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Required practicals and working scientifically – GCSE Combined Science Physics

Foundation and Higher · Grade 5

Jump to the questions and answers

How to plan a fair test, take careful measurements, spot errors and draw conclusions, using the Physics practicals on heating, resistance, springs, motion and waves.

Objectives

Specification references this topic covers: AQA 6.1.1, 6.3.2; Edexcel P2.11, P2.19, P4.17, P6.26, P10.4, P10.7, P10.12, P10.13.

Key equations

force applied to a spring = spring constant × extension
F=k eF = k\,e
  • F force (N)
  • k spring constant (N/m)
  • e extension (m)
wave speed = frequency × wavelength
v=f λv = f\,\lambda
  • v wave speed (m/s)
  • f frequency (Hz)
  • λ wavelength (m)

Required practicals and working scientifically explained

Start here

Imagine you and a friend argue about which trainers grip best. You slide one pair down a wet ramp and the other pair down a dry ramp, and the dry one wins. Is that fair?

Of course it is not. The ramp was different, so you cannot tell whether the trainers made the difference. A Physics practical is just a careful way of answering a question without fooling yourself.

Think of baking. If you change the oven temperature, the flour and the time all at once, you will never know which one burnt the cake. Change one thing, keep the rest the same, and the result tells you something.

In GCSE Physics you meet a set of practical investigations. They cover heating a block, resistance and I-V graphs, springs, motion and waves. You do not need to learn a perfect script, you need to understand why each step is there.

Exam questions may ask about a practical you have done, or about a new one that uses the same skills. So the skills matter more than memorising one method. The chart shows the path that almost every practical follows, and each step in this lesson fits somewhere along it.

By the end you will be able to do three things. You will name the variables in an investigation and plan a fair test. You will take careful measurements and spot errors, and you will use graphs to reach a conclusion.

ask a questionplan a fairtestmeasure andrepeatplot a graphwrite aconclusion
A flow chart of a practical: ask a question, plan a fair test, measure and repeat, plot a graph, then write a conclusion.

Planning a fair test

A fair test changes only one thing, so that you know what caused any difference you see. A variable is anything that can change in an investigation. The variable you change on purpose is called the independent variable.

The variable you measure to see what happened is the dependent variable. It depends on the independent variable, which is an easy way to remember it. In the trainers test, the independent variable is the type of trainer and the dependent variable is how far each one slides.

Everything else must stay the same, and these are called the control variables. If a control variable changes, it might be the real cause of your result and you would never know. For the ramp, the controls would be the slope, the surface, how wet it is and the size of the push.

Now a Physics example. To test how the length of a wire affects its resistance, the length is independent and the resistance is dependent. The wire material, the thickness and the temperature are all controlled, because each of them would change the resistance too.

Before you start, write a prediction, which is what you think will happen and why, because a guess with a reason is much more useful than a guess on its own. A good prediction has a reason, such as a longer wire gives the electrons more to collide with, so the resistance will be bigger. Afterwards you can say whether the results agree with it.

Measuring well

Repeat each measurement at least three times and find the mean. The mean is the total of the readings divided by the number of readings, so for 4 s, 5 s and 6 s it is 15 ÷ 3 = 5 s. Repeating helps to cancel out random errors, which are small unpredictable differences, such as your reaction time on a stopwatch.

A reading that is far from the others is called an anomaly. In the bars, three readings are close to 4 s and one is 6.9 s, so the 6.9 s was probably a slip, such as starting the stopwatch late. Leave the anomaly out of the mean, and say why you left it out.

Choose the right instrument. A measuring cylinder is better than a beaker for a volume, and a ruler marked in millimetres is better than one marked in centimetres. The smallest change an instrument can show is called its resolution, and a smaller resolution gives you more detail.

Some errors push every reading the same way, and this is called a systematic error. A balance that reads 2 g with nothing on it has a zero error, so every mass you measure is 2 g too big, however many times you repeat it. Checking that the balance reads zero before you start fixes this.

Read scales at eye level, because looking from the side makes the reading wrong. Write the unit with every number, such as seconds for time and newtons for force. A number without a unit is only half an answer, and it often loses the mark.

4.1reading 14.2reading 26.9reading 34reading 4
Bars showing four timing readings. Three are close together at about 4 seconds and the third reading is much larger, at 6.9 seconds.

Specific heat capacity

Different materials need different amounts of energy to warm up. A kettle of water takes a long time to heat, but a metal spoon in it warms up quickly. Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1 °C, so water has a large one and metals have smaller ones.

