IB · PHYSICS SL

Physics: Standard Level

Space, time and motion — Theme A

Name: ____________________Date: October 10, 2026

1.
A 2.0 kg block starts from rest on a horizontal frictionless surface. A constant resultant force of 3.0 N acts on it for 4.0 s. Determine the acceleration, the final speed and the displacement of the block.
[5 marks]
Answer explanation
Draft walkthroughs are based on marking guidance, not independently verified derivations.
1. The force is the resultant force, so Newton's second law gives a = F/m = 3.0/2.0 = 1.5 m/s². Constant force and mass give constant acceleration.
2. The block starts from rest: u = 0. After 4.0 s, v = u + at = 0 + 1.5(4.0) = 6.0 m/s.
3. Displacement is s = ut + ½at² = ½(1.5)(4.0²) = 12 m. A check uses mean velocity (0 + 6)/2 = 3 m/s and s = 3(4) = 12 m.
Marking points
- Uses a = F/m = 3.0/2.0.
- Obtains a = 1.5 m/s².
- Uses v = u + at with u = 0.
- Obtains v = 6.0 m/s.
- Uses s = ut + ½at² (or equivalent) to obtain s = 12 m.
Examiner tip: With constant acceleration from rest, any of the SUVAT equations can be used once acceleration is known — pick whichever needs the fewest extra steps.
2.
Marking analysis: A learner attempts the following task: “A 2.0 kg block starts from rest on a horizontal frictionless surface. A constant resultant force of 3.0 N acts on it for 4.0 s. Determine the acceleration, the final speed and the displacement of the block.” Their response addresses only this point: “Uses a = F/m = 3.0/2.0.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 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 a = F/m = 3.0/2.0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Obtains a = 1.5 m/s². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Uses v = u + at with u = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains v = 6.0 m/s. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Requirement 5: Identifies the missing requirement: Uses s = ut + ½at² (or equivalent) to obtain s = 12 m. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
7. 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 a = F/m = 3.0/2.0.
- Identifies the missing requirement: Obtains a = 1.5 m/s².
- Identifies the missing requirement: Uses v = u + at with u = 0.
- Identifies the missing requirement: Obtains v = 6.0 m/s.
- Identifies the missing requirement: Uses s = ut + ½at² (or equivalent) to obtain s = 12 m.
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.
A ball is thrown horizontally at 15 m/s from the top of a 20 m cliff. Calculate the time to reach the ground and the horizontal distance travelled. Use g = 9.8 m/s².
[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. Work through this mathematical step: States that vertical and horizontal motion are independent, and uses the vertical equation 20 = ½(9.8)t² to find time. 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: Obtains t ≈ 2.02 s. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Work through this mathematical step: Uses horizontal distance = horizontal speed × time. 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 horizontal distance ≈ 30.3 m. 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: Projectile motion always splits into two fully independent 1D problems: vertical motion sets the time of flight, horizontal motion (constant velocity) then uses that time.
Marking points
- States that vertical and horizontal motion are independent, and uses the vertical equation 20 = ½(9.8)t² to find time.
- Obtains t ≈ 2.02 s.
- Uses horizontal distance = horizontal speed × time.
- Obtains horizontal distance ≈ 30.3 m.
Examiner tip: Projectile motion always splits into two fully independent 1D problems: vertical motion sets the time of flight, horizontal motion (constant velocity) then uses that time.
4.
Marking analysis: A learner attempts the following task: “A ball is thrown horizontally at 15 m/s from the top of a 20 m cliff. Calculate the time to reach the ground and the horizontal distance travelled. Use g = 9.8 m/s².” Their response addresses only this point: “States that vertical and horizontal motion are independent, and uses the vertical equation 20 = ½(9.8)t² to find time.” 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: States that vertical and horizontal motion are independent, and uses the vertical equation 20 = ½(9.8)t² to find time. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Obtains t ≈ 2.02 s. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Uses horizontal distance = horizontal speed × time. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains horizontal distance ≈ 30.3 m. 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: States that vertical and horizontal motion are independent, and uses the vertical equation 20 = ½(9.8)t² to find time.
