AQA A-Level Physics Paper 3 (3BC), June 2025: Question 2

7 marks · Medium difficulty · Extended Answer

Derive the formula for the linear velocity of a rolling object from energy conservation, explain the effect of mass distribution on moment of inertia, and deduce the order four rolling objects reach the bottom of an incline.

Practise this question

Question

Question 02 begins with the formula for the moment of inertia I = A m r squared, where A is a shape constant. Figure 4 illustrates an object rolling without slipping down an inclined plane through a vertical height h, with linear velocity v and angular velocity omega. Question 02.1 asks to show that v = sqrt(2gh / (1 + A)) by considering energy transfers, for 3 marks. Table 1 lists four objects of equal mass m and outer radius r: solid sphere P with I = 2/5 m r squared, hollow sphere Q with I = 2/3 m r squared, solid cylinder R with I = 1/2 m r squared, and hollow cylinder S with I = m r squared. Question 02.2 asks to explain why hollow sphere Q has a greater moment of inertia than solid sphere P, for 1 mark. Figure 5 depicts the four objects released together from the top of the slope. Question 02.3 asks to deduce the order in which the objects reach the bottom of the slope, for 3 marks.
Question text

02.1 The moment of inertia I of any rolling object about its axis of rotation is given by

I = Amr2

where A is a constant that depends on the shape of the object

m is the mass of the object

r is the radius of the object.

Figure 4 shows an object that starts from rest and rolls down a slope without slipping.

Figure 4

The object falls a vertical distance h. It then has a linear velocity v and angular

velocity ω.

For a rolling object the linear velocity v of the centre of mass is related to the

angular velocity ω by v = rω.

Show, by considering the energy transfers that occur, that

2gh

v =

1 + A

[3 marks]

Table 1 gives the moments of inertia of a solid sphere P, a hollow sphere Q, a solid

cylinder R and a hollow cylinder S about their axes of rotation.

P, Q, R and S have the same mass m and outer radius r.

Table 1

Object Moment of inertia

solid sphere P mr

hollow sphere Q mr

solid cylinder R mr

hollow cylinder S mr2

02.2 Explain why the hollow sphere Q has a greater moment of inertia about its axis

of rotation than the solid sphere P.

*05* [1 mark]

02.3 Figure 5 shows all four objects at rest at the top of a slope. They are released at the

same time and roll down the slope without slipping.

Figure 5

Deduce the order in which the objects reach the bottom of the slope.

*06* [3 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 02: Part 02.1 awards 3 marks sequentially for equating gravitational potential energy to linear kinetic energy plus rotational kinetic energy (mgh = 1/2 m v^2 + 1/2 I omega^2), substituting I = A m r^2 and v = r omega to eliminate r and omega, and cancelling m and rearranging to the final expression. Part 02.2 gives 1 mark for the concept that mass in the hollow sphere is distributed at a greater distance from the rotational axis. Part 02.3 awards 3 marks for determining the velocity expression or A-values (A = 2/5, 1/2, 2/3, 1), recognizing that a smaller A gives a greater linear velocity v, and concluding the descending order is P, R, Q, S.

Question Answers Additional comments/Guidance Mark AO

02.1 The award of each mark is contingent on the 3 1× AO1

award of the previous mark.

2× AO2

MP1: Accept words or symbols:

Equates EP to EKlinear plus EKrot ✓ mgh = ½mv2 + ½ I ω2

v2

2 MP2: mgh = ½mv2 + ½(Amr2)

Substitutes Amr for I and v = r in some form to 2

r

eliminate r and ω ✓ m cancels, r2 cancels

Cancels m and rearranges ✓

Condone missing ½mv2 in MP1 and MP2 but

withhold MP2 if required answer is (incorrectly)

derived.

2𝑔ℎ

(expect to see 𝑣 = √ )

𝐴

02.2 Idea that more of the mass is at greater distance from the Accept idea that (for hollow sphere) mass is 1 AO1

axis (in the hollow sphere) (so moment of inertia greater) distributed further from axis

✓

Do not allow idea that Q has more mass or greater

radius than P.

Treat references to density as neutral.

– A-LEVEL PHYSICS – –

Derives expression for v for any one object ✓1a Expect to see:

02.3 3 2× AO2

10gh

✓2a P solid sphere v = (leading to: 1× AO3

Repeats for all other objects 7

m s-1)

v = 3.7 h

(Order of descending v =) P, R, Q, S ✓3a

6gh

Q hollow sphere v = (leading to:

m s-1)

v = 3.4 h

4gh

OR R solid cylinder v = (leading to:

22 1 ✓ m s-1)

A = , , ,1 1b v = 3.6 h

53 2

S hollow cylinder v = gh (leading to:

v = 3.1 h m s-1)

condone missing g and/or h in their calculations.

In alternative ✓2b:

Idea that the greater the value of A, the smaller the allow reverse argument;

value of v ✓2b condone idea that the greater value of I, the smaller the

value of v.

