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

The light slows down when it enters the glass, so it refracts towards the normal. It speeds up when it leaves, so it refracts away from the normal. Different colours (wavelengths) are refracted by different amounts, so the white light is dispersed into a spectrum. Red is refracted least and violet is refracted most. The diagram should show a triangular prism, the incident white ray, a bent ray inside the prism, and a fan of rays after the prism labelled red at one edge and violet at the other (red nearer the original direction of the beam).

Mark 1: ray bends towards the normal on entering the prism and away from the normal on leaving, with a correct overall deviation towards the thick base of the prism. ALLOW normals not drawn if the bending is clearly in the correct direction. IGNORE small reflected rays.

Mark 2: emerging light spread into a spectrum with red and violet labelled at the correct ends (red deviated least, violet deviated most). ALLOW 'R' and 'V' if clear. ALLOW labels for intermediate colours. DO NOT ALLOW the colours labelled the wrong way round.

Assessment objectives: AO1.1 (1 mark), AO1.2 (1 mark).

If the beam is shown bending the wrong way at both surfaces, award 0 for mark 1 but mark 2 may still be awarded.

The sledge moves at constant velocity, so by Newton's first law the resultant force is zero.
Vertically: weight (downwards, \( W = mg = 24 \times 10 = 240\) N) is balanced by the normal contact force from the snow (upwards).
Horizontally: the pull of the rope (to the right) is balanced by friction between sledge and snow (to the left).
So the two vertical arrows are equal in length and opposite, and the two horizontal arrows are equal in length and opposite.

Any three from the following (AO1.2 / AO2.1):
1. Four arrows drawn from the dot, in the correct directions: weight/gravity downwards, normal contact/reaction force upwards, pull/tension to the right, friction/drag to the left. (1 mark; award only if all four directions are correct and no extra forces are drawn.)
2. Each arrow labelled correctly. (1 mark) ALLOW 'weight', 'gravitational force', 'force of gravity'; 'normal contact force', 'normal reaction', 'reaction'; 'tension', 'pull', 'force from rope', 'applied force'; 'friction', 'resistive force'. DO NOT ALLOW 'upthrust', 'gravity' alone, 'push', 'air resistance' alone, 'force' alone.
3. Vertical arrows equal in length and horizontal arrows equal in length (balanced forces, resultant zero). (1 mark) ALLOW tolerance of about 2 mm. Horizontal arrows need not be the same length as the vertical ones.

IGNORE arrows not starting at the dot if all else is correct for marks 2 and 3, but mark 1 requires arrows starting at (or clearly acting through) the dot. Maximum 2 marks if a fifth force (e.g. 'motion' or 'momentum') is drawn.

1. At constant temperature the particle speed is unchanged. Halving the volume means the particles collide with the walls twice as often, so the pressure doubles (\( pV = \text{constant} \)). This is B. D is the distractor, as it describes a temperature rise.
2. Pressure in a liquid is given by \( p = \rho g h \). Greater depth \( h \) gives a greater pressure, because there is more water above the diver. This is C.
3. Pressure increases with depth, so the upward force on the lower surface of the bottle is greater than the downward force on the upper surface. The resultant force (upthrust) is upwards. This is A.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | B | 1 | AO1.1 | DO NOT ALLOW D |
| 2 | C | 1 | AO1.1 | |
| 3 | A | 1 | AO2.1 | |

Total 3 marks. Award one mark for each correct letter. If a letter is used more than once, award no mark for the repeated use unless it is the correct answer to that row. If two letters are given in one box, award no mark for that row.

Take the change in speed from 0.0 s to 2.0 s: \( 1.20 - 0.00 = 1.20\ \mathrm{m/s} \).
Time taken = 2.0 s.
\( a = \dfrac{1.20}{2.0} = 0.60\ \mathrm{m/s^2} \).
The speed rises by 0.30 m/s every 0.5 s, so the acceleration is constant and any pair of readings gives the same value.

| Question | Expected answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | Correct substitution of a pair of table values, e.g. \( \dfrac{1.20 - 0.00}{2.0} \) or \( \dfrac{0.30}{0.5} \) | 1 | AO2.1 (M1c, M3c, M4b) | ALLOW any valid pair of readings from the table. Change in speed and time interval must correspond. DO NOT ALLOW speed ÷ time using a time interval that does not match the speed change (e.g. 0.30 ÷ 1.0). |
| 1 | 0.60 m/s² | 1 | AO2.1 | Value 0.6 is acceptable. ALLOW m/s/s or m s⁻². DO NOT ALLOW 0.60 with an incorrect or missing unit. Correct answer with no working gains 2 marks. Unit must be correct for the second mark. |

(i) \(100 \times 60 = 6000\), \(150 \times 40 = 6000\), \(200 \times 30 = 6000\), \(250 \times 24 = 6000\). The product is the same each time, so \(pV\) is constant.

(ii) \(pV = \text{constant}\), so \(400 \times V = 6000\). \(V = 6000 \div 400 = 15\) cm\(^3\).

| Question | Expected answer | Marks | AO | Guidance |
|---|---|---|---|---|
| (i) | Calculates pressure \(\times\) volume for at least two (ideally all) pairs of readings and shows the products are equal (6000 kPa cm\(^3\)) | 1 | AO3.1a | ALLOW a comparison of any two pairs, e.g. \(100 \times 60 = 150 \times 40\). DO NOT ALLOW a statement of constancy without calculated values. |
| (ii) | Correct substitution into \(p_1V_1 = p_2V_2\) or \(pV = 6000\), e.g. \(400 \times V = 6000\) (or \(100 \times 60 = 400 \times V\)) | 1 | AO2.1 | Method mark. ALLOW any pair of table values as the starting point. |
| (ii) | \(V = 15\) (cm\(^3\)) | 1 | AO2.1 | Accuracy mark. Unit is printed on the answer line. If the final answer is correct with no working, award both marks. ALLOW ECF from an incorrect constant in (i) only if used consistently. |

\(p = \rho g h\)
\(p = 1000 \times 9.8 \times 15\)
\(p = 147\,000\) Pa \(= 1.47 \times 10^{5}\) Pa

Question | Expected answer | Marks | AO | Guidance
1 | 147 000 (Pa) or \(1.47 \times 10^{5}\) (Pa) | 1 | AO2.1 | Correct substitution of 1000 × 9.8 × 15 with correct evaluation gives the mark. ALLOW 150 000 or \(1.5 \times 10^{5}\) (Pa) or 147 kPa. ALLOW 1.47 × 10^5 without unit if the answer line already prints Pa. DO NOT ALLOW an answer with a wrong power of ten or a wrong unit (e.g. N, N/m). If the answer is incorrect, there is no mark for working alone because this is a 1-mark question.