- Identifies the missing requirement: Obtains t ≈ 2.02 s.
- Identifies the missing requirement: Uses horizontal distance = horizontal speed × time.
- Identifies the missing requirement: Obtains horizontal distance ≈ 30.3 m.
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 linear momentum, and state the principle of conservation of momentum.
[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. Work through this mathematical step: Defines momentum as the product of an object's mass and velocity (p = mv), a vector quantity. 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 the total momentum of an isolated system remains constant, provided no external resultant force acts on it. 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: Momentum conservation applies to the total (vector) momentum of a system — always assign a positive direction and keep signs consistent when objects move oppositely.
Marking points
- Defines momentum as the product of an object's mass and velocity (p = mv), a vector quantity.
- States that the total momentum of an isolated system remains constant, provided no external resultant force acts on it.
Examiner tip: Momentum conservation applies to the total (vector) momentum of a system — always assign a positive direction and keep signs consistent when objects move oppositely.
6.
Marking analysis: A learner attempts the following task: “Define linear momentum, and state the principle of conservation of momentum.” Their response addresses only this point: “Defines momentum as the product of an object's mass and velocity (p = mv), a vector quantity.” 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 momentum as the product of an object's mass and velocity (p = mv), a vector quantity. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that the total momentum of an isolated system remains constant, provided no external resultant force acts on it. 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 momentum as the product of an object's mass and velocity (p = mv), a vector quantity.
- Identifies the missing requirement: States that the total momentum of an isolated system remains constant, provided no external resultant force acts on it.
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.
A 0.50 kg ball moving at 8.0 m/s collides head-on with a stationary 1.5 kg ball, and they stick together. Calculate their common velocity after the collision.
[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 conservation of momentum: total momentum before = total momentum after. 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: Calculates initial momentum = 0.50 × 8.0 = 4.0 kg m/s, and total mass after = 2.0 kg. 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 common velocity = 4.0/2.0 = 2.0 m/s. 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: A collision where objects stick together conserves momentum but not kinetic energy — never assume KE is conserved unless the collision is explicitly stated to be elastic.
Marking points
- Uses conservation of momentum: total momentum before = total momentum after.
- Calculates initial momentum = 0.50 × 8.0 = 4.0 kg m/s, and total mass after = 2.0 kg.
- Obtains common velocity = 4.0/2.0 = 2.0 m/s.
Examiner tip: A collision where objects stick together conserves momentum but not kinetic energy — never assume KE is conserved unless the collision is explicitly stated to be elastic.
8.
Marking analysis: A learner attempts the following task: “A 0.50 kg ball moving at 8.0 m/s collides head-on with a stationary 1.5 kg ball, and they stick together. Calculate their common velocity after the collision.” Their response addresses only this point: “Uses conservation of momentum: total momentum before = total momentum after.” 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 conservation of momentum: total momentum before = total momentum after. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Calculates initial momentum = 0.50 × 8.0 = 4.0 kg m/s, and total mass after = 2.0 kg. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains common velocity = 4.0/2.0 = 2.0 m/s. 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 conservation of momentum: total momentum before = total momentum after.
- Identifies the missing requirement: Calculates initial momentum = 0.50 × 8.0 = 4.0 kg m/s, and total mass after = 2.0 kg.
- Identifies the missing requirement: Obtains common velocity = 4.0/2.0 = 2.0 m/s.
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.
State the work-energy theorem, and use it to explain why a car's braking distance depends on the square of its initial speed.
[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: States that the work done by the resultant force on an object equals its change in kinetic energy. 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 that kinetic energy is proportional to the square of speed (KE = ½mv²). 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: Explains that since the braking force does work equal to the kinetic energy lost, and that energy scales with v², doubling speed requires four times the braking distance for the same braking force. 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 v² dependence of braking distance is a direct, testable consequence of the work-energy theorem — a classic application question.
