Acceleration

1 Plan and risk assessment

A resultant force makes a mass accelerate: resultant force = mass × acceleration. Double the force on the same mass and the acceleration doubles; double the mass with the same force and it halves. The hanging weight provides the pull; the resultant force also accounts for friction and the runway slope. The mass is everything the string accelerates: the trolley, the hanger and every mass on either.

F = m a

Prediction

2 Apparatus

• ✓Dynamics trolley The object being accelerated.
• ✓10 cm card on the trolley Cuts the light gate beams.
• ✓Runway (1.3 m) A smooth, straight track for the trolley.
• ✓Wooden block Tilts the runway to compensate for friction.
• ✓Bench pulley and clamp Turns the hanging weight into a horizontal pull.
• ✓String Joins the trolley to the mass hanger.
• ✓20 g mass hanger and four 20 g slotted masses Their weight is the accelerating force (0.2–1.0 N).
• ✓Four 200 g trolley masses Change the mass of the trolley in part 2.
• ✓Two light gates and a data logger (resolution 0.01 m/s²) Measure the speed at each gate and the acceleration.
• ✓Stopwatch (resolution 0.01 s) The other way to time the trolley.
• ✓Metre rule (resolution 1 mm) Measures the card and the distances along the runway.
• ✓Digital balance (resolution 0.1 g) Measures the mass of the trolley.
• ✓Box of sand and an end stop Catch the falling masses and the trolley.

3 Method: your choices

• Which version are you doing?: AQA required practical 7: force, mass and accelerationA trolley pulled by a falling mass: change the force, then the mass, and find F = m a.
• Measure the card: DoneThe logger works out each speed as card length ÷ the time the card blocks the beam. A wrong length makes every speed, and every acceleration, wrong by the same proportion.
• Compensate for friction: DoneOn the slope a small part of the trolley’s weight acts down the runway. It balances the friction, so the resultant force on the trolley is just the pull of the string.
• Choose how to measure the acceleration: Light gates and a data loggerThe card cuts each beam: the logger times it to 0.1 ms, works out the speed at each gate and the acceleration between them. No reaction time, so repeats agree to a few hundredths of a m/s².
• Place the second light gate: 55 cm from the startThe trolley passes both gates while the string is still pulling, so the logger measures the acceleration the force gives.
• Load the trolley: Move masses from trolley to hangerThe total mass being accelerated stays the same, so only the force changes: a fair test.
• Sand box and end stopSafety: DoneFalling masses can land on feet, and a trolley running off the end can fall off the bench.
• Release from rest: DoneThe method assumes the trolley starts from rest at the same place each time.

• 0.0 Struck through: the anomaly, retaken; the new reading is beside it.

Anomaly: Acceleration, 80 g on the hanger, trial 1, retaken. Likely cause: the string rubbed on the side of the pulley in that run, so there was more friction.

6 Analysis

1. The force with 60 g on the hanger

   weight = mass × gravitational field strength (W = m g)

   With 60 g on the hanger, the force pulling the trolley is the weight of the hanging masses. Calculate it.

   0.59 N Right first time

2. The acceleration with 100 g on the hanger

   acceleration = change in velocity ÷ time taken (a = Δv ÷ t)

   Check the logger’s arithmetic for your first run with 100 g on the hanger: work out the acceleration from the two speeds and the time between the gates.

   1.26 m/s² Right first time

3. Gradient (acceleration per newton)

   gradient = change in y ÷ change in x

   1.28 1/kg Right first time

4. Mass estimate (1 ÷ gradient, constant-mass model)

   0.783 kg Right first time

5. Where does your line cross?

   Your line meets the force axis at about 0.00 N. What does that show?

   It goes through the origin (within the scatter): acceleration is directly proportional to force, so friction was compensated. Right first time

6. Where did the pull go?

   3 of 3 right first time Right first time

7. 1 ÷ mass

   total mass = trolley + masses added + hanger and its masses (0.100 kg)

   For the run with 400 g added to the trolley, work out the total mass being accelerated in kg, then 1 ÷ total mass.

   0.840 1/kg Right first time

8. Shape of the graph

   What does your graph show?

   A straight line: acceleration is inversely proportional to the mass (a ∝ 1 ÷ m). Right first time

9. Gradient (the accelerating force)

   gradient = change in y ÷ change in x

   0.987 N Right first time

7 Conclusion and evaluation

Conclusion

Sources of error

Improvement

Repeatable or reproducible

9 Exam questions on your data

On your own readings, self-marked against the mark scheme, with error carried forward where it applies. Total 30/30.