Use \( p = \rho g h \).

\( p = 1025 \times 9.8 \times 8.5 = 85\,382.5 \) Pa

To 2 significant figures: \( p = 8.5 \times 10^{4} \) Pa.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | Correct substitution: 1025 × 9.8 × 8.5 (M1) | 1 | AO2.1 | ALLOW correct equation p = ρgh followed by correct substitution. ALLOW substitution of 1.025 × 10^3. |
| | Answer: 85 000 Pa / 8.5 × 10^4 Pa (A1) | 1 | AO2.1 | ALLOW 85 382.5 or 85 380 or 85 400 seen with the final answer given to 2 s.f. for full marks; if the final answer is not given to 2 s.f. but unrounded value is correct, award 2 marks only if 2 s.f. value is also stated, otherwise 1 mark. ALLOW 85 kPa. DO NOT ALLOW wrong unit. Correct answer with no working scores 2 marks. |

Total: 2 marks. Method mark (substitution) 1; accuracy mark (evaluation and rounding) 1. No ECF applies (single-step calculation).

Use \(p = \rho g h\).
\(p = 1030 \times 9.8 \times 30 = 302\,820\) Pa.
To 2 significant figures: \(3.0 \times 10^{5}\) Pa.

Substitution into p = ρgh: 1030 × 9.8 × 30 (1 mark)
Evaluation: 302 820 (Pa) (1 mark)
Rounding to 2 s.f.: 3.0 × 10^5 (Pa) (1 mark)
ALLOW 300 000 Pa for the final mark. ALLOW 1 mark for the equation if the substitution is incorrect. Correct final answer with no working scores 3 marks. DO NOT ALLOW the final mark without a correct rounding of the unrounded value (ECF from the evaluation mark). Ignore units on the working lines.

Momentum before = \(0.80 \times 1.5 = 1.2\) kg m/s.
Momentum is conserved, so momentum after = 1.2 kg m/s.
Total mass = 0.80 + 0.40 = 1.20 kg.
\(v = 1.2 \div 1.20 = 1.0\) m/s.

Momentum before collision = 0.80 × 1.5 = 1.2 (kg m/s) (1 mark)
Uses conservation of momentum with total mass 1.2 kg: v = 1.2 ÷ 1.2 (1 mark)
v = 1.0 (m/s) (1 mark)
Correct final answer with no working scores 3 marks. ALLOW ECF for the final mark from an incorrect momentum calculation. DO NOT ALLOW use of 0.80 kg alone as the mass after the collision (gives 1.5 m/s, 0 for the second and third marks).

Use \(F = BIl\).
\(F = 0.060 \times 4.5 \times 0.12 = 0.0324\) N.
To 2 significant figures: 0.032 N.

Substitution into F = BIl: 0.060 × 4.5 × 0.12 (1 mark)
Evaluation: 0.0324 (N) (1 mark)
Rounding to 2 s.f.: 0.032 (N) (1 mark)
ALLOW 3.2 × 10^-2 N. Correct final answer with no working scores 3 marks. ALLOW ECF for rounding of an incorrect evaluation. DO NOT ALLOW answers with a unit error in the final answer line, e.g. 32 N.

Time elapsed = 24 hours.
Number of half-lives = \(24 \div 6.0 = 4\).
Activity: 960 → 480 → 240 → 120 → 60 Bq.

Number of half-lives = 24 ÷ 6.0 = 4 (1 mark)
Activity halved the correct number of times, e.g. 960 ÷ 2^4 or 960 ÷ 16, or successive halving shown (1 mark)
60 (Bq) (1 mark)
Correct final answer with no working scores 3 marks. ALLOW ECF from an incorrect number of half-lives for the second and third marks if the halving is done correctly.

Distance travelled by the pulse = speed × time = \(1500 \times 0.42 = 630\) m.
This is the journey down and back, so depth = \(630 \div 2 = 315\) m.

Distance = speed × time: 1500 × 0.42 (1 mark)
= 630 (m) (1 mark)
Divides by 2 because the pulse travels to the sea bed and back: 315 (m) (1 mark)
Correct final answer with no working scores 3 marks. ALLOW 320 m (2 s.f.). ALLOW 1 mark for 630 m with no halving, i.e. the first two marks only.

(a) \(p = \rho g h = 1000 \times 9.8 \times 18 = 176\,400\) Pa.
(b) \(F = pA = 176\,400 \times 0.0020 = 352.8\) N, which is 350 N to 2 significant figures.

(a) 1 mark: substitution of 1000, 9.8 and 18 into \(p=\rho g h\) (equation may be implied). 1 mark: 176 400 Pa. ALLOW 1.8 × 10^5 Pa or 176 000 Pa. Correct answer with no working gains 2 marks. (b) 1 mark: \(F = pA\) with substitution of (a) and 0.0020 (ECF from (a)). 1 mark: 350 N (352.8 N rounded to 2 s.f.). ALLOW 3.5 × 10^2 N. DO NOT ALLOW unrounded 352.8 N for the final mark. Unit mark is not separately awarded; unit must be correct (Pa, N) or the final mark in each part is withheld.

(a) Momentum before = \(0.80 \times 1.5 = 1.2\) kg m/s. Momentum is conserved, so \(1.2 = (0.80+1.2)\,v\), giving \(v = 0.60\) m/s.
(b) Change in momentum of the stationary trolley = \(1.2 \times 0.60 = 0.72\) kg m/s. \(F = \Delta p / t = 0.72 / 0.15 = 4.8\) N.

(a) 1 mark: total momentum before = 1.2 kg m/s (0.80 × 1.5), or equating momentum before and after with total mass 2.0 kg. 1 mark: 0.60 m/s. Correct answer with no working gains 2 marks. (b) 1 mark: change in momentum of the 1.2 kg trolley = 0.72 kg m/s (1.2 × 0.60; ECF from (a)). 1 mark: \(F = 0.72/0.15 = 4.8\) N. ALLOW ECF for a wrong velocity from (a). DO NOT ALLOW use of the 0.80 kg trolley's momentum change unless the candidate shows the same value (0.80 × 0.9 = 0.72 is acceptable).