Marking points
- States that the work done by the resultant force on an object equals its change in kinetic energy.
- States that kinetic energy is proportional to the square of speed (KE = ½mv²).
- Explains that since the braking force does work equal to the kinetic energy lost, and that energy scales with v², doubling speed requires four times the braking distance for the same braking force.
Examiner tip: The v² dependence of braking distance is a direct, testable consequence of the work-energy theorem — a classic application question.
10.
Marking analysis: A learner attempts the following task: “State the work-energy theorem, and use it to explain why a car's braking distance depends on the square of its initial speed.” Their response addresses only this point: “States that the work done by the resultant force on an object equals its change in kinetic energy.” 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 work done by the resultant force on an object equals its change in kinetic energy. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that kinetic energy is proportional to the square of speed (KE = ½mv²). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Explains that since the braking force does work equal to the kinetic energy lost, and that energy scales with v², doubling speed requires four times the braking distance for the same braking force. 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 work done by the resultant force on an object equals its change in kinetic energy.
- Identifies the missing requirement: States that kinetic energy is proportional to the square of speed (KE = ½mv²).
- Identifies the missing requirement: Explains that since the braking force does work equal to the kinetic energy lost, and that energy scales with v², doubling speed requires four times the braking distance for the same braking force.
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 crane lifts a 500 kg load through a vertical height of 12 m in 20 s at constant speed. Calculate the useful power output of the crane. Use g = 9.8 m/s².
[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. Work through this mathematical step: Calculates the weight lifted: 500 × 9.8 = 4900 N. 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: Uses work done = force × distance = 4900 × 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 work done = 58 800 J. 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: Uses power = work/time = 58 800/20 to obtain 2940 W. 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: At constant speed, the lifting force equals the weight exactly, since the resultant force (and hence acceleration) is zero.
Marking points
- Calculates the weight lifted: 500 × 9.8 = 4900 N.
- Uses work done = force × distance = 4900 × 12.
- Obtains work done = 58 800 J.
- Uses power = work/time = 58 800/20 to obtain 2940 W.
Examiner tip: At constant speed, the lifting force equals the weight exactly, since the resultant force (and hence acceleration) is zero.
12.
Marking analysis: A learner attempts the following task: “A crane lifts a 500 kg load through a vertical height of 12 m in 20 s at constant speed. Calculate the useful power output of the crane. Use g = 9.8 m/s².” Their response addresses only this point: “Calculates the weight lifted: 500 × 9.8 = 4900 N.” 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: Calculates the weight lifted: 500 × 9.8 = 4900 N. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Uses work done = force × distance = 4900 × 12. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains work done = 58 800 J. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Uses power = work/time = 58 800/20 to obtain 2940 W. 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: Calculates the weight lifted: 500 × 9.8 = 4900 N.
- Identifies the missing requirement: Uses work done = force × distance = 4900 × 12.
- Identifies the missing requirement: Obtains work done = 58 800 J.
- Identifies the missing requirement: Uses power = work/time = 58 800/20 to obtain 2940 W.
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.
A car of mass 1200 kg travelling at 20 m/s brakes and comes to rest over a distance of 40 m. Calculate the average braking force, using the work-energy theorem.
[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. Work through this mathematical step: Calculates the initial kinetic energy: ½ × 1200 × 20² = 240 000 J. 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 the work done by the braking force equals the kinetic energy lost (the car comes to rest). Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Work through this mathematical step: Uses work = force × distance, so 240 000 = force × 40. 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: Obtains braking force = 6000 N. 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: Equating work done to the change in kinetic energy is a shortcut that avoids calculating acceleration or time separately.
Marking points
- Calculates the initial kinetic energy: ½ × 1200 × 20² = 240 000 J.
- States that the work done by the braking force equals the kinetic energy lost (the car comes to rest).