Here is the method. Put a metal block on a balance to find its mass, in kilograms. Push a heater and a thermometer into the holes in the block, with a little oil in the thermometer hole so that heat passes well from the block to the thermometer.

Wrap the block in insulation, which is a material that slows down heat escaping. Switch on the heater and write down the starting temperature. After a known time, record the final temperature and work out the temperature rise.

Find the energy supplied from a joule meter, or use energy = power × time. Then specific heat capacity = energy ÷ (mass × temperature rise), written c = ΔE ÷ (m × Δθ). The unit is joules per kilogram per degree Celsius, J/kg °C.

Some energy escapes to the surroundings, so the block warms less than it should. This makes the temperature rise too small and the answer for c too big. Insulation cuts the loss, and you should wait for the temperature to stop rising before you read the highest value, because the heat takes a moment to spread through the block.

weigh theblockfit heater andthermometerwrap ininsulationheat for aknown timework out thetemperaturerise
A flow chart of the specific heat capacity practical: weigh the block, fit the heater and thermometer, wrap in insulation, heat for a known time, then work out the temperature rise.

Resistance and meters

Resistance tells you how hard it is for current to flow, a bit like a narrow pipe makes it hard for water to flow. A big resistance means a small current for the same push from the battery. We measure resistance in ohms, Ω.

In the circuit, the component goes in series with an ammeter and a variable resistor. Series means they are in one loop, one after another, so the same current goes through each. The ammeter measures that current in amps, A, and the variable resistor lets you change it in small steps.

The voltmeter goes in parallel, which means on its own branch across the component. It measures the potential difference, which is the voltage across the component, in volts, V. Mixing up series and parallel is one of the commonest mistakes, so say it to yourself: ammeter in series, voltmeter in parallel.

Change the variable resistor to alter the current, and record a pair of readings each time. Then work out resistance = potential difference ÷ current, or R = V ÷ I. Use it whenever you know the volts and the amps, for example 9.0 V and 0.50 A gives 9.0 ÷ 0.50 = 18 Ω.

To test a wire, change its length instead and keep the current small. A wire that heats up changes its resistance, and that would spoil the test. Switch off between readings so that the wire stays cool, and wait a moment for each reading to settle.

batteryvariable resistorAammetercomponentVvoltmeter
A circuit with a battery, variable resistor, ammeter and a component in one loop. A voltmeter is connected in parallel across the component.

I-V graphs and conclusions

Plot current up the side and potential difference along the bottom, with the independent variable along the bottom. The shape of the graph tells you how the component behaves, which is why these graphs are called characteristics. A resistor at constant temperature gives a straight line through the origin.

A straight line through the origin means the current is directly proportional to the potential difference. Double the potential difference and the current doubles too, so the resistance stays the same. A steeper line means a smaller resistance.

A filament lamp gives a curve. The thin wire gets very hot, the metal ions vibrate more, and the electrons collide with them more often. So the resistance goes up and the line flattens.

A diode lets current flow in one direction only, which is why it is used to protect circuits from being connected backwards. If the potential difference is reversed, the current is almost zero. The right way round, it flows once the potential difference is big enough.

Draw a smooth line of best fit through the points, not a zigzag that joins every dot. A good conclusion matches the data and says what range was tested. It must not claim more than the evidence shows, so do not say what happens far outside the readings you took.

000.80.81.61.62.42.43.23.244potential difference (arbitrary units)current (arbitrary units)resistorfilament lampdiode
Three I-V graphs. The resistor is a straight line through the origin, the filament lamp is a curve that flattens, and the diode stays near zero and then rises steeply.

Springs, motion and waves

Springs: hang a spring from a clamp and measure its natural length, which is its length with nothing on it. Add a 100 g mass at a time, measure the new length, and work out extension = new length − natural length. The force on the spring is the weight of the masses, and a ruler fixed next to the spring helps you read the length.

Weight = mass × g, so a 200 g mass is 0.2 kg, and 0.2 × 10 = 2 N. Take g = 10 N/kg. Plot force against extension, and for a spring within its limit the graph is a straight line through the origin.

That straight line means force = spring constant × extension, or F = k × e. The spring constant, k, is in newtons per metre, N/m, and it is the gradient of the straight part of the graph. A stiff spring has a large k and a stretchy one has a small k, and a spring stretched too far no longer goes back to its first length.

Motion: a trolley is pulled along a track by a falling weight, and light gates measure its acceleration, which is how quickly its speed changes. Move masses from the trolley to the hanger to change the pulling force while the total mass stays the same. You can then show that acceleration is directly proportional to force, which fits force = mass × acceleration.

