IB · PHYSICS SL

Physics: Standard Level

Fields — Theme D

Name: ____________________Date: October 10, 2026

1.
State Newton's law of gravitation, and use it to explain why gravitational force decreases with the square of separation.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Work through this mathematical step: States F = Gm₁m₂/r², where G is the gravitational constant. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Develop this part of the answer: States that gravitational force is directly proportional to the product of the two masses. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that force is inversely proportional to the square of the separation r, so doubling r reduces the force to a quarter, explaining the inverse-square dependence. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The inverse-square law appears throughout physics (gravity, electric field, radiation intensity) because it reflects how a fixed quantity spreads over the surface of an expanding sphere (area ∝ r²).
Marking points
- States F = Gm₁m₂/r², where G is the gravitational constant.
- States that gravitational force is directly proportional to the product of the two masses.
- States that force is inversely proportional to the square of the separation r, so doubling r reduces the force to a quarter, explaining the inverse-square dependence.
Examiner tip: The inverse-square law appears throughout physics (gravity, electric field, radiation intensity) because it reflects how a fixed quantity spreads over the surface of an expanding sphere (area ∝ r²).
2.
Marking analysis: A learner attempts the following task: “State Newton's law of gravitation, and use it to explain why gravitational force decreases with the square of separation.” Their response addresses only this point: “States F = Gm₁m₂/r², where G is the gravitational constant.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States F = Gm₁m₂/r², where G is the gravitational constant. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that gravitational force is directly proportional to the product of the two masses. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that force is inversely proportional to the square of the separation r, so doubling r reduces the force to a quarter, explaining the inverse-square dependence. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States F = Gm₁m₂/r², where G is the gravitational constant.
- Identifies the missing requirement: States that gravitational force is directly proportional to the product of the two masses.
- Identifies the missing requirement: States that force is inversely proportional to the square of the separation r, so doubling r reduces the force to a quarter, explaining the inverse-square dependence.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
3.
Calculate the gravitational field strength at the surface of a planet of mass 4.0 × 10²⁴ kg and radius 5.0 × 10⁶ m. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses g = GM/r². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes (6.67 × 10⁻¹¹ × 4.0 × 10²⁴)/(5.0 × 10⁶)². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Develop this part of the answer: Obtains g ≈ 10.7 N kg⁻¹. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Gravitational field strength g = GM/r² is the gravitational force per unit mass, and can be applied at the surface of any planet, not just Earth.
Marking points
- Uses g = GM/r².
- Substitutes (6.67 × 10⁻¹¹ × 4.0 × 10²⁴)/(5.0 × 10⁶)².
- Obtains g ≈ 10.7 N kg⁻¹.
Examiner tip: Gravitational field strength g = GM/r² is the gravitational force per unit mass, and can be applied at the surface of any planet, not just Earth.
4.
Marking analysis: A learner attempts the following task: “Calculate the gravitational field strength at the surface of a planet of mass 4.0 × 10²⁴ kg and radius 5.0 × 10⁶ m. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².” Their response addresses only this point: “Uses g = GM/r².” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses g = GM/r². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes (6.67 × 10⁻¹¹ × 4.0 × 10²⁴)/(5.0 × 10⁶)². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains g ≈ 10.7 N kg⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses g = GM/r².
- Identifies the missing requirement: Substitutes (6.67 × 10⁻¹¹ × 4.0 × 10²⁴)/(5.0 × 10⁶)².
- Identifies the missing requirement: Obtains g ≈ 10.7 N kg⁻¹.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
5.
Define electric field strength, and state its direction relative to a positive test charge.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: Defines electric field strength as the electric force per unit positive charge experienced at a point. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that the field direction is the direction of the force on a positive test charge placed at that point. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Electric field lines always point away from positive charges and towards negative charges, consistent with the force direction on a positive test charge.
Marking points
- Defines electric field strength as the electric force per unit positive charge experienced at a point.
- States that the field direction is the direction of the force on a positive test charge placed at that point.
Examiner tip: Electric field lines always point away from positive charges and towards negative charges, consistent with the force direction on a positive test charge.
6.
Marking analysis: A learner attempts the following task: “Define electric field strength, and state its direction relative to a positive test charge.” Their response addresses only this point: “Defines electric field strength as the electric force per unit positive charge experienced at a point.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Defines electric field strength as the electric force per unit positive charge experienced at a point. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that the field direction is the direction of the force on a positive test charge placed at that point. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Defines electric field strength as the electric force per unit positive charge experienced at a point.