- Uses work = force × distance, so 240 000 = force × 40.
- Obtains braking force = 6000 N.
Examiner tip: Equating work done to the change in kinetic energy is a shortcut that avoids calculating acceleration or time separately.
14.
Marking analysis: A learner attempts the following task: “A car of mass 1200 kg travelling at 20 m/s brakes and comes to rest over a distance of 40 m. Calculate the average braking force, using the work-energy theorem.” Their response addresses only this point: “Calculates the initial kinetic energy: ½ × 1200 × 20² = 240 000 J.” 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: Calculates the initial kinetic energy: ½ × 1200 × 20² = 240 000 J. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that the work done by the braking force equals the kinetic energy lost (the car comes to rest). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Uses work = force × distance, so 240 000 = force × 40. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains braking force = 6000 N. 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: Calculates the initial kinetic energy: ½ × 1200 × 20² = 240 000 J.
- Identifies the missing requirement: States that the work done by the braking force equals the kinetic energy lost (the car comes to rest).
- Identifies the missing requirement: Uses work = force × distance, so 240 000 = force × 40.
- Identifies the missing requirement: Obtains braking force = 6000 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.
15.
An object moves in a horizontal circle of radius 0.50 m at a constant speed of 4.0 m/s. Calculate its centripetal acceleration and state the direction of the resultant force causing it.
[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 centripetal acceleration a = v²/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 4.0²/0.50 to obtain a = 32 m/s². 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: States that the resultant (centripetal) force points towards the centre of the circle. 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: Centripetal force is not a new type of force — it is simply the name given to whichever resultant force (tension, gravity, friction, etc.) happens to point towards the centre and maintain circular motion.
Marking points
- Uses centripetal acceleration a = v²/r.
- Substitutes 4.0²/0.50 to obtain a = 32 m/s².
- States that the resultant (centripetal) force points towards the centre of the circle.
Examiner tip: Centripetal force is not a new type of force — it is simply the name given to whichever resultant force (tension, gravity, friction, etc.) happens to point towards the centre and maintain circular motion.
16.
Marking analysis: A learner attempts the following task: “An object moves in a horizontal circle of radius 0.50 m at a constant speed of 4.0 m/s. Calculate its centripetal acceleration and state the direction of the resultant force causing it.” Their response addresses only this point: “Uses centripetal acceleration a = v²/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 centripetal acceleration a = v²/r. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes 4.0²/0.50 to obtain a = 32 m/s². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that the resultant (centripetal) force points towards the centre of the circle. 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 centripetal acceleration a = v²/r.
- Identifies the missing requirement: Substitutes 4.0²/0.50 to obtain a = 32 m/s².
- Identifies the missing requirement: States that the resultant (centripetal) force points towards the centre of the circle.
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.
State Newton's third law of motion, and use it to explain the forces acting when a swimmer pushes against the water to move forward.
[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: States Newton's third law: for every action force, there is an equal and opposite reaction force acting on a different object. 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 swimmer pushes water backward (the action force). 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 the water pushes the swimmer forward with an equal and opposite reaction force, propelling them through the water. 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: Action and reaction forces always act on two different objects, never on the same object, and are always equal in magnitude and opposite in direction.
Marking points
- States Newton's third law: for every action force, there is an equal and opposite reaction force acting on a different object.
- States that the swimmer pushes water backward (the action force).
- States that the water pushes the swimmer forward with an equal and opposite reaction force, propelling them through the water.
Examiner tip: Action and reaction forces always act on two different objects, never on the same object, and are always equal in magnitude and opposite in direction.
18.
Marking analysis: A learner attempts the following task: “State Newton's third law of motion, and use it to explain the forces acting when a swimmer pushes against the water to move forward.” Their response addresses only this point: “States Newton's third law: for every action force, there is an equal and opposite reaction force acting on a different object.” 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 Newton's third law: for every action force, there is an equal and opposite reaction force acting on a different object. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that the swimmer pushes water backward (the action force). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that the water pushes the swimmer forward with an equal and opposite reaction force, propelling them through the water. 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 Newton's third law: for every action force, there is an equal and opposite reaction force acting on a different object.