(a) \(F = BIl = 0.25 \times 3.2 \times 0.060 = 0.048\) N.
(b) \(I = F/(Bl) = 0.090 / (0.25 \times 0.060) = 6.0\) A.

(a) 1 mark: substitution into \(F = BIl\) (0.25 × 3.2 × 0.060). 1 mark: 0.048 N. ALLOW 4.8 × 10^-2 N. (b) 1 mark: correct rearrangement \(I = F/(Bl)\) or substitution 0.090 / (0.25 × 0.060). 1 mark: 6.0 A. ALLOW 6 A. Correct answer with no working gains full marks for the part. Do not ECF from (a) in (b), as the parts are independent.

Pressure in a fluid depends on depth, density and gravitational field strength (\( p = \rho g h \)). As depth increases, more water lies above the diver, so the pressure increases. Pressure in a fluid acts in all directions, and the width of the lake is irrelevant.

1 mark: first box ticked only.
DO NOT ALLOW more than one box ticked.
AO1.1

An interaction pair acts on two different objects, is of the same type and has equal magnitude in opposite directions. The Earth pulls the book, so the pair is the book pulling the Earth with a gravitational force. The table's contact force is a different type of force and merely balances the weight.

1 mark: second box ticked only.
DO NOT ALLOW more than one box ticked.
AO1.2

\( p = mv = 1500 \times 12 = 18\,000 \) kg m/s.

1 mark: third box ticked only.
DO NOT ALLOW more than one box ticked.
AO2.1

An alpha particle is 2 protons and 2 neutrons (helium nucleus). The mass number falls by 4, the proton number falls by 2 (not 4), and the neutron number falls by 2.

1 mark: all three rows correct (True, False, True).
DO NOT ALLOW the mark if any row has both or neither box ticked.
AO1.1

\( F = BIl \), so force is directly proportional to current. Doubling the current from 2.0 A to 4.0 A doubles the force. Reversing the current reverses the force direction only; halving the length or flux density halves the force.

1 mark: first box ticked only.
DO NOT ALLOW more than one box ticked.
AO2.1

Pressure in a fluid is given by \( p = \rho g h \). Pressure therefore increases with depth (true) and with density (sea water is denser than fresh water, so true). Fluid pressure acts normal to any surface in all directions, not only downwards (false). The pressure depends on depth, density and gravitational field strength, not on the width of the lake (false).

All four rows correct = 2 marks (AO1.1). Three rows correct = 1 mark. Two or fewer rows correct = 0 marks.
Correct answers: row 1 True; row 2 True; row 3 False; row 4 False.
If both boxes are ticked in a row, or neither, that row is incorrect.

Isotopes differ only in neutron number, so the proton number is the same (true). In beta decay a neutron changes to a proton, so the mass number stays the same (false). An alpha particle contains 2 protons and 2 neutrons, so the proton number falls by 2 (true). After two half-lives the fraction remaining is \( \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4} \), not one half (false).

All four rows correct = 2 marks (AO1.1). Three rows correct = 1 mark. Two or fewer rows correct = 0 marks.
Correct answers: row 1 True; row 2 False; row 3 True; row 4 False.
If both boxes are ticked in a row, or neither, that row is incorrect.

The temperature is constant, so the average particle speed is unchanged. A smaller volume means particles hit a given area of wall more often, so the pressure rises (consistent with \(pV = \text{constant}\)).

Sentence 1: 'stays the same' (1 mark).
Sentence 2: 'increases' for collisions AND 'increases' for pressure, both needed (1 mark).
DO NOT ALLOW the second mark if only one of the two blanks in the second sentence is correct.
IGNORE any extra words written in the blanks if the chosen list word is clear.

An alpha particle has mass number 4 and proton number 2, so the new nucleus has mass number \(222 - 4 = 218\) and proton number \(86 - 2 = 84\). In beta-minus decay a neutron becomes a proton, so the mass number is unchanged at 14 and the proton number rises by 1 to 7.

Decay 1: 218 and 84 both correct (1 mark).
Decay 2: 14 and 7 both correct (1 mark).
No mark if only one number in a row is correct.
ALLOW answers in nuclide notation if the numbers are in the correct positions.

\(p = \rho g h = 1000 \times 10 \times 4.0 = 40\,000\) Pa

40 000 (Pa) [1 mark, accuracy for correct substitution and evaluation]. ALLOW 4 × 10^4 Pa. ALLOW 1 mark for 40 kPa. DO NOT ALLOW the answer without a correct value; ignore omitted unit if the unit is printed on the answer line.

\(p = mv = 2.5 \times 1.6 = 4.0\) kg m/s

4.0 (kg m/s) [1 mark, correct substitution and evaluation]. ALLOW 4. IGNORE unit if printed on the answer line. DO NOT ALLOW answers from using weight in place of mass.

18 hours is 3 half-lives. \(800 \to 400 \to 200 \to 100\) Bq.

100 (Bq) [1 mark, for recognising 3 half-lives and halving correctly]. ALLOW 800 ÷ 2^3. IGNORE unit if printed on the answer line.

\(F = BIl = 0.040 \times 3.0 \times 0.25 = 0.030\) N

0.030 (N) [1 mark, correct substitution and evaluation]. ALLOW 0.03 or 3.0 × 10^-2. IGNORE unit if printed on the answer line.

\(p = F \div A = 240 \div 0.0060 = 40\,000\) Pa.

Substitution / correct evaluation of 240 ÷ 0.0060 = 1 mark. Answer 40 000 with correct unit Pa (or N/m\(^2\)) = 1 mark. ALLOW 4.0 × 10\(^4\) and 40 kPa. ALLOW N/m2. DO NOT ALLOW the unit mark for N/m alone. Answer without unit scores 1 mark. AO2.1.

At constant temperature \(pV\) is constant: \(100 \times 120 = p \times 40\), so \(p = 12\,000 \div 40 = 300\) kPa.

Correct substitution into \(p_1V_1 = p_2V_2\) (100 × 120 = p × 40) = 1 mark. 300 (kPa) = 1 mark. ALLOW 300 000 Pa. ALLOW 1 mark for 33 or 0.33 (inverted ratio). AO2.1.

Momentum before = \(0.50 \times 1.2 = 0.60\) kg m/s. Momentum is conserved, so \(0.60 = (0.50 + 0.30)v\) and \(v = 0.60 \div 0.80 = 0.75\) m/s.