Waves: in a ripple tank, count the waves passing a point in a measured time to find the frequency, because frequency is the number of waves each second. Measure the wavelength with a ruler held against a picture of the waves. Then use wave speed = frequency × wavelength, or v = f × λ, with frequency in hertz, wavelength in metres and speed in m/s.

200 gextensionweightspring forceF = k × e, and weight = m × g
A spring hanging from a clamp with a 200 gram load. The extension is marked beside it, with the spring force and the weight arrows.

Recap

A fair test changes one variable, measures one variable and controls the rest. Repeat readings, find the mean, and leave out any anomaly with a reason. Insulate a heated block to reduce energy lost.

Specific heat capacity needs energy, mass and temperature rise: c = ΔE ÷ (m × Δθ). Resistance: R = V ÷ I, with the ammeter in series and the voltmeter in parallel. A resistor gives a straight I-V line, a lamp gives a curve, and a diode lets current through one way only.

Springs: F = k × e, and the spring constant is the gradient. Trolley: keep the total mass the same while you change the force. Ripple tank: v = f × λ.

On a graph, a straight line through the origin means directly proportional. A conclusion should match the data and stay within the range you tested. The table pulls the practicals together.

How this is asked in an exam: a method question wants steps in order, saying what you change, what you measure and what you keep the same. A suggestion question wants one clear improvement and the reason it helps, such as insulation to cut energy loss. A calculation wants the equation, the numbers and the unit.

Planning questions are often the easiest marks to win, because you can use common sense. Name what you change, what you measure and what you keep the same, and you have already earned most of the marks. Then add one sentence about repeating the readings.

Try this before you leave: choose any practical and say aloud the independent, dependent and control variables. If you can do that for each one, you are well prepared for the planning questions.

Key points

Worked examples

Example 1

A student times a trolley three times and gets 2.4 s, 2.5 s and 2.3 s. Calculate the mean time.

  1. Check for an anomaly: 2.4 s, 2.5 s and 2.3 s are all close together, so keep all three.
  2. Add the readings: 2.4 + 2.5 + 2.3 = 7.2 s.
  3. Divide by the number of readings: 7.2 ÷ 3.
  4. The mean is 2.4 s.

Answer: 2.4 s

Common slips
  • Slip: Throw away 2.5 s because it is the biggest. 2.5 s is close to the other readings, so it is not an anomaly. Keep it.
  • Slip: 2.4 + 2.5 + 2.3 = 6.2 s. Check the addition: the total is 7.2.
  • Slip: Divide by 2 because there are two decimal places. Divide by the number of readings, which is 3.
  • Slip: The mean is 7.2 s. 7.2 s is the total. The mean is the total divided by 3.

Example 2

A heater supplies 6300 J of energy to 0.30 kg of water. The temperature rises by 5.0 °C. Calculate the specific heat capacity of the water.

  1. Write the equation: specific heat capacity = energy ÷ (mass × temperature rise).
  2. Multiply the mass by the temperature rise: 0.30 × 5.0 = 1.5.
  3. Divide the energy by this number: 6300 ÷ 1.5.
  4. Write the answer with its unit: 4200 J/kg °C.

Answer: 4200 J/kg °C

Common slips
  • Slip: c = energy × mass × temperature rise. Divide the energy by mass × temperature rise. Multiplying them all together is wrong.
  • Slip: 0.30 + 5.0 = 5.3 The mass and the temperature rise are multiplied, not added.
  • Slip: c = 6300 ÷ 5.0 = 1260 The mass must be in the division too. Divide by 1.5, not 5.0.
  • Slip: c = 4200 J Specific heat capacity is in J/kg °C, not just joules.

Example 3

A student measures 3.0 V across a resistor and 0.25 A through it. Calculate the resistance, then explain why the resistance of a filament lamp rises as the lamp gets hotter.

  1. Write the equation: resistance = potential difference ÷ current.
  2. Substitute: 3.0 ÷ 0.25 = 12 Ω.
  3. In a hot filament the metal ions vibrate more.
  4. The electrons collide with the ions more often, so the resistance increases.

Answer: 12 Ω, and the resistance rises because the electrons collide with the vibrating ions more often

Common slips
  • Slip: resistance = current ÷ potential difference. This is upside down. Divide the potential difference by the current.
  • Slip: 3.0 × 0.25 = 0.75 Ω Divide the potential difference by the current. Do not multiply.
  • Slip: In a hot filament the ions vibrate less. Heating makes the ions vibrate more, not less.
  • Slip: The electrons move faster so they collide less. More vibrating ions mean more collisions, not fewer.