- Identifies the missing requirement: States that the swimmer pushes water backward (the action force).
- Identifies the missing requirement: States that the water pushes the swimmer forward with an equal and opposite reaction force, propelling them through the water.
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.
A ball of mass 0.20 kg is dropped from rest and hits the ground 1.5 s later. Calculate its momentum just before impact, and the impulse delivered to it by gravity during the fall. Use g = 9.8 m/s².
[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. Work through this mathematical step: Uses v = gt = 9.8 × 1.5 to find the speed just before impact. 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: Obtains v ≈ 14.7 m/s. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Work through this mathematical step: Uses momentum p = mv to obtain p ≈ 2.94 kg m/s. 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: States that the impulse delivered equals the change in momentum, so impulse ≈ 2.94 N s (since the ball started with zero momentum). 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: Impulse is defined as the change in momentum (or equivalently, force × time) — for an object starting from rest, the impulse simply equals its final momentum.
Marking points
- Uses v = gt = 9.8 × 1.5 to find the speed just before impact.
- Obtains v ≈ 14.7 m/s.
- Uses momentum p = mv to obtain p ≈ 2.94 kg m/s.
- States that the impulse delivered equals the change in momentum, so impulse ≈ 2.94 N s (since the ball started with zero momentum).
Examiner tip: Impulse is defined as the change in momentum (or equivalently, force × time) — for an object starting from rest, the impulse simply equals its final momentum.
20.
Marking analysis: A learner attempts the following task: “A ball of mass 0.20 kg is dropped from rest and hits the ground 1.5 s later. Calculate its momentum just before impact, and the impulse delivered to it by gravity during the fall. Use g = 9.8 m/s².” Their response addresses only this point: “Uses v = gt = 9.8 × 1.5 to find the speed just before impact.” 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: Uses v = gt = 9.8 × 1.5 to find the speed just before impact. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Obtains v ≈ 14.7 m/s. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Uses momentum p = mv to obtain p ≈ 2.94 kg m/s. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: States that the impulse delivered equals the change in momentum, so impulse ≈ 2.94 N s (since the ball started with zero momentum). 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: Uses v = gt = 9.8 × 1.5 to find the speed just before impact.
- Identifies the missing requirement: Obtains v ≈ 14.7 m/s.
- Identifies the missing requirement: Uses momentum p = mv to obtain p ≈ 2.94 kg m/s.
- Identifies the missing requirement: States that the impulse delivered equals the change in momentum, so impulse ≈ 2.94 N s (since the ball started with zero momentum).
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.
A 1.0 kg trolley moving at 6.0 m/s collides elastically with a stationary 1.0 kg trolley. Determine the velocities of both trolleys after the collision, and verify that both momentum and kinetic energy are conserved.
[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. Work through this mathematical step: States that momentum conservation requires m₁u₁ = m₁v₁ + m₂v₂. 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 conservation of kinetic energy (elastic collision) requires ½m₁u₁² = ½m₁v₁² + ½m₂v₂². 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 solving these simultaneously for equal masses gives v₁ = 0 (the first trolley stops) and v₂ = 6.0 m/s (the second trolley moves off with the first trolley's initial speed). 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: Verifies momentum before (6.0 kg m/s) equals momentum after (0 + 6.0 = 6.0 kg m/s), and kinetic energy before (18 J) equals kinetic energy after (0 + 18 = 18 J). 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: When equal masses collide elastically in one dimension with one initially at rest, the incoming object always stops completely and the target takes on its exact velocity — a useful special case worth recognising.
Marking points
- States that momentum conservation requires m₁u₁ = m₁v₁ + m₂v₂.