- Identifies the missing requirement: States that the field direction is the direction of the force on a positive test charge placed at that point.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
7.
Two point charges of +3.0 μC and −3.0 μC are separated by 0.20 m. Calculate the magnitude of the electric force between them. Use k = 8.99 × 10⁹ N m² C⁻².
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses Coulomb's law F = kq₁q₂/r². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes (8.99 × 10⁹)(3.0 × 10⁻⁶)(3.0 × 10⁻⁶)/(0.20)². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Develop this part of the answer: Obtains F ≈ 2.02 N. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Coulomb's law has exactly the same inverse-square mathematical form as Newton's law of gravitation, but describes electric force between charges rather than gravitational force between masses.
Marking points
- Uses Coulomb's law F = kq₁q₂/r².
- Substitutes (8.99 × 10⁹)(3.0 × 10⁻⁶)(3.0 × 10⁻⁶)/(0.20)².
- Obtains F ≈ 2.02 N.
Examiner tip: Coulomb's law has exactly the same inverse-square mathematical form as Newton's law of gravitation, but describes electric force between charges rather than gravitational force between masses.
8.
Marking analysis: A learner attempts the following task: “Two point charges of +3.0 μC and −3.0 μC are separated by 0.20 m. Calculate the magnitude of the electric force between them. Use k = 8.99 × 10⁹ N m² C⁻².” Their response addresses only this point: “Uses Coulomb's law F = kq₁q₂/r².” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses Coulomb's law F = kq₁q₂/r². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes (8.99 × 10⁹)(3.0 × 10⁻⁶)(3.0 × 10⁻⁶)/(0.20)². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains F ≈ 2.02 N. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses Coulomb's law F = kq₁q₂/r².
- Identifies the missing requirement: Substitutes (8.99 × 10⁹)(3.0 × 10⁻⁶)(3.0 × 10⁻⁶)/(0.20)².
- Identifies the missing requirement: Obtains F ≈ 2.02 N.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
9.
Explain why a charged particle moving parallel to a uniform magnetic field experiences no magnetic force, while one moving perpendicular to the field does.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Work through this mathematical step: States that the magnetic force on a moving charge is given by F = qvB sinθ, where θ is the angle between velocity and field. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: States that when the particle moves parallel to the field, θ = 0°, so sinθ = 0 and the force is zero. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: States that when the particle moves perpendicular to the field, θ = 90°, so sinθ = 1 and the force is at its maximum value. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The magnetic force on a moving charge depends on the sine of the angle between velocity and field — it is zero when parallel and maximum when perpendicular.
Marking points
- States that the magnetic force on a moving charge is given by F = qvB sinθ, where θ is the angle between velocity and field.
- States that when the particle moves parallel to the field, θ = 0°, so sinθ = 0 and the force is zero.
- States that when the particle moves perpendicular to the field, θ = 90°, so sinθ = 1 and the force is at its maximum value.
Examiner tip: The magnetic force on a moving charge depends on the sine of the angle between velocity and field — it is zero when parallel and maximum when perpendicular.
10.
Marking analysis: A learner attempts the following task: “Explain why a charged particle moving parallel to a uniform magnetic field experiences no magnetic force, while one moving perpendicular to the field does.” Their response addresses only this point: “States that the magnetic force on a moving charge is given by F = qvB sinθ, where θ is the angle between velocity and field.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that the magnetic force on a moving charge is given by F = qvB sinθ, where θ is the angle between velocity and field. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that when the particle moves parallel to the field, θ = 0°, so sinθ = 0 and the force is zero. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that when the particle moves perpendicular to the field, θ = 90°, so sinθ = 1 and the force is at its maximum value. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that the magnetic force on a moving charge is given by F = qvB sinθ, where θ is the angle between velocity and field.
- Identifies the missing requirement: States that when the particle moves parallel to the field, θ = 0°, so sinθ = 0 and the force is zero.
- Identifies the missing requirement: States that when the particle moves perpendicular to the field, θ = 90°, so sinθ = 1 and the force is at its maximum value.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
11.
A charged particle moves in a circular path perpendicular to a uniform magnetic field. State the direction of the magnetic force at any instant, relative to the particle's velocity.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that the magnetic force is always perpendicular to the particle's velocity. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that this force acts as the centripetal force, directed towards the centre of the circular path. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Because the magnetic force is always perpendicular to velocity, it never does work on the particle — it changes direction but never speeds up or slows down the charge.