1. 01.1[1 mark]

   Write down the equation that links acceleration (a), mass (m) and resultant force (F).

   Your answer

   resultant force = mass × acceleration

   1/1 self-marked

   • resultant force = mass × acceleration, F = m a (or any correct rearrangement)

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: resultant force = mass × acceleration F = m a

2. 01.2[1 mark]

   In part 1 a student changed the force on the trolley by changing the number of masses on the hanger.

   Which variable should the student keep the same in part 1?

   Tick (✓) one box.

   • The mass on the hanger
   • The acceleration of the trolley
   • ✓ The combined mass of the trolley and the mass hanger with its masses (your choice)
   • The time between the light gates

   Your answer

   The combined mass of the trolley and the mass hanger with its masses

   1/1

   • The combined mass of the trolley and the mass hanger with its masses

   Full marks. That is exactly what the mark scheme is looking for.

   Model answer: The combined mass of the trolley and the mass hanger with its masses.

3. 01.3[2 marks]

   Before the investigation the student put a wooden block under one end of the runway so that it sloped slightly.

   Explain why the student tilted the runway.

   Your answer

   A component of the trolley’s weight acts down the slope. It balances the friction, so the resultant force on the trolley is just the pull of the string.

   2/2 self-marked

   • A component of the trolley’s weight acts down (parallel to) the runway
   • This balances friction, so the resultant force on the trolley is the pull of the string (or: on the trolley and hanging masses together, the weight of the hanging masses)

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: A component of the trolley’s weight acts down the slope. This balances the friction, so the resultant force on the trolley is just the pull of the string.

4. 01.4[2 marks]

   In part 1 you started with four 20 g masses on the trolley and moved one to the hanger for each new force.

   Explain why moving the masses from the trolley to the hanger gives a more valid result than taking new masses from the box.

   Your answer

   The total mass being accelerated stays the same, because the masses on the hanger are accelerated too. So only the force changes and it is a fair test.

   2/2 self-marked

   • The total mass (trolley, hanger and masses) stays the same
   • The hanging masses are accelerated too, so only the force changes (a fair test; mass is a control variable)

   Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

   Model answer: The total mass being accelerated stays the same, because the hanging masses are accelerated too. So only the force changes: it is a fair test.

5. 01.5[1 mark]

   You used two light gates and a data logger.

   Give one advantage of using light gates rather than a stopwatch to time the trolley.

   Your answer

   There is no reaction time because the timing is automatic.

   1/1 self-marked

   • No human reaction time (timing is automatic); or a higher resolution

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: There is no human reaction time: the light gates start and stop the timing automatically.

6. 01.6[1 mark]

   Give one hazard in this investigation and the harm it could cause.

   Your answer

   The masses could fall from the hanger and land on your feet, injuring your toes.

   1/1 self-marked

   • Falling masses (or the trolley falling off the bench) could injure feet or toes

   Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

   Model answer: The masses could fall off the hanger when it hits the floor and land on someone’s feet, bruising or breaking toes.

7. 02.1[1 mark]

   Table 1 shows your three trials with 80 g on the hanger.

   Which trial is anomalous?

   Select one row.

   Table 1

   Trial	Acceleration in m/s²
   1	0.72
   2	1.01
   3	0.98

   Your answer

   trial 1

   1/1

   • Trial 1: far from the other two, which agree

   Full marks. Your answer meets all the points for this question. Trial 1 (0.72) is far from the other two, which agree with each other.

   Model answer: Trial 1. It is far from the other two trials, which agree with each other.

8. 02.2[2 marks]

   Table 2 shows your repeat readings with 60 g on the hanger.

   Table 2

   Trial	Acceleration in m/s²
   1	0.73
   2	0.74

   Calculate the uncertainty in the mean acceleration.

   Your answer

   0.005

   Working: Range = 0.74 − 0.73 = 0.01 m/s² Uncertainty = 0.01 ÷ 2 = 0.005 m/s²

   2/2

   • Range = 0.74 − 0.73 = 0.01 m/s²
   • Uncertainty = ±0.005 m/s² (half the range)

   Full marks. Clear and precise.

   Model answer: Range = 0.74 − 0.73 = 0.01 m/s² Uncertainty = ±0.01 ÷ 2 = ±0.005 m/s²

9. 02.3[3 marks]

   With 80 g on the hanger the force on the trolley was 0.78 N. Your mean acceleration was 0.99 m/s².