Do not accept argument based on angular acceleration

✓3b or angular velocity without support in terms of v or t.

(Order of descending v = ) P, R, Q, S

Total 15

How to answer it

Rolling Down an Incline & Moment of Inertia

What this question tests

This question assesses your understanding of rotational dynamics in rolling motion: equating gravitational potential energy to the sum of translational and rotational kinetic energy, physical factors affecting moment of inertia (mass distribution relative to the axis), and deducing acceleration/final velocities based on geometric shape constants.

Question 02.1

Derivation of Linear Velocity for a Rolling Body (3 Marks)

📐 Step-by-Step Derivation

Step 1: State Conservation of Energy
Loss in GPE = Gain in translational KE + Gain in rotational KE
mgh = ½mv² + ½Iω²
Step 2: Substitute given formulas
Substitute I = Amr² and rolling condition ω = v / r :
mgh = ½mv² + ½(Amr²)(v / r)²
mgh = ½mv² + ½Amr²(v² / r²)
Step 3: Simplify and cancel terms
Notice that r² cancels in the rotational term:
mgh = ½mv² + ½Amv²
Divide all terms by m :
gh = ½v²(1 + A)
Step 4: Rearrange for v
Multiply by 2 and divide by (1 + A) :
v² = 2gh / (1 + A)
v = √[ 2gh / (1 + A) ]
Mark Scheme Breakdown:
• Mark 1: Equates Ep to Ek,linear + Ek,rot.
• Mark 2: Correctly substitutes Amr² and eliminates ω using v = rω .
• Mark 3: Cancels mass m and algebraically rearranges to the final given expression. (Each mark is contingent on earning the previous mark).

💡 Key Knowledge

  • When an object rolls without slipping, total kinetic energy is split into two components:
    Ek,total = ½mv² + ½Iω²
  • The relationship between linear velocity at the centre of mass and angular velocity is strictly v = rω (or ω = v / r ).
  • Notice that neither outer radius r nor mass m appears in the final velocity formula; it depends only on height h and shape factor A .

❌ Common Errors

  • Forgetting translational KE: Setting mgh = ½Iω² leads to v = √(2gh/A) , completely losing subsequent marks.
  • Failing to show intermediate algebraic steps: In "show that" questions, examiners look closely at how terms cancel. Show the cancellation of r² and m clearly.

Question 02.2

Explaining Differences in Moment of Inertia (1 Mark)

✅ Correct Answer

In the hollow sphere (Q), more of the mass is distributed further away / at a greater distance from the axis of rotation compared to the solid sphere (P).

1 Mark: Clear statement that mass in Q is located at a greater average distance from the axis of rotation.

🧠 Exam Technique & Guidance

  • Definition anchor: Always refer back to I = Σmr² . Moment of inertia depends on both the amount of mass and where that mass is positioned relative to the axis.
  • Watch out: Both spheres have identical total mass m and outer radius r (given in the question). Do not say Q is heavier or larger!
  • Examiners accept phrases like: "mass is distributed further from the axis". References to density are treated as neutral.

Question 02.3

Deducing the Order of Arrival at the Bottom (3 Marks)

📐 Method Comparison & Values

From the formula v = √[ 2gh / (1 + A) ] , the smaller the constant A , the larger the linear velocity v .

Values of shape constant A:
• P (solid sphere): A = 2/5 = 0.40
• R (solid cylinder): A = 1/2 = 0.50
• Q (hollow sphere): A = 2/3 ≈ 0.67
• S (hollow cylinder): A = 1 = 1.00
Alternatively, calculate final velocities:
• vP = √(10gh / 7) ≈ 1.20√(gh) = 3.7√h m s⁻¹
• vR = √(4gh / 3) ≈ 1.15√(gh) = 3.6√h m s⁻¹
• vQ = √(6gh / 5) ≈ 1.10√(gh) = 3.4√h m s⁻¹
• vS = √(gh) = 3.1√h m s⁻¹

✅ Correct Deduction & Final Order

Since initial velocity is zero and distance down the slope is identical for all objects, the object with the greatest linear velocity has the greatest average acceleration and arrives first.

1st: P  →  2nd: R  →  3rd: Q  →  4th: S
Mark Scheme Breakdown:
• Mark 1: Identifies A values (2/5, 2/3, 1/2, 1) OR calculates v for at least one object.
• Mark 2: Clear link: smaller A (or smaller fraction of energy to rotation) means greater v OR calculates all four velocities.
• Mark 3: Correct sequence: P, R, Q, S.

❌ Common Pitfalls

  • Confusing the ranking order: Ranking from largest A to smallest gives slowest-to-fastest (S, Q, R, P) which reaches the bottom last.
  • Vague acceleration arguments: Examiners do not accept arguments solely based on angular acceleration without linking directly to time or linear velocity v .

Topics

Optional topics · 3.11 Engineering physics (A-level only)

Question and mark scheme from the AQA A-Level Physics examination, Paper 3 (3BC), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.