Marking points
- States that the magnetic force is always perpendicular to the particle's velocity.
- States that this force acts as the centripetal force, directed towards the centre of the circular path.
Examiner tip: Because the magnetic force is always perpendicular to velocity, it never does work on the particle — it changes direction but never speeds up or slows down the charge.
12.
Marking analysis: A learner attempts the following task: “A charged particle moves in a circular path perpendicular to a uniform magnetic field. State the direction of the magnetic force at any instant, relative to the particle's velocity.” Their response addresses only this point: “States that the magnetic force is always perpendicular to the particle's velocity.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that the magnetic force is always perpendicular to the particle's velocity. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that this force acts as the centripetal force, directed towards the centre of the circular path. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that the magnetic force is always perpendicular to the particle's velocity.
- Identifies the missing requirement: States that this force acts as the centripetal force, directed towards the centre of the circular path.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
13.
Define gravitational potential energy in a radial field, and explain why it is defined to be negative for a mass at a finite distance from another mass.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: Defines gravitational potential energy as the work done to bring a mass from infinity to a point in the gravitational field. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that gravitational potential energy is taken to be zero at infinity, the point of maximum (least negative) potential energy. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Develop this part of the answer: States that since gravity is always attractive, work must be done ON the mass by an external agent to move it away to infinity, meaning energy is released (a negative value results) as it moves closer, hence potential energy at any finite separation is negative relative to infinity. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: The negative sign convention for gravitational potential energy is a direct consequence of choosing zero potential energy at infinity, combined with gravity always being an attractive force.
Marking points
- Defines gravitational potential energy as the work done to bring a mass from infinity to a point in the gravitational field.
- States that gravitational potential energy is taken to be zero at infinity, the point of maximum (least negative) potential energy.
- States that since gravity is always attractive, work must be done ON the mass by an external agent to move it away to infinity, meaning energy is released (a negative value results) as it moves closer, hence potential energy at any finite separation is negative relative to infinity.
Examiner tip: The negative sign convention for gravitational potential energy is a direct consequence of choosing zero potential energy at infinity, combined with gravity always being an attractive force.
14.
Marking analysis: A learner attempts the following task: “Define gravitational potential energy in a radial field, and explain why it is defined to be negative for a mass at a finite distance from another mass.” Their response addresses only this point: “Defines gravitational potential energy as the work done to bring a mass from infinity to a point in the gravitational field.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Defines gravitational potential energy as the work done to bring a mass from infinity to a point in the gravitational field. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that gravitational potential energy is taken to be zero at infinity, the point of maximum (least negative) potential energy. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that since gravity is always attractive, work must be done ON the mass by an external agent to move it away to infinity, meaning energy is released (a negative value results) as it moves closer, hence potential energy at any finite separation is negative relative to infinity. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Defines gravitational potential energy as the work done to bring a mass from infinity to a point in the gravitational field.
- Identifies the missing requirement: States that gravitational potential energy is taken to be zero at infinity, the point of maximum (least negative) potential energy.
- Identifies the missing requirement: States that since gravity is always attractive, work must be done ON the mass by an external agent to move it away to infinity, meaning energy is released (a negative value results) as it moves closer, hence potential energy at any finite separation is negative relative to infinity.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
15.
A satellite orbits Earth in a circular orbit of radius r. Show that its orbital period T satisfies T² ∝ r³, using Newton's law of gravitation and the centripetal force equation.
[4 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Equates gravitational force to the required centripetal force: GMm/r² = mv²/r (or equivalently mω²r). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes v = 2πr/T (or ω = 2π/T) into the equation. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Simplifies to obtain GM/r² = 4π²r/T². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Work through this mathematical step: Rearranges to T² = (4π²/GM)r³, showing T² ∝ r³ since 4π²/GM is constant for a given central mass. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: This derivation (Kepler's third law from Newton's law of gravitation) is a classic 'show that' question — always keep symbols algebraic throughout and only substitute numbers if the question asks for a numerical answer.
Marking points
- Equates gravitational force to the required centripetal force: GMm/r² = mv²/r (or equivalently mω²r).
- Substitutes v = 2πr/T (or ω = 2π/T) into the equation.
- Simplifies to obtain GM/r² = 4π²r/T².
- Rearranges to T² = (4π²/GM)r³, showing T² ∝ r³ since 4π²/GM is constant for a given central mass.