   Assume friction was fully compensated.

   Calculate the total mass being accelerated.

   Your answer

   0.788

   Working: 0.78 = m × 0.99 m = 0.78 ÷ 0.99 m = 0.788 kg

   3/3

   • Substitution: 0.78 = m × 0.99
   • Rearrangement: m = 0.78 ÷ 0.99
   • m = 0.788 kg

   Full marks. Clear and precise.

   Model answer: 0.78 = m × 0.99 m = 0.78 ÷ 0.99 m = 0.788 kg

10. 02.4[2 marks]

    Another student did not tilt the runway. Their line of acceleration against force is straight, but it crosses the force axis at 0.13 N instead of passing through the origin.

    Explain why the line does not pass through the origin.

    Your answer

    Friction acts on the trolley. A force of 0.13 N is needed to overcome friction before the trolley accelerates.

    2/2 self-marked

    • Friction acts on the trolley (between the wheels and the runway, and at the pulley)
    • So a force of about 0.13 N is needed to overcome friction before the trolley accelerates (the resultant force is less than the weight on the hanger)

    Self-marked: 2 of 2 marks. This is your own judgement against the mark scheme.

    Model answer: Friction acts on the trolley, so a force of about 0.13 N is needed to overcome friction before the trolley accelerates. The resultant force is less than the weight on the hanger.

11. 03.1[3 marks]

    Another student did not have light gates. They released the trolley from rest and timed it over a distance of 0.50 m with a stopwatch.

    The time was 1.47 s. The trolley accelerated uniformly the whole way.

    Calculate the acceleration of the trolley.

    Your answer

    0.463

    Working: mean velocity = 0.50 ÷ 1.47 = 0.340 m/s final velocity = 2 × 0.340 = 0.68 m/s a = 0.68 ÷ 1.47 = 0.463 m/s²

    3/3

    • Mean velocity = 0.50 ÷ 1.47 = 0.340 m/s
    • Final velocity = 2 × mean velocity = 0.68 m/s
    • a = 0.68 ÷ 1.47 = 0.463 m/s²

    Full marks. That is exactly what the mark scheme is looking for.

    Model answer: mean velocity = 0.50 ÷ 1.47 = 0.340 m/s final velocity = 2 × 0.340 = 0.68 m/s a = 0.68 ÷ 1.47 = 0.463 m/s²

12. 03.2[1 mark]

    The resultant force on the trolley stays the same while the hanger falls.

    Describe the motion of the trolley while the hanger is falling.

    Your answer

    It has a constant acceleration: its velocity increases at a steady rate.

    1/1 self-marked

    • Uniform (constant, steady) acceleration; or velocity increasing at a constant rate

    Self-marked: 1 of 1 mark. This is your own judgement against the mark scheme.

    Model answer: It accelerates uniformly: its velocity increases at a constant rate. (Do not accept "accelerating at a constant speed".)

13. 03.3[4 marks]

    The total mass of a trolley, hanger and masses is 0.79 kg. The resultant force on them is 0.59 N. The trolley starts from rest.

    v² − u² = 2 a s

    Calculate the speed of the trolley after it has travelled 0.6 m.

    Use the Physics Equations Sheet. Give your answer to 2 significant figures.

    Your answer

    0.95

    Working: a = F ÷ m = 0.59 ÷ 0.79 = 0.747 m/s^2 v^2 - 0^2 = 2 × 0.747 × 0.6 v^2 = 0.896 v = 0.95 m/s

    4/4

    • a = 0.59 ÷ 0.79 = 0.747 m/s² (the later marks need F = m a to have been used)
    • v² = 2 × 0.747 × 0.6 = 0.896
    • v = 0.947 m/s (allow an answer consistent with their acceleration)
    • Answer given to 2 significant figures

    Full marks. Well done.

    Model answer: a = F ÷ m = 0.59 ÷ 0.79 = 0.747 m/s² v² − 0² = 2 × 0.747 × 0.6 v² = 0.896 v = 0.95 m/s

14. 04[6 marks]

    A student wants to investigate how the acceleration of a trolley depends on the resultant force on it.

    The student has a trolley, a runway, a bench pulley, string, a mass hanger, slotted masses, two light gates and a data logger.

    Describe a method the student could use. Your answer should include how the mass is kept constant.

