 0.025 = 0.20 \) m/s.

(c) The speed is constant and \( v = f \lambda \). If the frequency rises, the wavelength must fall, so the crests are closer together.

Q(a) transverse AND wavelength (both needed) [1] AO1.1. DO NOT ALLOW if more than one word is given in a gap.

Q(b) Check the answer line first. Correct answer 0.20 (m/s) OR 0.2 = 2 marks. Otherwise: substitution of 8.0 and 0.025 (or 8.0 × 2.5 giving 20) into speed = frequency × wavelength [1]; correct evaluation with unit m/s, 0.20 [1]. ALLOW 20 cm/s for full marks. If the wavelength is not converted, 20 m/s scores 1 mark (ECF for substitution only). AO2.1 (method), AO2.2 (answer).

Q(c) Distance (between crests) decreases OR gets smaller OR wavelength decreases [1] AO2.1. IGNORE 'waves get closer' on its own if it is clearly about crests; DO NOT ALLOW 'waves get slower'.