Examiner tip: This derivation (Kepler's third law from Newton's law of gravitation) is a classic 'show that' question — always keep symbols algebraic throughout and only substitute numbers if the question asks for a numerical answer.
16.
Marking analysis: A learner attempts the following task: “A satellite orbits Earth in a circular orbit of radius r. Show that its orbital period T satisfies T² ∝ r³, using Newton's law of gravitation and the centripetal force equation.” Their response addresses only this point: “Equates gravitational force to the required centripetal force: GMm/r² = mv²/r (or equivalently mω²r).” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Equates gravitational force to the required centripetal force: GMm/r² = mv²/r (or equivalently mω²r). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes v = 2πr/T (or ω = 2π/T) into the equation. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Simplifies to obtain GM/r² = 4π²r/T². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Rearranges to T² = (4π²/GM)r³, showing T² ∝ r³ since 4π²/GM is constant for a given central mass. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Equates gravitational force to the required centripetal force: GMm/r² = mv²/r (or equivalently mω²r).
- Identifies the missing requirement: Substitutes v = 2πr/T (or ω = 2π/T) into the equation.
- Identifies the missing requirement: Simplifies to obtain GM/r² = 4π²r/T².
- Identifies the missing requirement: Rearranges to T² = (4π²/GM)r³, showing T² ∝ r³ since 4π²/GM is constant for a given central mass.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
17.
A parallel plate capacitor has a uniform electric field of 2000 V m⁻¹ between its plates, which are 0.050 m apart. Calculate the potential difference between the plates.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses E = V/d for a uniform field between parallel plates. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Rearranges to V = Ed. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Substitutes 2000 × 0.050 to obtain V = 100 V. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: E = V/d only applies to the uniform field between parallel plates — it is not the general formula for electric field strength around a point charge (which follows an inverse-square law instead).
Marking points
- Uses E = V/d for a uniform field between parallel plates.
- Rearranges to V = Ed.
- Substitutes 2000 × 0.050 to obtain V = 100 V.
Examiner tip: E = V/d only applies to the uniform field between parallel plates — it is not the general formula for electric field strength around a point charge (which follows an inverse-square law instead).
18.
Marking analysis: A learner attempts the following task: “A parallel plate capacitor has a uniform electric field of 2000 V m⁻¹ between its plates, which are 0.050 m apart. Calculate the potential difference between the plates.” Their response addresses only this point: “Uses E = V/d for a uniform field between parallel plates.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses E = V/d for a uniform field between parallel plates. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Rearranges to V = Ed. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Substitutes 2000 × 0.050 to obtain V = 100 V. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses E = V/d for a uniform field between parallel plates.
- Identifies the missing requirement: Rearranges to V = Ed.
- Identifies the missing requirement: Substitutes 2000 × 0.050 to obtain V = 100 V.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
19.
Compare the direction of the gravitational force between two masses with the direction of the electric force between two like (same-sign) charges.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Break the command into its requested parts. For each part, connect a relevant fact or observation to the conclusion it supports. Describing what happens and explaining why it happens are different tasks.
2. Develop this part of the answer: States that the gravitational force between two masses is always attractive. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Develop this part of the answer: States that the electric force between two like charges is repulsive, unlike gravity, since gravitational 'charge' (mass) has only one sign while electric charge can be positive or negative. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Gravity is always attractive because mass is always positive; electric force can be attractive or repulsive because charge can be either sign — this is the key structural difference between the two inverse-square force laws.
Marking points
- States that the gravitational force between two masses is always attractive.
- States that the electric force between two like charges is repulsive, unlike gravity, since gravitational 'charge' (mass) has only one sign while electric charge can be positive or negative.
Examiner tip: Gravity is always attractive because mass is always positive; electric force can be attractive or repulsive because charge can be either sign — this is the key structural difference between the two inverse-square force laws.
20.
Marking analysis: A learner attempts the following task: “Compare the direction of the gravitational force between two masses with the direction of the electric force between two like (same-sign) charges.” Their response addresses only this point: “States that the gravitational force between two masses is always attractive.” Evaluate the response against the complete 2-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[2 marks] · no calculator
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: States that the gravitational force between two masses is always attractive. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that the electric force between two like charges is repulsive, unlike gravity, since gravitational 'charge' (mass) has only one sign while electric charge can be positive or negative. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: States that the gravitational force between two masses is always attractive.
