Required Practicals: Mechanics & Materials

This lesson covers the AQA Required Practicals relevant to the Mechanics and Materials section: determination of g by free fall, Young's modulus experiment, resolving forces, and investigation of springs. These practicals are frequently assessed in AQA exams.

Spec mapping (AQA 7408 Required Practicals 1 and 2): This lesson covers RP1 (determination of the acceleration due to gravity, g, by a free-fall method, linking to §3.4.1.2 kinematics) and RP2 (determination of Young's modulus by stretching a wire, linking to §3.4.2.2 mechanical properties of bulk solids). Both practicals are formally assessed under the CPAC criteria (Common Practical Assessment Criteria) for the practical-endorsement element, and the underpinning techniques (graph-plotting with linearisation, uncertainty propagation, error analysis) feature in the written papers. The lesson also covers a Hooke's-law spring investigation and a coplanar-force-equilibrium experiment that develop the same practical skills in the §3.4.1.1 and §3.4.2.1 contexts. (Refer to the official AQA specification document for exact wording.)

Synoptic links: Practical skills here transfer across the A-Level physics course. (i) Kinematics (Lesson 1, §3.4.1.2) is the theoretical foundation of RP1: every g-by-free-fall calculation routes through h = ½gt², and the linearisation by plotting h against t² generalises to all "two-variable function" experiments. (ii) Young's modulus (Lesson 7, §3.4.2.2) is the theoretical foundation of RP2: the practical generates the F-ΔL data needed to compute E, and the diameter-squared dependence makes precision in the diameter measurement the dominant source of uncertainty. (iii) All later practicals (Year 13) — Required Practicals 3-12 — use the same skill set: take repeats, identify random and systematic errors, compute mean and uncertainty, plot a straight-line graph after linearising, and extract the physical quantity from the gradient or intercept. The methodological habits formed here govern the rest of A-Level practical work. (iv) University laboratory work in the first year of physics, engineering, or any natural science course uses the same approach but with more sophisticated tools (oscilloscopes, data loggers, computer-controlled apparatus).

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1 Required Practical: Determination of g by Free Fall

Aim

To measure the acceleration due to gravity using an electromagnetic release and electronic timer.

Equipment

• Electromagnet and steel ball bearing
• Two light gates (or an electromagnetic release with a timer mat)
• Electronic timer or data logger
• Metre rule
• Clamp stand

Method

1. Set up the electromagnet at the top of a clamp stand. Place a timer mat (or lower light gate) at a measured distance h below.
2. Hold the ball bearing with the electromagnet. When the electromagnet is switched off, the timer starts (the circuit is broken) and the ball falls freely.
3. The timer stops when the ball hits the timer mat.
4. Record the time t for each height h.
5. Repeat for at least 6 different heights (e.g., 0.20 m to 1.20 m in 0.20 m intervals).
6. For each height, repeat 3 times and calculate the mean time.

Analysis

From s = ut + ½at² with u = 0:

h = ½gt²

Rearranging: h = ½g × t²

Plot h (y-axis) against t² (x-axis). The graph should be a straight line through the origin.

Gradient = ½g, so g = 2 × gradient.

Worked Example 1

The following data are obtained:

h (m)	t₁ (s)	t₂ (s)	t₃ (s)	t_mean (s)	t² (s²)
0.20	0.201	0.203	0.200	0.201	0.0404
0.40	0.286	0.285	0.287	0.286	0.0818
0.60	0.350	0.351	0.349	0.350	0.1225
0.80	0.404	0.405	0.403	0.404	0.1632
1.00	0.451	0.452	0.450	0.451	0.2034
1.20	0.495	0.494	0.496	0.495	0.2450

Gradient = Δh / Δt² = (1.20 − 0.20) / (0.2450 − 0.0404) = 1.00 / 0.2046 = 4.89 m s⁻²

g = 2 × gradient = 2 × 4.89 = 9.78 m s⁻²

This is within 0.3% of the accepted value of 9.81 m s⁻².

Sources of Uncertainty

Source	Effect	How to Minimise
Reaction time (manual timer)	Random error in t	Use electronic timer
Measuring height	±1 mm uncertainty	Use a metre rule carefully; measure from the bottom of the ball
Air resistance	Slightly reduces g	Use a dense, small ball to minimise drag
Electromagnetic delay	Ball released slightly late	Use a consistent method; the systematic error cancels in the gradient

Exam Tip: AQA often asks why we plot h vs t² rather than h vs t. Answer: h = ½gt² is a linear relationship between h and t², giving a straight line whose gradient directly gives g/2. Plotting h vs t would give a curve, making it harder to determine g accurately.

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2 Required Practical: Determination of Young's Modulus

Aim

To determine the Young's modulus of a material (usually copper or steel wire).