Momentum before collision = 0.60 kg m/s, or \(0.50 \times 1.2 = (0.50+0.30)v\) = 1 mark. 0.75 (m/s) = 1 mark. ALLOW 1 mark for a correct momentum equation with total mass 0.80 kg used, if the final answer is wrong. AO2.1.

The pair of a gravitational force is of the same type, equal in size, opposite in direction and acts on the other object. The book pulls the Earth upwards with 8.0 N.

Acts on the Earth (the force the book exerts on the Earth) = 1 mark. 8.0 N and upwards (opposite direction) = 1 mark. ALLOW gravitational pull of book on Earth. DO NOT ALLOW the normal contact force of the table on the book (acts on the same object, different type). Both size and direction needed for the second mark. AO1.2.

24 hours is \(24 \div 6.0 = 4\) half-lives. \(480 \to 240 \to 120 \to 60 \to 30\).

4 half-lives identified (or halving sequence shown to at least 60) = 1 mark. 30 (counts per second) = 1 mark. ALLOW 1 mark for 60 (3 half-lives) or 15 (5 half-lives). AO2.1.

\(F = BIl = 0.040 \times 3.5 \times 0.12 = 0.0168\) N.

Correct substitution 0.040 × 3.5 × 0.12 = 1 mark. 0.0168 N = 1 mark. ALLOW 0.017 or 1.7 × 10\(^{-2}\). ALLOW 1 mark for a correct substitution with a power-of-ten error in the answer. AO2.1.

Distance for the round trip = \(1500 \times 0.080 = 120\) m. The depth is half of this: 60 m.

1500 × 0.080 = 120 m = 1 mark. Halved to give 60 (m) = 1 mark. ALLOW 1 mark for 120 m only. AO2.1.

\(v^2 - u^2 = 400 - 64 = 336\). \(a = 336 \div (2 \times 84) = 336 \div 168 = 2.0\) m/s\(^2\).

Correct substitution or rearrangement, e.g. \(20^2 - 8.0^2 = 2 \times a \times 84\) = 1 mark. 2.0 (m/s\(^2\)) = 1 mark. ALLOW 2. Using (20 − 8)² = 144 scores 0 for the first mark. AO2.1.

The deeper the diver, the more water lies above and its weight presses down on each unit of area. Pressure = \(\rho g h\), so it increases in proportion to depth.

Greater depth means more (a larger column of) water above the diver / greater weight of water above = 1 mark. This produces a greater force on each unit area (more particle collisions per unit area) so pressure is greater = 1 mark. ALLOW pressure = density × g × depth, so pressure increases with depth, for 1 mark only. IGNORE references to temperature. AO1.1, AO2.2.

\(p = \rho g h = 1030 \times 9.8 \times 15 = 151\,410\) Pa.
To 2 significant figures: \(1.5 \times 10^5\) Pa.

Marks (AO2.1 / M3b, M3c):
1. Correct substitution: 1030 \(\times\) 9.8 \(\times\) 15 (or 151 410 seen) [1 - method mark].
2. Final answer 150 000 Pa (1.5 \(\times\) 10\(^5\) Pa) given to 2 significant figures [1 - accuracy mark].

Guidance:
ALLOW 151 410 (or 151 000, 1.5141 \(\times\) 10\(^5\)) for 1 mark only if not rounded to 2 s.f.
ALLOW 1 mark for a correct unrounded answer in a different number of significant figures.
DO NOT ALLOW an answer without a unit or with a wrong unit for the second mark, unless the unit is printed on the answer line.
IGNORE any addition of atmospheric pressure; if added (about 2.5 \(\times\) 10\(^5\) Pa) award 1 mark for the correct substitution only.
Full marks for the correct final answer without working.

\(p = \rho g h\)
\(p = 1000 \times 9.8 \times 12 = 117\,600\) Pa
To 2 significant figures: \(1.2 \times 10^5\) Pa.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | Correct substitution: 1000 \(\times\) 9.8 \(\times\) 12 (1) | 1 | AO2.1 | Equation selection may be shown in words or symbols; ALLOW this mark if the equation is stated and substitution is correct. |
| | Evaluation: 117 600 (Pa) (1) | 1 | AO2.1 | ALLOW 117 000 to 118 000 if g = 9.8 used. |
| | Rounded correctly to 2 s.f.: \(1.2 \times 10^5\) (Pa) (1) | 1 | AO2.1 | ECF from an incorrect evaluation for rounding only. ALLOW 120 000 Pa. DO NOT ALLOW 1.2 without the power of ten or 12 \(\times\) 10^4 unless equivalent. Correct answer with no working scores 3. |

Momentum before = \(0.80 \times 1.5 = 1.2\) kg m/s (stationary trolley has zero momentum).
Momentum is conserved, so momentum after = 1.2 kg m/s.
Total mass = \(0.80 + 1.2 = 2.0\) kg.
\(v = 1.2 \div 2.0 = 0.60\) m/s.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 2 | Initial momentum = 0.80 \(\times\) 1.5 = 1.2 (kg m/s) (1) | 1 | AO2.1 | |
| | Momentum after = momentum before and total mass 2.0 kg used (1) | 1 | AO2.1 | Method mark for \(1.2 = 2.0 \times v\) or equivalent. |
| | v = 0.60 (m/s) (1) | 1 | AO2.1 | ALLOW 0.6. Unit m/s needed for the final mark. Correct answer with no working scores 3. ECF from an incorrect initial momentum. DO NOT ALLOW use of 0.80 kg or 1.2 kg alone as the mass after collision. |

Number of half-lives = \(12 \div 3.0 = 4\).
Activity: 640 \(\to\) 320 \(\to\) 160 \(\to\) 80 \(\to\) 40.
Activity after 12 hours = 40 counts per minute.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 3 | 4 half-lives (1) | 1 | AO2.1 | May be shown by 12 \(\div\) 3. |
| | Activity halved the correct number of times, or 640 \(\div 2^4\) (1) | 1 | AO2.1 | ALLOW 640 \(\div\) 16. ECF for halving the stated number of half-lives. |
| | 40 (counts per minute) (1) | 1 | AO2.1 | Correct answer with no working scores 3. ALLOW ECF from a wrong number of half-lives. |

\(F = BIl\)
\(F = 0.20 \times 3.0 \times 0.15 = 0.090\) N.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 4 | Correct substitution: 0.20 \(\times\) 3.0 \(\times\) 0.15 (1) | 1 | AO2.1 | Equation selection may be shown in words or symbols. |
| | 0.090 (1) | 1 | AO2.1 | ALLOW 0.09. |
| | Unit: N (newton) (1) | 1 | AO1.2 | Independent mark. DO NOT ALLOW 'n' or 'Nm'. Correct numerical answer with unit and no working scores 3. |