- Identifies the missing requirement: States that the electric force between two like charges is repulsive, unlike gravity, since gravitational 'charge' (mass) has only one sign while electric charge can be positive or negative.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
21.
Calculate the electric potential at a point 0.30 m from a point charge of +5.0 μC. Use k = 8.99 × 10⁹ N m² C⁻².
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses V = kQ/r. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes (8.99 × 10⁹ × 5.0 × 10⁻⁶)/0.30. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Develop this part of the answer: Obtains V ≈ 1.50 × 10⁵ V. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Electric potential due to a point charge is a scalar quantity and follows an inverse (not inverse-square) relationship with distance, unlike electric field strength.
Marking points
- Uses V = kQ/r.
- Substitutes (8.99 × 10⁹ × 5.0 × 10⁻⁶)/0.30.
- Obtains V ≈ 1.50 × 10⁵ V.
Examiner tip: Electric potential due to a point charge is a scalar quantity and follows an inverse (not inverse-square) relationship with distance, unlike electric field strength.
22.
Marking analysis: A learner attempts the following task: “Calculate the electric potential at a point 0.30 m from a point charge of +5.0 μC. Use k = 8.99 × 10⁹ N m² C⁻².” Their response addresses only this point: “Uses V = kQ/r.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses V = kQ/r. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes (8.99 × 10⁹ × 5.0 × 10⁻⁶)/0.30. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains V ≈ 1.50 × 10⁵ V. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses V = kQ/r.
- Identifies the missing requirement: Substitutes (8.99 × 10⁹ × 5.0 × 10⁻⁶)/0.30.
- Identifies the missing requirement: Obtains V ≈ 1.50 × 10⁵ V.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
23.
Define capacitance, and calculate the capacitance of a capacitor that stores a charge of 6.0 × 10⁻⁴ C when the potential difference across it is 12 V.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Defines capacitance as the charge stored per unit potential difference: C = Q/V. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes (6.0 × 10⁻⁴)/12. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Obtains C = 5.0 × 10⁻⁵ F. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Capacitance is a fixed property of a capacitor (determined by its physical construction), so it remains the same value regardless of how much charge is currently stored or what voltage is applied.
Marking points
- Defines capacitance as the charge stored per unit potential difference: C = Q/V.
- Substitutes (6.0 × 10⁻⁴)/12.
- Obtains C = 5.0 × 10⁻⁵ F.
Examiner tip: Capacitance is a fixed property of a capacitor (determined by its physical construction), so it remains the same value regardless of how much charge is currently stored or what voltage is applied.
24.
Marking analysis: A learner attempts the following task: “Define capacitance, and calculate the capacitance of a capacitor that stores a charge of 6.0 × 10⁻⁴ C when the potential difference across it is 12 V.” Their response addresses only this point: “Defines capacitance as the charge stored per unit potential difference: C = Q/V.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Defines capacitance as the charge stored per unit potential difference: C = Q/V. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes (6.0 × 10⁻⁴)/12. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains C = 5.0 × 10⁻⁵ F. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Defines capacitance as the charge stored per unit potential difference: C = Q/V.
- Identifies the missing requirement: Substitutes (6.0 × 10⁻⁴)/12.
- Identifies the missing requirement: Obtains C = 5.0 × 10⁻⁵ F.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
25.
Calculate the energy stored in a 5.0 × 10⁻⁵ F capacitor when it is charged to a potential difference of 12 V.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses energy = ½CV². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes ½ × (5.0 × 10⁻⁵) × 12². Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Obtains energy = 3.6 × 10⁻³ J. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Because energy stored depends on V², doubling the charging voltage of a capacitor quadruples the energy it stores, not merely doubles it.
Marking points
- Uses energy = ½CV².
- Substitutes ½ × (5.0 × 10⁻⁵) × 12².
- Obtains energy = 3.6 × 10⁻³ J.
Examiner tip: Because energy stored depends on V², doubling the charging voltage of a capacitor quadruples the energy it stores, not merely doubles it.
26.
Marking analysis: A learner attempts the following task: “Calculate the energy stored in a 5.0 × 10⁻⁵ F capacitor when it is charged to a potential difference of 12 V.” Their response addresses only this point: “Uses energy = ½CV².” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses energy = ½CV². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes ½ × (5.0 × 10⁻⁵) × 12². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains energy = 3.6 × 10⁻³ J. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses energy = ½CV².
- Identifies the missing requirement: Substitutes ½ × (5.0 × 10⁻⁵) × 12².