- States that conservation of kinetic energy (elastic collision) requires ½m₁u₁² = ½m₁v₁² + ½m₂v₂².
- States that solving these simultaneously for equal masses gives v₁ = 0 (the first trolley stops) and v₂ = 6.0 m/s (the second trolley moves off with the first trolley's initial speed).
- Verifies momentum before (6.0 kg m/s) equals momentum after (0 + 6.0 = 6.0 kg m/s), and kinetic energy before (18 J) equals kinetic energy after (0 + 18 = 18 J).
Examiner tip: When equal masses collide elastically in one dimension with one initially at rest, the incoming object always stops completely and the target takes on its exact velocity — a useful special case worth recognising.
22.
Marking analysis: A learner attempts the following task: “A 1.0 kg trolley moving at 6.0 m/s collides elastically with a stationary 1.0 kg trolley. Determine the velocities of both trolleys after the collision, and verify that both momentum and kinetic energy are conserved.” Their response addresses only this point: “States that momentum conservation requires m₁u₁ = m₁v₁ + m₂v₂.” 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: States that momentum conservation requires m₁u₁ = m₁v₁ + m₂v₂. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that conservation of kinetic energy (elastic collision) requires ½m₁u₁² = ½m₁v₁² + ½m₂v₂². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States that solving these simultaneously for equal masses gives v₁ = 0 (the first trolley stops) and v₂ = 6.0 m/s (the second trolley moves off with the first trolley's initial speed). Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Verifies momentum before (6.0 kg m/s) equals momentum after (0 + 6.0 = 6.0 kg m/s), and kinetic energy before (18 J) equals kinetic energy after (0 + 18 = 18 J). 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: States that momentum conservation requires m₁u₁ = m₁v₁ + m₂v₂.
- Identifies the missing requirement: States that conservation of kinetic energy (elastic collision) requires ½m₁u₁² = ½m₁v₁² + ½m₂v₂².
- Identifies the missing requirement: States that solving these simultaneously for equal masses gives v₁ = 0 (the first trolley stops) and v₂ = 6.0 m/s (the second trolley moves off with the first trolley's initial speed).
- Identifies the missing requirement: Verifies momentum before (6.0 kg m/s) equals momentum after (0 + 6.0 = 6.0 kg m/s), and kinetic energy before (18 J) equals kinetic energy after (0 + 18 = 18 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.
23.
A ball is kicked with an initial speed of 20 m/s at an angle of 30° above the horizontal. Calculate the time taken to reach maximum height, and the total time of flight, assuming it lands at the same height from which it was kicked. Use g = 9.8 m/s².
[5 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: Resolves the initial vertical velocity component: uᵧ = 20 sin30° = 10.0 m/s. 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: Uses v = u − gt at maximum height, where the vertical velocity v = 0. 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: Rearranges t = uᵧ/g = 10.0/9.8 to obtain t ≈ 1.02 s to reach maximum height. 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: States that, by symmetry, the total time of flight is twice the time to reach maximum height. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
6. Develop this part of the answer: Obtains total time of flight ≈ 2.04 s. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
7. Check the complete task again, including restrictions, units, precision and supporting evidence when relevant. Specific caution: Only the vertical component of the initial velocity affects time of flight — the horizontal component (constant throughout) only affects the range, not how long the projectile stays in the air.
Marking points
- Resolves the initial vertical velocity component: uᵧ = 20 sin30° = 10.0 m/s.
- Uses v = u − gt at maximum height, where the vertical velocity v = 0.
- Rearranges t = uᵧ/g = 10.0/9.8 to obtain t ≈ 1.02 s to reach maximum height.
- States that, by symmetry, the total time of flight is twice the time to reach maximum height.
- Obtains total time of flight ≈ 2.04 s.
Examiner tip: Only the vertical component of the initial velocity affects time of flight — the horizontal component (constant throughout) only affects the range, not how long the projectile stays in the air.
24.