The block is at rest, so the forces on it are balanced and the resultant force is zero. The Earth pulls the block downwards with its weight. The bench pushes upwards on the block with a normal contact force. For the resultant to be zero these two forces must be equal in size, so the arrows have equal lengths and point in opposite directions.

| Question | Expected answer | Marks | AO | Guidance |
|---|---|---|---|---|
| Diagram | Downward vertical arrow labelled weight AND upward vertical arrow labelled normal contact force, both starting at the dot, arrows equal in length | 1 | AO1.2 | ALLOW 'gravity', 'force of gravity', 'W', 'mg' for weight. ALLOW 'reaction', 'normal reaction', 'support force', 'R', DO NOT ALLOW 'gravity' arrow drawn upwards. DO NOT ALLOW if the arrows are clearly unequal in length. DO NOT ALLOW 'friction' or 'push' as the upward label. IGNORE small differences in length if the arrows look equal. Both arrows and both labels are needed for the single mark. |

Pressure = density × g × depth, so pressure increases with depth. The bottom face of a submerged object is deeper than the top face, so the upward force on the bottom face is greater than the downward force on the top face. The resultant of these forces is the upthrust.

Cube B (fully submerged): pressure difference between faces = \(1000 \times 10 \times 0.10 = 1000\) Pa. Area of a face = \(0.10 \times 0.10 = 0.010\) m\(^2\). Upthrust = \(1000 \times 0.010 = 10\) N.
Volume = 0.0010 m\(^3\). Mass of B = \(1200 \times 0.0010 = 1.2\) kg. Weight of B = \(1.2 \times 10 = 12\) N. Weight (12 N) is greater than the maximum upthrust (10 N), so B sinks.

Cube A: mass = \(600 \times 0.0010 = 0.60\) kg, weight = 6.0 N. The maximum upthrust (10 N) exceeds the weight, so A rises until it floats. It floats when upthrust = weight = 6.0 N, which needs a pressure difference of \(6.0 / 0.010 = 600\) Pa, i.e. a submerged depth of \(600/(1000 \times 10) = 0.060\) m. This is less than 0.10 m, so the top face stays above the surface.

Level of response (6 marks).

Level 3 (5–6 marks): Clear, correct explanation that upthrust is due to pressure increasing with depth (force on bottom face greater than on top face), with correct calculations for both cubes (upthrust or pressure difference and weights), and a clear comparison of weight and upthrust leading to the conclusion for both A and B. The reasoning is logical and complete.

Level 2 (3–4 marks): Explains the origin of upthrust OR makes correct calculations for both cubes with a conclusion, but with some gaps in reasoning or missing links (e.g. does not explain why A floats at equilibrium, or calculation for only one cube plus a correct explanation).

Level 1 (1–2 marks): Some relevant statement, e.g. pressure increases with depth, or one correct calculation (weight or upthrust), with little or no connection to floating or sinking.

Level 0 (0 marks): No creditable response.

Indicative content (guide, not a checklist):
- pressure = \(\rho g h\); pressure increases with depth
- bottom face deeper so greater force upwards than downwards on top face; resultant is upthrust
- pressure difference for 0.10 m = 1000 Pa; force = pressure × area = 10 N (area 0.010 m\(^2\))
- mass = density × volume: B = 1.2 kg, A = 0.60 kg; weights 12 N and 6.0 N
- B: weight 12 N > upthrust 10 N so sinks
- A: weight 6.0 N < 10 N so rises; floats when upthrust = weight, submerged depth 0.060 m (< 0.10 m)

ALLOW ECF from an arithmetic slip. ALLOW upthrust = weight of water displaced (10 N) as an alternative route to the 10 N value. IGNORE general statements that A is 'less dense' without any supporting force or pressure argument (not enough for Level 3). DO NOT ALLOW 'upthrust is caused by water pushing up only'. DO NOT ALLOW use of gravity cancelling upthrust for B.

Thinking distance = speed × reaction time = \(20 \times 0.70 = 14\) m (the car moves at constant speed; same on both roads).

Dry road: deceleration \(a = F/m = 6000/1200 = 5.0\) m/s\(^2\). Using \(v^2 - u^2 = 2as\): \(0 - 20^2 = 2 \times (-5.0) \times s\), so braking distance \(s = 400/10 = 40\) m. Stopping distance = \(14 + 40 = 54\) m.

Wet road: \(a = 4000/1200 = 3.33\) m/s\(^2\). Braking distance \(= 400/(2 \times 3.33) = 60\) m. Stopping distance = \(14 + 60 = 74\) m.

The wet road gives a stopping distance 20 m longer. The smaller braking force gives a smaller deceleration, so the car takes longer and travels further while slowing from the same speed. (Check using work done: \(Fs = \tfrac{1}{2}mv^2 = 240\,000\) J, so \(s = 240\,000/6000 = 40\) m and \(240\,000/4000 = 60\) m.) Reaction time only changes thinking distance, which does not depend on the road surface; a longer reaction time would increase both stopping distances by the same amount.

Level of response (6 marks).

Level 3 (5–6 marks): Correct calculations of thinking and braking distances for both roads (total 54 m and 74 m), with a clear comparison, and a clear explanation of how both reaction time (thinking distance, same on both roads) and braking force (deceleration, braking distance) affect stopping distance. Logical and complete.

Level 2 (3–4 marks): Correct calculation of thinking distance and of braking distance for at least one road, with some explanation; OR full correct calculations with limited or no explanation of the effects of reaction time and braking force.

Level 1 (1–2 marks): One correct step (e.g. thinking distance = 14 m, or deceleration = 5.0 m/s\(^2\)) or a simple statement that wet roads give longer stopping distances, with little or no supporting calculation or reasoning.

Level 0 (0 marks): No creditable response.