- Identifies the missing requirement: Obtains energy = 3.6 × 10⁻³ J.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
27.
Define gravitational potential at a point, and calculate the gravitational potential at the surface of a planet of mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Develop this part of the answer: Defines gravitational potential as the work done per unit mass to bring a small test mass from infinity to that point. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
3. Work through this mathematical step: States Vg = −GM/r. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Substitutes −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴)/(6.4 × 10⁶). Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Develop this part of the answer: Obtains Vg ≈ −6.25 × 10⁷ J kg⁻¹. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Gravitational potential is always negative (since gravitational potential energy is negative and potential is energy per unit mass) and increases (becomes less negative) with distance from the mass, reaching zero at infinity.
Marking points
- Defines gravitational potential as the work done per unit mass to bring a small test mass from infinity to that point.
- States Vg = −GM/r.
- Substitutes −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴)/(6.4 × 10⁶).
- Obtains Vg ≈ −6.25 × 10⁷ J kg⁻¹.
Examiner tip: Gravitational potential is always negative (since gravitational potential energy is negative and potential is energy per unit mass) and increases (becomes less negative) with distance from the mass, reaching zero at infinity.
28.
Marking analysis: A learner attempts the following task: “Define gravitational potential at a point, and calculate the gravitational potential at the surface of a planet of mass 6.0 × 10²⁴ kg and radius 6.4 × 10⁶ m. Use G = 6.67 × 10⁻¹¹ N m² kg⁻².” Their response addresses only this point: “Defines gravitational potential as the work done per unit mass to bring a small test mass from infinity to that point.” Evaluate the response against the complete 4-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[4 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Defines gravitational potential as the work done per unit mass to bring a small test mass from infinity to that point. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States Vg = −GM/r. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Substitutes −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴)/(6.4 × 10⁶). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains Vg ≈ −6.25 × 10⁷ J kg⁻¹. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Defines gravitational potential as the work done per unit mass to bring a small test mass from infinity to that point.
- Identifies the missing requirement: States Vg = −GM/r.
- Identifies the missing requirement: Substitutes −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴)/(6.4 × 10⁶).
- Identifies the missing requirement: Obtains Vg ≈ −6.25 × 10⁷ J kg⁻¹.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
29.
A straight wire of length 0.50 m carries a current of 4.0 A perpendicular to a uniform magnetic field of flux density 0.20 T. Calculate the magnetic force on the wire.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. List the given quantities and the requested unknown. Choose the relation that connects them, state any required assumptions, then substitute before rounding. Preserve exact expressions when the task asks for an exact result.
2. Work through this mathematical step: Uses F = BIL for a wire perpendicular to the field. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
3. Work through this mathematical step: Substitutes 0.20 × 4.0 × 0.50. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
4. Work through this mathematical step: Obtains F = 0.40 N. Write the intermediate operation, keep the units consistent where applicable, and check the relation against the quantities given in the question.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: F = BIL applies only when the current-carrying wire is perpendicular to the magnetic field — for other angles, the force scales with sinθ, exactly as it does for a single moving charge.
Marking points
- Uses F = BIL for a wire perpendicular to the field.
- Substitutes 0.20 × 4.0 × 0.50.
- Obtains F = 0.40 N.
Examiner tip: F = BIL applies only when the current-carrying wire is perpendicular to the magnetic field — for other angles, the force scales with sinθ, exactly as it does for a single moving charge.
30.
Marking analysis: A learner attempts the following task: “A straight wire of length 0.50 m carries a current of 4.0 A perpendicular to a uniform magnetic field of flux density 0.20 T. Calculate the magnetic force on the wire.” Their response addresses only this point: “Uses F = BIL for a wire perpendicular to the field.” Evaluate the response against the complete 3-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[3 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. Separate the learner's stated response from the complete task. Credit only what their response demonstrates, then identify each missing requirement; do not assume unstated working.
2. Requirement 1: Recognises credit for the stated point: Uses F = BIL for a wire perpendicular to the field. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes 0.20 × 4.0 × 0.50. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains F = 0.40 N. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.
Marking points
- Recognises credit for the stated point: Uses F = BIL for a wire perpendicular to the field.
- Identifies the missing requirement: Substitutes 0.20 × 4.0 × 0.50.
- Identifies the missing requirement: Obtains F = 0.40 N.
Examiner tip: Treat each marking point as a separate requirement. Do not award the same idea twice, and do not infer work the learner did not show.