Marking analysis: A learner attempts the following task: “A ball is kicked with an initial speed of 20 m/s at an angle of 30° above the horizontal. Calculate the time taken to reach maximum height, and the total time of flight, assuming it lands at the same height from which it was kicked. Use g = 9.8 m/s².” Their response addresses only this point: “Resolves the initial vertical velocity component: uᵧ = 20 sin30° = 10.0 m/s.” Evaluate the response against the complete 5-mark task. Identify what earns credit and state every additional requirement needed for full marks.
[5 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: Resolves the initial vertical velocity component: uᵧ = 20 sin30° = 10.0 m/s. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Uses v = u − gt at maximum height, where the vertical velocity v = 0. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Rearranges t = uᵧ/g = 10.0/9.8 to obtain t ≈ 1.02 s to reach maximum height. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: States that, by symmetry, the total time of flight is twice the time to reach maximum height. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
6. Requirement 5: Identifies the missing requirement: Obtains total time of flight ≈ 2.04 s. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
7. 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: Resolves the initial vertical velocity component: uᵧ = 20 sin30° = 10.0 m/s.
- Identifies the missing requirement: Uses v = u − gt at maximum height, where the vertical velocity v = 0.
- Identifies the missing requirement: Rearranges t = uᵧ/g = 10.0/9.8 to obtain t ≈ 1.02 s to reach maximum height.
- Identifies the missing requirement: States that, by symmetry, the total time of flight is twice the time to reach maximum height.
- Identifies the missing requirement: Obtains total time of flight ≈ 2.04 s.
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.
A spring with spring constant 200 N/m is stretched by 0.15 m within its elastic limit. State Hooke's law, and calculate the elastic potential energy stored in the spring.
[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. Work through this mathematical step: States Hooke's law: F = kx, valid within the elastic limit of the spring. 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 elastic potential energy = ½kx². 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 ½ × 200 × 0.15². 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: Obtains elastic potential energy = 2.25 J. 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: Elastic potential energy depends on the square of the extension, so doubling the extension of a spring quadruples the energy stored in it, not merely doubles it.
Marking points
- States Hooke's law: F = kx, valid within the elastic limit of the spring.
- States that elastic potential energy = ½kx².
- Substitutes ½ × 200 × 0.15².
- Obtains elastic potential energy = 2.25 J.
Examiner tip: Elastic potential energy depends on the square of the extension, so doubling the extension of a spring quadruples the energy stored in it, not merely doubles it.
26.
Marking analysis: A learner attempts the following task: “A spring with spring constant 200 N/m is stretched by 0.15 m within its elastic limit. State Hooke's law, and calculate the elastic potential energy stored in the spring.” Their response addresses only this point: “States Hooke's law: F = kx, valid within the elastic limit of the spring.” 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: States Hooke's law: F = kx, valid within the elastic limit of the spring. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: States that elastic potential energy = ½kx². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Substitutes ½ × 200 × 0.15². Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Obtains elastic potential energy = 2.25 J. 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: States Hooke's law: F = kx, valid within the elastic limit of the spring.
- Identifies the missing requirement: States that elastic potential energy = ½kx².
- Identifies the missing requirement: Substitutes ½ × 200 × 0.15².
- Identifies the missing requirement: Obtains elastic potential energy = 2.25 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.
A 5.0 kg block rests on a frictionless slope inclined at 30° to the horizontal. Calculate the component of the block's weight acting parallel to the slope, and hence its acceleration down the slope. Use g = 9.8 m/s².
[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. Work through this mathematical step: Calculates the weight of the block: 5.0 × 9.8 = 49 N. 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: Resolves the component of weight parallel to the slope: 49 × sin30°. Show which detail or principle supports it and how it addresses the command; equivalent supported wording is acceptable.
4. Work through this mathematical step: Obtains the parallel component = 24.5 N. 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: Uses a = F/m = 24.5/5.0 to obtain a = 4.9 m/s² (equivalently, a = g sinθ). 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: On a frictionless slope, the block's acceleration is always g sinθ, regardless of its mass — the mass cancels out exactly, just as it does for objects in free fall.