Indicative content (guide, not a checklist):
- thinking distance = speed × time = 14 m (constant speed during reaction time; same on both roads)
- \(a = F/m\): 5.0 m/s\(^2\) (dry) and 3.3 m/s\(^2\) (wet)
- \(v^2 - u^2 = 2as\) gives braking distance 40 m (dry) and 60 m (wet); or work done: \(Fs = \tfrac{1}{2}mv^2\)
- total stopping distance: 54 m (dry), 74 m (wet); wet is 20 m longer
- smaller braking force means smaller deceleration so greater distance to lose the same kinetic energy
- reaction time affects only thinking distance; longer reaction time increases stopping distance equally on both roads

ALLOW answers to 2 significant figures or more; ALLOW 3.3 m/s\(^2\) or 3.33 m/s\(^2\); ALLOW ECF from an incorrect deceleration. ALLOW a kinetic energy / work done method for braking distance. IGNORE statements about tyre wear or car type. DO NOT ALLOW 'the car goes slower on wet roads' as an explanation.

\(p = \rho g h = 1000 \times 9.8 \times 3.0 = 29\,400\) Pa

| Question | Expected answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | 29 400 (Pa) | 1 | AO2.1 | ALLOW 29 000 or 2.94 × 10^4. ALLOW 30 000 (1 s.f.). DO NOT ALLOW an answer with no value substituted correctly, e.g. 2940. Correct answer alone gains the mark. |

Distance travelled = 11.0 − 3.0 = 8.0 m. Time taken = 6.0 − 2.0 = 4.0 s.

\(v = \dfrac{8.0}{4.0} = 2.0\) m/s

| Question | Expected answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 2 | 2.0 (m/s) | 1 | AO2.1 | Correct answer alone gains the mark. DO NOT ALLOW 2.75 (11 ÷ 4) or 1.83 (11 ÷ 6). ALLOW 2. |

The activity halves from 800 to 400 counts per second in 3 hours, and halves again in each further 3 hours. The half-life is 3 hours.

| Question | Expected answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 3 | 3 (hours) | 1 | AO3.1a | ALLOW 3.0. DO NOT ALLOW 9 or 6. |

\(v = f \lambda = 8.0 \times 1.5 = 12\) cm/s

| Question | Expected answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 4 | 12 (cm/s) | 1 | AO2.1 | ALLOW 12.0. ALLOW 0.12 if the unit m/s is clearly given. DO NOT ALLOW 12 with an incorrect unit. |

Use \(p = \rho g h\).
\(p = 1000 \times 9.8 \times 15 = 147\,000\) Pa (\(1.5 \times 10^5\) Pa to 2 s.f.).

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | substitution: \(1000 \times 9.8 \times 15\) (1); \(147\,000\) (Pa) (1) | 2 | AO2.1 | ALLOW 1.5 × 10^5 or 150 000 (Pa). ALLOW 1 mark for 147 or 147 kPa if unit error only. Correct answer with no working gets 2 marks. DO NOT ALLOW answers in N or N/m without pascal equivalence (N/m\(^2\) is allowed). |

Use \(F = BIl\).
\(F = 0.25 \times 3.2 \times 0.060 = 0.048\) N.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 2 | substitution: \(0.25 \times 3.2 \times 0.060\) (1); \(0.048\) (N) (1) | 2 | AO2.1 | ALLOW 4.8 × 10^-2 (N). ALLOW 1 mark for 48 if length not converted or unit error (e.g. mN without conversion). Correct answer with no working gets 2 marks. |

Change in momentum: \(\Delta p = mv = 1200 \times 20 = 24\,000\) kg m/s.
Use \(\Delta p = Ft\), so \(F = \frac{24\,000}{4.0} = 6000\) N.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 3 | change in momentum \(= 1200 \times 20 = 24\,000\) (1); \(F = 24\,000 \div 4.0 = 6000\) (N) (1) | 2 | AO2.1 | ALLOW alternative route: \(a = 20 \div 4.0 = 5.0\) m/s\(^2\) (1), \(F = 1200 \times 5.0 = 6000\) (N) (1). ALLOW 6.0 × 10^3 or 6 kN. Correct answer with no working gets 2 marks. ALLOW ECF for a wrong momentum value used correctly. |

Pressure: \(p = \rho g h = 1025 \times 9.8 \times 35 = 351\,575\) Pa.

Force: \(F = pA = 351\,575 \times 0.20 = 70\,315\) N.

To 2 s.f. the force is \(7.0 \times 10^{4}\) N.

Mark 1: correct substitution into p = ρgh (1025 × 9.8 × 35) or p = 351 575 Pa seen.
Mark 2: use of F = pA (pressure × 0.20) with ECF from the pressure value.
Mark 3: final answer 7.0 × 10^4 N (ALLOW 70 000 or 70 315 for the first two marks but the final mark requires 2 s.f.).
Full marks for the correct answer with no working.
DO NOT ALLOW use of F = pA with an unprocessed depth (e.g. 35 as the pressure).
IGNORE unit errors in working if the final unit is N.

Momentum before: \(p = mv = 2.4 \times 1.5 = 3.6\) kg m/s.

Momentum is conserved, so after the collision \(3.6 = (2.4 + 0.60) \times v\).

\(v = 3.6 \div 3.0 = 1.2\) m/s.

Mark 1: momentum before = 2.4 × 1.5 = 3.6 kg m/s.
Mark 2: total mass after = 3.0 kg and momentum after = momentum before (conservation stated or used).
Mark 3: v = 1.2 m/s.
Full marks for the correct answer with no working.
DO NOT ALLOW use of 2.4 + 0.60 as the mass before the collision (gives 0.8 m/s or similar).
ALLOW ECF for a wrong momentum in mark 1.

Rearrange: \(I = \dfrac{F}{Bl}\).

\(I = \dfrac{0.027}{0.15 \times 0.12} = \dfrac{0.027}{0.018} = 1.5\) A.

Mark 1: correct rearrangement I = F / (B l).
Mark 2: correct substitution 0.027 / (0.15 × 0.12).
Mark 3: I = 1.5 A.
Full marks for the correct answer with no working.
ALLOW 1 mark for 0.027 ÷ 0.15 ÷ 0.12 only if evaluated correctly as 1.5.
DO NOT ALLOW answers with an incorrect unit in the final answer for mark 3.

640 → 320 → 160 → 80 → 40 → 20 Bq.

This is 5 half-lives (640 ÷ 20 = 32 = 2^5).

Time = 5 × 6.0 = 30 hours.

Mark 1: activity halved at least once correctly (e.g. 320 or 160 seen) or 640 ÷ 20 = 32 seen.
Mark 2: 5 half-lives identified.
Mark 3: 30 hours.
Full marks for the correct answer with no working.
ALLOW ECF for mark 3 from a wrong number of half-lives × 6.0.
ALLOW 1.25 days.

Total distance travelled by the pulse: \(s = vt = 1500 \times 0.16 = 240\) m.