Equipment

• Long thin wire (at least 2 m) clamped at the top
• Metre rule or tape measure
• Micrometer screw gauge
• Masses and hanger
• Marker and ruler (or Searle's apparatus)
• Safety glasses (wire may snap under tension)

Method

1. Measure the original length L of the wire from the clamp to the marker using a metre rule. Record L ± uncertainty.

2. Measure the diameter d of the wire at several points (at least 6 readings) using a micrometer. Calculate the mean diameter. Find the cross-sectional area: A = π(d/2)².

3. Apply loads in equal increments (e.g., 1.0 N steps using 100 g masses). For each load, measure the total extension ΔL using the marker against the ruler.

4. Record loading and unloading readings to check for elastic behaviour and calculate mean extensions.

5. Plot a graph of stress σ (= F/A) against strain ε (= ΔL/L). Alternatively, plot force F against extension ΔL.

Analysis

Method A (stress–strain graph):

Young's modulus E = gradient of the linear region of the stress–strain graph.

Method B (force–extension graph):

Gradient of graph = F/ΔL = k (effective stiffness)

E = k × L/A = gradient × L/A

Worked Example 2

A copper wire: L = 2.50 m, diameter = 0.28 mm (measured as mean of 6 readings).

Force F (N)	Extension ΔL (mm)
0	0
2.0	0.32
4.0	0.63
6.0	0.96
8.0	1.28
10.0	1.61

A = π(0.14 × 10⁻³)² = π × 1.96 × 10⁻⁸ = 6.16 × 10⁻⁸ m²

Gradient of F vs ΔL graph: Using two widely-spaced points:

Gradient = (10.0 − 0) / (1.61 × 10⁻³ − 0) = 10.0 / 1.61 × 10⁻³ = 6211 N m⁻¹

E = gradient × L/A = 6211 × 2.50 / (6.16 × 10⁻⁸) = 15 528 / (6.16 × 10⁻⁸) = 252 × 10⁹ Pa ≈ 252 GPa

Hmm, this is higher than the textbook value of ~130 GPa for copper. Let's recheck: actually, let me recalculate the gradient more carefully.

Gradient = (10.0 − 2.0) / ((1.61 − 0.32) × 10⁻³) = 8.0 / (1.29 × 10⁻³) = 6202 N m⁻¹

E = 6202 × 2.50 / (6.16 × 10⁻⁸) = 15 505 / (6.16 × 10⁻⁸) = 2.52 × 10¹¹ Pa

This large value suggests the wire diameter may have been measured incorrectly in this example (perhaps it was 0.38 mm rather than 0.28 mm), or it was a steel wire. In practice, careful diameter measurement is the largest source of error.

If d = 0.38 mm: A = π(0.19 × 10⁻³)² = 1.134 × 10⁻⁷ m²

E = 6202 × 2.50 / (1.134 × 10⁻⁷) = 15 505 / (1.134 × 10⁻⁷) = 1.37 × 10¹¹ Pa = 137 GPa ≈ 130 GPa ✓ (copper)

This demonstrates the importance of accurate diameter measurement and the propagating effect of the d² term in the area formula.

Worked Example 2b — Estimating g With Realistic Random Error

Consider a more realistic dataset where each timing has random scatter. Suppose a student takes 5 repeated measurements for h = 1.00 m:

Trial	Time t (s)
1	0.448
2	0.456
3	0.452
4	0.449
5	0.453

Mean t = (0.448 + 0.456 + 0.452 + 0.449 + 0.453)/5 = 2.258/5 = 0.4516 s.

Range = 0.456 − 0.448 = 0.008 s. Half-range as uncertainty: ±0.004 s.

Percentage uncertainty in t: 0.004/0.4516 × 100 = 0.89%.

Since g is derived from g = 2h/t², the percentage uncertainty in g picks up twice the percentage uncertainty in t (because of the squared term) plus the percentage uncertainty in h.

If h is measured as 1.00 ± 0.005 m, percentage uncertainty in h: 0.5%. Total percentage uncertainty in g: 0.5 + 2 × 0.89 = 2.28%.

Computed g = 2 × 1.00 / (0.4516)² = 2.00/0.2039 = 9.81 m s⁻². So g = 9.81 ± 0.22 m s⁻², where the absolute uncertainty 0.22 = 2.28% × 9.81.

The accepted value 9.81 m s⁻² lies within the uncertainty bounds — the experiment is consistent with the accepted value, which is the goal.

Sources of Uncertainty

Source	Effect	How to Minimise
Diameter measurement	Largest % uncertainty (enters as d²)	Take ≥6 readings at different points and angles; use a micrometer
Extension measurement	Difficult for small extensions	Use a long wire (increases ΔL); use a vernier scale or travelling microscope
Original length	Small % uncertainty for long wires	Use a long wire (≥2 m)
Zero error on micrometer	Systematic error in d	Check and record zero error before starting
Wire not straight	Initial curvature gives false extension	Apply a small initial load to straighten the wire

Key Point: The diameter d is squared when calculating the area (A = πd²/4). A 5% error in d leads to a ~10% error in E. This is why the diameter measurement is the most critical.

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3 Required Practical: Investigation of Springs (Hooke's Law)

Aim

To investigate the relationship between force and extension for a spring, and to determine the spring constant.