Questions and answers

Required practicals and working scientifically practice questions for GCSE Combined Science Physics. Try each one, then open the answer. Easy questions come first.

Question 1 Easy

Name the variable that you change on purpose in an investigation.

Show answer

Answer: the independent variable

  1. You choose its values.
  2. This is the independent variable.

Question 2 Easy

Three readings of a length are 12 cm, 14 cm and 13 cm. Calculate the mean.

Show answer

Answer: 13 cm

  1. 12 + 14 + 13 = 39.
  2. 39 ÷ 3 = 13 cm.

Question 3 Easy

Which meter is connected in series with a component?

Show answer

Answer: ammeter

  1. The same current flows through the component and the meter.
  2. This is the ammeter.

Question 4 Easy

Which meter is connected in parallel across a component?

Show answer

Answer: voltmeter

  1. It measures the potential difference across the component.
  2. This is the voltmeter.

Question 5 Medium

A heater supplies 9000 J to a 0.50 kg metal block. The temperature rises by 10 °C. Calculate the specific heat capacity.

Show answer

Answer: 1800 J/kg °C

  1. 0.50 × 10 = 5.
  2. 9000 ÷ 5 = 1800 J/kg °C.

Question 6 Medium

A student measures 6.0 V across a resistor and 0.50 A through it. Calculate the resistance.

Show answer

Answer: 12 Ω

  1. resistance = potential difference ÷ current.
  2. 6.0 ÷ 0.50 = 12 Ω.

Question 7 Medium

Four readings are 3.1 s, 3.2 s, 5.8 s and 3.0 s. State the anomalous reading.

Show answer

Answer: 5.8 s

  1. Three readings are close to 3 s.
  2. The reading far from the others is 5.8 s.

Question 8 Medium

A spring extends by 0.040 m when a force of 2.0 N is applied. Calculate the spring constant.

Show answer

Answer: 50 N/m

  1. spring constant = force ÷ extension.
  2. 2.0 ÷ 0.040 = 50 N/m.

Exam-style questions

Longer questions in the style of a written paper, with the marking points. These are Yumio practice questions.

Exam-style question 14 marks

A student measures the specific heat capacity of a metal block. The heater supplies 4800 J to a 0.40 kg block and the temperature rises by 12.0 °C. (a) Calculate the specific heat capacity of the metal. [3 marks] (b) Suggest one way to make the result more accurate. [1 mark]

Show answer and marking points

Answer: (a) 1000 J/kg °C (b) Insulate the block to reduce energy lost to the surroundings.

  • (a) specific heat capacity = energy ÷ (mass × temperature rise) (1)
  • (a) 4800 ÷ (0.40 × 12.0) (1)
  • (a) 1000 J/kg °C (1)
  • (b) insulate the block (1)

Exam-style question 23 marks

Describe how to investigate how the current through a lamp changes with the potential difference across it. [3 marks]

Show answer and marking points

Answer: Put an ammeter in series with the lamp and a voltmeter across it. Use a variable resistor to change the current. Record pairs of readings and plot a graph of current against potential difference.

  • ammeter in series with the lamp and voltmeter in parallel across it (1)
  • change the current with a variable resistor and take several readings (1)
  • plot current against potential difference (1)

Common mistakes

  1. Changing more than one variable at the same time

    Change one variable and control the rest, so you know what caused the result.

  2. Including an anomalous result in the mean

    Leave the anomaly out, say why, then find the mean of the rest.

  3. Connecting the voltmeter in series

    The voltmeter goes in parallel across the component. The ammeter goes in series.

  4. Drawing a line of best fit that joins every point

    Draw a smooth or straight line that passes close to the points, with points on both sides.

  5. Using grams in F = m × g without changing to kilograms

    Change grams to kilograms first. 200 g is 0.2 kg, so the weight is 2 N.

Questions students ask

Is Required practicals and working scientifically on the Foundation or Higher tier?

Required practicals and working scientifically is on both Foundation and Higher papers for GCSE Combined Science Physics, aimed at around grade 5.

What is a common mistake on Required practicals and working scientifically?

Changing more than one variable at the same time. Change one variable and control the rest, so you know what caused the result.

Where are the answers to the Required practicals and working scientifically questions?

Open Show answer under each question on this page. The questions are Yumio practice questions, not past exam questions.

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