    Your answer

    1. Put the trolley on a runway with a pulley clamped at the end of the bench. Tie a string to the trolley, pass it over the pulley and attach a mass hanger. 2. Tilt the runway with a block until the trolley rolls down at a steady speed when pushed, to compensate for friction. 3. Clamp two light gates over the runway, connected to a data logger, so the card on the trolley passes through both before the hanger hits the floor. 4. Put all the slotted masses on the trolley. Release the trolley from rest at the same starting line and record the acceleration from the data logger. 5. Move one mass from the trolley to the hanger to increase the force, keeping the total mass constant. The force is the weight of the hanging masses, W = m g. 6. Repeat each force and calculate a mean acceleration. 7. Plot a graph of acceleration against force.

    6/6 self-marked

    • Trolley on a runway, string over a pulley to a mass hanger
    • Tilt (raise one end of) the runway so the trolley rolls at a steady speed, to compensate for friction
    • Measure the acceleration with light gates and a data logger (or time over a measured distance)
    • Change the force by changing the masses on the hanger (force = weight, W = m g)
    • Keep the total mass constant: move masses from the trolley to the hanger
    • Release the trolley from rest from the same starting position each time
    • Repeat for each force and calculate a mean
    • Plot a graph of acceleration against force
    • Place the second gate so the trolley passes it before the hanger hits the floor

    Self-marked: Level 3, 6 of 6 marks. You ticked 9 of 9 indicative points. This is your own judgement against the mark scheme.

    Model answer: 1. Put the trolley on a runway with a pulley clamped at the end of the bench. Tie a string to the trolley, pass it over the pulley and attach a mass hanger. 2. Tilt the runway with a block until the trolley rolls down at a steady speed when pushed, to compensate for friction. 3. Clamp two light gates over the runway, connected to a data logger, so the card on the trolley passes through both before the hanger hits the floor. 4. Put all the slotted masses on the trolley. Release the trolley from rest at the same starting line and record the acceleration from the data logger. 5. Move one mass from the trolley to the hanger to increase the force, keeping the total mass constant. The force is the weight of the hanging masses, W = m g. 6. Repeat each force and calculate a mean acceleration. 7. Plot a graph of acceleration against force.

Unseen data: the same skills, new apparatus

1. 01[2 marks]

   A separate simulated trolley moves in the same direction through two light gates. Use acceleration = change in velocity ÷ time.

   Data logger readings

   Velocity at first gate in m/s	Velocity at second gate in m/s	Time between gates in s
   0.621	1.036	0.426

   Calculate the average acceleration between the gates.

   Your answer

   0.974

   Working: 1.036 - 0.621 = 0.415 0.415 / 0.426

   2/2

   • Change in velocity = 0.415 m/s.
   • 0.974 m/s².

   Full marks. Well done.

   Model answer: Change in velocity = 1.036 − 0.621 = 0.415 m/s. Acceleration = 0.415 ÷ 0.426 = 0.974 m/s².

2. 02[2 marks]

   In a separate simulated investigation, a checked apparatus problem affected one run. That reading is marked rejected below. Calculate using the retained repeats.

   Acceleration repeats

   Trial	Acceleration in m/s²	Use in mean
   1	0.52	Rejected
   2	0.71	Yes
   3	0.71	Yes

   Calculate the mean acceleration of the retained readings.

   Your answer

   0.71

   Working: 0.71 + 0.71 = 1.42 1.42 / 2

   2/2

   • Sum of retained accelerations = 1.42 m/s².
   • 0.71 m/s².

   Full marks. Your answer meets all the points for this question.

   Model answer: Retain trials 2 and 3. Mean = 1.42 ÷ 2 = 0.71 m/s².

3. 03[1 mark]

   A different student adds each new hanging mass from a box. Friction is compensated. These are two settings in the investigation.

   Method choices

   Hanging mass in g	Combined moving mass in g
   20	764.9
   40	784.9

   Which change would let the next investigation isolate the effect of pulling force on acceleration? Tick (✓) one box.

   • ✓ Begin with the spare masses on the trolley and move them to the hanger, keeping total moving mass constant. (your choice)
   • Repeat the same method more times while continuing to add new masses from the box.
   • Give the trolley a larger push at each new force.
   • Keep only the bare trolley mass constant; the hanging masses do not count.

   Your answer

   Begin with the spare masses on the trolley and move them to the hanger, keeping total moving mass constant.

   1/1

   • Begin with the spare masses on the trolley and move them to the hanger, keeping total moving mass constant.

   Full marks. Clear and precise.

   Model answer: Both trolley and hanging masses accelerate. Transfer masses within that system to vary the pulling force without adding to its total mass. Repeats alone do not fix a changing control variable.