This is the distance down and back, so depth = \(240 \div 2 = 120\) m.

Mark 1: distance = speed × time = 1500 × 0.16 (or 240 m seen).
Mark 2: halving because the pulse travels to the seabed and back.
Mark 3: 120 m.
Full marks for the correct answer with no working.
ALLOW 1500 × 0.08 for marks 1 and 2.
If the halving is omitted, answer 240 m scores 1 mark only.

(i) \(p = \rho g h = 1030 \times 9.8 \times 35 = 353\,290\ \text{Pa}\).
(ii) \(F = pA = 353\,290 \times 0.020 = 7065.8\ \text{N} \approx 7100\ \text{N}\) (2 s.f.).

(i) Substitution into \(p = \rho g h\): \(1030 \times 9.8 \times 35\) [1]; \(353\,290\) Pa [1]. ALLOW \(3.5\times10^5\) Pa or 350 000 Pa. ALLOW 360 000 if g = 10 used with substitution shown (1 mark only for method).
(ii) Use of \(F = pA\) (or rearrangement from \(p = F/A\)) with \(353\,290 \times 0.020\) [1]; 7100 N (2 s.f.) [1]. ECF from (i). ALLOW 7066 or 7065.8 for the evaluation mark only if the rounding is also given; DO NOT ALLOW unrounded answer as the final mark if 2 s.f. is not given. Correct answer with no working = 4 marks.

(i) \(p = mv = 0.80 \times 1.5 + 0.60 \times 0 = 1.2\ \text{kg m/s}\).
(ii) Momentum is conserved, so \(1.2 = (0.80 + 0.60) \times v\), giving \(v = 1.2 \div 1.4 = 0.857\ \text{m/s} \approx 0.86\ \text{m/s}\).

(i) \(0.80 \times 1.5\) [1]; 1.2 kg m/s [1]. Trolley B contributes zero; do not penalise if omitted.
(ii) Total momentum after = total momentum before / \(1.2 = 1.4 v\) (total mass 1.4 kg used) [1]; 0.86 m/s [1]. ALLOW 0.857 if rounding to 2 s.f. is also shown. ECF from (i) for the final mark. DO NOT ALLOW use of 0.80 kg or 0.60 kg alone as the mass after collision. Correct answer with no working = 4 marks.

(i) \(F = BIl = 0.35 \times 4.5 \times 0.12 = 0.189\ \text{N}\).
(ii) \(I = F \div (Bl) = 0.30 \div (0.35 \times 0.12) = 0.30 \div 0.042 = 7.14\ \text{A} \approx 7.1\ \text{A}\).

(i) Substitution \(0.35 \times 4.5 \times 0.12\) [1]; 0.189 N [1]. ALLOW 0.19 N.
(ii) Rearrangement \(I = F/(Bl)\) or substitution \(0.30 \div (0.35 \times 0.12)\) [1]; 7.1 A [1]. ALLOW 7.14 if 2 s.f. is also given. Part (ii) is independent of part (i) (no ECF needed). Correct answer with no working = 4 marks.

Pressure in a liquid increases with depth (\( p = \rho g h \)) because there is a greater column of water, with greater weight, above the diver. Density and the mass of particles do not change.

Third box ticked only: 1 mark.
DO NOT ALLOW more than one box ticked.
Other options are distractors: decrease (wrong direction), 'stays the same' (ignores depth), 'particles become heavier' (particle mass is constant).

\( F = BIl \), so force is directly proportional to current. Doubling \( I \) doubles \( F \). Halving \( l \) or \( B \) halves the force; reversing the current changes only the direction of the force.

First box ticked only: 1 mark.
DO NOT ALLOW more than one box ticked.

In beta decay a neutron changes into a proton and an electron, and the electron is emitted. The total number of nucleons is unchanged, so the mass number stays the same, while the number of protons rises by one.

Second box ticked only: 1 mark.
DO NOT ALLOW more than one box ticked.
The first option describes alpha decay.

Velocity is a vector, so a change in direction is a change in velocity even at constant speed. A changing velocity means the car accelerates, so a resultant force acts on it (towards the centre of the circle, at right angles to its motion).

Third box ticked only: 1 mark.
DO NOT ALLOW more than one box ticked.

Pressure due to a fluid is \(p = \rho g h\), so it is directly proportional to depth \(h\) and to density \(\rho\).

Row 1: 8 m is deeper than 4 m, so the pressure is greater: true.
Row 2: seawater has the greater density (1030 > 1000 kg/m³), so at the same depth seawater gives the greater pressure and freshwater the smaller: false.
Row 3: \(p \propto h\), so doubling the depth doubles the pressure due to the water: true.

| Question | Expected answer | Marks | AO | Guidance |
|---|---|---|---|---|
| Row 1 | True ticked | 1 | AO1.1 | Mark each row independently. |
| Row 2 | False ticked | 1 | AO2.1 | No mark if both boxes or neither box is ticked in that row. |
| Row 3 | True ticked | 1 | AO2.1 | DO NOT ALLOW a row with two boxes ticked. |

Total: 3 marks.

The same particles occupy half the volume, so they hit the walls more often. The temperature is constant, so the average particle speed is unchanged. More frequent collisions give a larger force per unit area, so the pressure increases. Since \(pV\) is constant, halving \(V\) doubles \(p\) (to 200 kPa).

(i) increases [1]. ALLOW 'more' as the idea of an increase.
(ii) stays the same [1]. AO1.1
(iii) increases [1]. AO1.2
(iv) doubles [1]. ALLOW 'increases' only if followed by 200 kPa; DO NOT ALLOW 'increases' alone. AO2.1
If more than one word is given for a blank, no mark for that blank.
Total 4 marks.

Mass is conserved, so A = 3.0 kg. Before: \(p = mv = 2.0 \times 3.0 = 6.0\) kg m/s (the stationary trolley has zero momentum), so B = 6.0. Momentum is conserved in the collision, so C = 6.0 kg m/s. Then \(v = p/m = 6.0 \div 3.0 = 2.0\) m/s, so D = 2.0 m/s.

A = 3.0 [1]
B = 6.0 [1]. AO2.1 (2.0 x 3.0; stationary trolley contributes zero)
C = 6.0 [1]. Momentum conserved. ALLOW ECF from B. AO1.2
D = 2.0 [1]. Correct use of v = p/m. ALLOW ECF from C and A. AO2.1
Units are printed in the table so are not required.
Total 4 marks.