Marking points
- Calculates the weight of the block: 5.0 × 9.8 = 49 N.
- Resolves the component of weight parallel to the slope: 49 × sin30°.
- Obtains the parallel component = 24.5 N.
- Uses a = F/m = 24.5/5.0 to obtain a = 4.9 m/s² (equivalently, a = g sinθ).
Examiner tip: On a frictionless slope, the block's acceleration is always g sinθ, regardless of its mass — the mass cancels out exactly, just as it does for objects in free fall.
28.
Marking analysis: A learner attempts the following task: “A 5.0 kg block rests on a frictionless slope inclined at 30° to the horizontal. Calculate the component of the block's weight acting parallel to the slope, and hence its acceleration down the slope. Use g = 9.8 m/s².” Their response addresses only this point: “Calculates the weight of the block: 5.0 × 9.8 = 49 N.” 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: Calculates the weight of the block: 5.0 × 9.8 = 49 N. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Resolves the component of weight parallel to the slope: 49 × sin30°. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: Obtains the parallel component = 24.5 N. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
5. Requirement 4: Identifies the missing requirement: Uses a = F/m = 24.5/5.0 to obtain a = 4.9 m/s² (equivalently, a = g sinθ). 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: Calculates the weight of the block: 5.0 × 9.8 = 49 N.
- Identifies the missing requirement: Resolves the component of weight parallel to the slope: 49 × sin30°.
- Identifies the missing requirement: Obtains the parallel component = 24.5 N.
- Identifies the missing requirement: Uses a = F/m = 24.5/5.0 to obtain a = 4.9 m/s² (equivalently, a = g sinθ).
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.
An electric motor is supplied with 400 J of electrical energy and does 260 J of useful work lifting a load against gravity. Calculate the efficiency of the motor, and state one reason why the efficiency is less than 100%.
[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 efficiency = (useful energy output ÷ total energy input) × 100%. 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 (260/400) × 100 to obtain efficiency = 65%. 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: States a valid reason, e.g. some of the input electrical energy is dissipated as thermal energy due to friction or electrical resistance within the motor. 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: No real machine achieves 100% efficiency, since some energy is always converted into non-useful forms (usually heat) through friction or resistance at some stage.
Marking points
- Uses efficiency = (useful energy output ÷ total energy input) × 100%.
- Substitutes (260/400) × 100 to obtain efficiency = 65%.
- States a valid reason, e.g. some of the input electrical energy is dissipated as thermal energy due to friction or electrical resistance within the motor.
Examiner tip: No real machine achieves 100% efficiency, since some energy is always converted into non-useful forms (usually heat) through friction or resistance at some stage.
30.
Marking analysis: A learner attempts the following task: “An electric motor is supplied with 400 J of electrical energy and does 260 J of useful work lifting a load against gravity. Calculate the efficiency of the motor, and state one reason why the efficiency is less than 100%.” Their response addresses only this point: “Uses efficiency = (useful energy output ÷ total energy input) × 100%.” 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 efficiency = (useful energy output ÷ total energy input) × 100%. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
3. Requirement 2: Identifies the missing requirement: Substitutes (260/400) × 100 to obtain efficiency = 65%. Compare this requirement with the supplied learner response; missing evidence cannot earn credit.
4. Requirement 3: Identifies the missing requirement: States a valid reason, e.g. some of the input electrical energy is dissipated as thermal energy due to friction or electrical resistance within the motor. 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 efficiency = (useful energy output ÷ total energy input) × 100%.
- Identifies the missing requirement: Substitutes (260/400) × 100 to obtain efficiency = 65%.
- Identifies the missing requirement: States a valid reason, e.g. some of the input electrical energy is dissipated as thermal energy due to friction or electrical resistance within the motor.
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.