Equipment

• Helical spring
• Clamp stand, boss, and clamp
• Masses and hanger
• Metre rule (or ruler with mm divisions)
• Set square (to avoid parallax)

Method

1. Suspend the spring from the clamp. Attach the mass hanger. Record the initial length of the spring (with the hanger only) using a ruler.
2. Add masses in equal increments (e.g., 50 g = 0.49 N).
3. For each load, record the total length and calculate the extension (current length − original length).
4. Continue until the spring is noticeably stretched beyond the linear region (or until safe limit).
5. Unload in the same increments and record the length for each unloading step.

Analysis

Plot force (y-axis) against extension (x-axis).

• The linear region confirms Hooke's law (F ∝ x).
• Gradient = spring constant k (in N m⁻¹).
• The limit of proportionality is where the graph starts to curve.
• If loading and unloading curves coincide → elastic behaviour.
• If they don't coincide → the elastic limit has been exceeded (permanent deformation).

Worked Example 3

Mass (g)	Force (N)	Length (mm)	Extension (mm)
0 (hanger)	0.49	120	0
50	0.98	132	12
100	1.47	144	24
150	1.96	156	36
200	2.45	168	48
250	2.94	181	61
300	3.43	196	76

The first five points (0 to 200 g added) are linear. Using these:

Gradient = ΔF/Δx = (2.45 − 0.49) / ((48 − 0) × 10⁻³) = 1.96 / 0.048 = 40.8 N m⁻¹

The data for 250 g and 300 g deviate from linearity — the spring has passed its limit of proportionality.

The elastic potential energy stored at 200 g (within Hooke's law):

E = ½kx² = ½ × 40.8 × (0.048)² = ½ × 40.8 × 0.002304 = 0.047 J

Or: E = ½Fx = ½ × 1.96 × 0.048 = 0.047 J ✓

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4 Required Practical: Resolving Forces (Using a Force Board)

Aim

To verify that forces can be resolved and that the conditions for equilibrium can be predicted using vector addition.

Equipment

• Force board (vertical board with pulleys at the edges)
• Strings with knot at the centre
• Masses to provide known forces
• Protractor to measure angles
• Sheet of paper behind the strings to mark positions

Method

1. Arrange three strings from a central knot, each passing over a pulley at the edge of the board, with masses hanging from each string.
2. Allow the system to reach equilibrium.
3. Mark the position of each string on the paper behind.
4. Record the force (weight = mg) and angle for each string.
5. Draw a scale diagram: place the three force vectors nose-to-tail. If the system is in equilibrium, they form a closed triangle.

Analysis — Using Components

For three forces F₁, F₂, F₃ at angles θ₁, θ₂, θ₃ to the horizontal:

Equilibrium requires:

• ΣFₓ = F₁ cos θ₁ + F₂ cos θ₂ + F₃ cos θ₃ = 0
• ΣFᵧ = F₁ sin θ₁ + F₂ sin θ₂ + F₃ sin θ₃ = 0

Worked Example 4

Three forces are in equilibrium:

• F₁ = 3.0 N at 0° (horizontal, to the right)
• F₂ = 4.0 N at 90° (vertically upward)
• F₃ = unknown, at angle θ below the horizontal to the left

Find F₃ and θ.

Solution:

ΣFₓ = 0: 3.0 + F₃ cos(180° + θ) = 0 → 3.0 − F₃ cos θ = 0 → F₃ cos θ = 3.0

ΣFᵧ = 0: 4.0 + F₃ sin(180° + θ) = 0 → 4.0 − F₃ sin θ = 0 → F₃ sin θ = 4.0

F₃ = √(3.0² + 4.0²) = √(9 + 16) = √25 = 5.0 N

θ = tan⁻¹(4.0/3.0) = 53.1° below the horizontal to the left

This confirms F₃ is the resultant of F₁ and F₂ reversed in direction.

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5 Practical Skills — Uncertainties and Error Analysis

AQA exams frequently test practical skills. Here are the key concepts:

Types of Error

Type	Description	How to Deal With It
Random	Unpredictable fluctuations in readings	Take repeats and calculate mean
Systematic	Consistent offset in all readings	Identify and correct the cause
Zero error	Instrument does not read zero when it should	Record and subtract the zero error
Parallax	Reading taken from wrong angle	Use a set square or read at eye level

Calculating Percentage Uncertainty

% uncertainty = (absolute uncertainty / measured value) × 100%

For derived quantities:

• Addition/subtraction: Add absolute uncertainties
• Multiplication/division: Add percentage uncertainties
• Powers: Multiply percentage uncertainty by the power

Worked Example 5 — Propagating Uncertainties for Young's Modulus

Given: F = 10.0 ± 0.1 N, L = 2.500 ± 0.002 m, d = 0.38 ± 0.01 mm, ΔL = 1.60 ± 0.05 mm

E = FL/(AΔL) = 4FL/(πd²ΔL)

% uncertainty in F = 0.1/10.0 × 100 = 1.0%