\(p = \rho g h = 1000 \times 10 \times 12 = 120\,000\) Pa.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 1 | 120 000 Pa | 1 | AO2.1 | ALLOW 1.2 \(\times\) 10^5 Pa. ALLOW 120 kPa. ALLOW N/m^2 as the unit. Both correct value and unit are needed for the mark. DO NOT ALLOW a correct value with a wrong or missing unit. |

\(p = mv = 0.80 \times 1.5 = 1.2\) kg m/s.

| Question | Answer | Marks | AO | Guidance |
|---|---|---|---|---|
| 2 | 1.2 (kg m/s) | 1 | AO2.1 | ALLOW 1.20. ALLOW Ns as the unit. Unit is printed on the answer line so is not required. |

\(p = \rho g h = 1030 \times 9.8 \times 35 = 353\,290\) Pa, which is \(3.5 \times 10^{5}\) Pa to 2 s.f.

Substitution into p = ρgh: 1030 × 9.8 × 35 (1 mark, method). Answer 3.5 × 10^5 Pa (1 mark, accuracy). ALLOW 353 290 or 3.53 × 10^5 for 1 mark only if not in 2 s.f. DO NOT ALLOW missing/incorrect unit. Correct answer with no working scores 2.

Change in momentum = 0.16 × 25 = 4.0 kg m/s. \(F = \Delta p / t = 4.0 / 0.020 = 200\) N.

Change in momentum = 4.0 kg m/s or use of F = Δp/t with substitution (1 mark, method). Answer 200 N (1 mark, accuracy). ALLOW 2.0 × 10^2 N. Correct answer with no working scores 2.

Newton's third law pairs are the same type of force and act on two different objects. The pair of the Earth's pull on the book is the book's pull on the Earth. The table's force is a contact force acting on the same object as the weight.

The gravitational force / pull of the book on the Earth (1). The contact force from the table is a different type of force OR acts on the same object as the weight (not on a different object) (1). ALLOW equal and opposite force on the Earth. IGNORE statements that the two forces are equal in size only.

12 hours is 4 half-lives. Activity: 640 → 320 → 160 → 80 → 40 Bq.

Identifies 4 half-lives (1 mark, method). Answer 40 Bq (1 mark, accuracy). ECF from an incorrect number of half-lives if halving is applied correctly. Correct answer with no working scores 2.

\(F = BIl = 0.25 \times 3.2 \times 0.080 = 0.064\) N.

Substitution 0.25 × 3.2 × 0.080 (1 mark, method). Answer 0.064 N (1 mark, accuracy). ALLOW 6.4 × 10^-2 N. Correct answer with no working scores 2.

The pulse travels down and back in 0.80 s, so the one-way time is 0.40 s. Depth = 1500 × 0.40 = 600 m.

Use of one-way time 0.40 s OR distance 1500 × 0.80 = 1200 m then halved (1 mark, method). Answer 600 m (1 mark, accuracy). If 1200 m given with no halving, award 1 mark. Correct answer with no working scores 2.

\(v^2 - u^2 = 400 - 64 = 336\). \(a = 336 / (2 \times 84) = 2.0\) m/s\(^2\).

Correct rearrangement or substitution, e.g. a = (20² − 8.0²)/(2 × 84) (1 mark, method). Answer 2.0 m/s² (1 mark, accuracy). ALLOW 2 m/s². Correct answer with no working scores 2.

Fission of one nucleus releases more than one neutron. These neutrons can be absorbed by other unstable nuclei, causing further fission, which releases more neutrons, so the reaction sustains itself.

Fission releases (extra) neutrons (1). These neutrons cause / trigger further fission of other nuclei, repeating the process (1). IGNORE references to fusion. DO NOT ALLOW the nucleus 'burning' or atoms 'splitting' without neutrons being released.

\(p = \rho g h = 1000 \times 9.8 \times 12 = 117\,600\) Pa, which is \(1.2 \times 10^{5}\) Pa to 2 significant figures.

Mark 1: correct substitution \(1000 \times 9.8 \times 12\) (or evaluation 117 600). Mark 2: final answer \(1.2 \times 10^{5}\) Pa (accuracy and rounding to 2 s.f.). ALLOW 120 000 Pa. ALLOW 1 mark for 117 600 if not rounded (or rounded incorrectly). ALLOW 1.2 x 10^5 written as 1.2e5. DO NOT ALLOW a missing or wrong unit for the second mark. Correct answer with no working gets 2 marks.

Momentum before = \(0.80 \times 1.5 = 1.2\) kg m/s. Momentum is conserved, so \(1.2 = (0.80 + 0.40) \times v\), giving \(v = 1.2 \div 1.2 = 1.0\) m/s.

Mark 1: total momentum before = 1.2 kg m/s (or \(0.80 \times 1.5\)) with conservation used, i.e. \(1.2 = 1.2 \times v\) or equivalent. Mark 2: \(v = 1.0\) m/s. ALLOW ECF from an incorrect momentum calculation for the second mark. ALLOW 1 m/s. IGNORE direction. Correct answer with no working gets 2 marks.

Pressure due to the water: \(p = \rho g h = 1030 \times 9.8 \times 40 = 403\,760\) Pa.
Force on the window: \(F = pA = 403\,760 \times 0.020 = 8075\) N.
To 2 significant figures, \(F = 8100\) N (8.1 × 10\(^3\) N).

Question 1 | Expected answer | Marks | AO | Guidance
1 | Substitution into \(p = \rho g h\): \(1030 \times 9.8 \times 40\) (= 403 760 Pa) | 1 | AO2.1 | ALLOW correct use of p = hρg. ALLOW 4.0 × 10\(^5\) Pa or 404 000 Pa as an intermediate value.
2 | Use of \(F = pA\): \(403\,760 \times 0.020\) (= 8075 N) | 1 | AO2.1 | ALLOW ECF from an incorrect pressure. ALLOW use of rounded pressure, giving 8000 N or 8080 N.
3 | Answer 8100 N (8.1 × 10\(^3\) N) given to 2 significant figures | 1 | AO2.2 | Correct answer to 2 s.f. scores all 3 marks. ALLOW 8075 N or 8080 N for 2 marks only (not given to 2 s.f.). DO NOT ALLOW 8000 N unless the working is shown and the answer is consistent with a rounded pressure of 4.0 × 10\(^5\) Pa (then ALLOW as 3rd mark). IGNORE unit errors in working if the final unit N is shown. Three marks total.