Pearson Edexcel GCSE Combined Science · P15

Forces and matter

50 revision questions with answers for P15 Forces and matter in Pearson Edexcel GCSE Combined Science (spec 1SC0). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below.

All 50 questions and answers

Question 1 · Spec P15.1
Why does stretching an object require more than one force?
Answer:
Opposing forces must act at different points to change shape rather than simply move the whole object.
Question 2 · Spec P15.2
What is elastic distortion?
Answer:
A shape change that reverses when the deforming forces are removed.
Question 3 · Spec P15.2
What is inelastic distortion?
Answer:
A shape change that is not fully reversed when the forces are removed.
Question 4 · Spec P15.1
Name three ways forces can distort an object.
Answer:
Stretching, bending and compressing.
Question 5 · Spec P15.3
State the linear spring force-extension equation.
Answer:
F = kx, where x is extension and k is spring constant.
Question 6 · Spec P15.3
What units are used for spring constant and extension?
Answer:
N/m for spring constant and m for extension.
Question 7 · Spec P15.3
How does extension differ from total spring length?
Answer:
Extension = loaded length - unloaded length.
Question 8 · Spec P15.3
When is force proportional to extension?
Answer:
In the spring's linear region, before its limit of proportionality is exceeded.
Question 9 · Spec P15.3
Find force for k = 120 N/m and extension 0.050 m.
Answer:
F = 120 × 0.050 = 6.0 N.
Question 10 · Spec P15.3
A spring extends 0.040 m under 8.0 N. Find k.
Answer:
k = F/x = 8.0/0.040 = 200 N/m.
Question 11 · Spec P15.3
Find extension under 15 N for k = 300 N/m.
Answer:
x = F/k = 15/300 = 0.050 m.
Question 12 · Spec P15.3
Convert a 6.0 cm extension to metres.
Answer:
6.0 cm = 0.060 m.
Question 13 · Spec P15.3
What gives spring constant on a force-vertical, extension-horizontal graph?
Answer:
The gradient of the straight-line region.
Question 14 · Spec P15.5
What graph shape shows a linear force-extension relationship?
Answer:
A straight line through the origin when extension is zero at zero force.
Question 15 · Spec P15.5
What does a curved force-extension graph show?
Answer:
Force is not directly proportional to extension over that region.
Question 16 · Spec P15.5
Does a non-linear force-extension relationship necessarily mean permanent deformation?
Answer:
No. Some materials behave non-linearly but still return to their original shape.
Question 17 · Spec P15.2
How can you test whether deformation was elastic?
Answer:
Remove the force and check whether the original shape or length is restored.
Question 18 · Spec P15.1
Which opposing forces stretch a hanging spring?
Answer:
The support acts on one end and the attached load pulls on the other.
Question 19 · Spec P15.1
How can forces bend a ruler?
Answer:
Forces acting at different positions and directions change its shape between the supports or loading points.
Question 20 · Spec P15.1
Why does compressing a spring require opposing forces?
Answer:
The ends must be pushed towards each other to shorten the spring.
Question 21 · Spec P15.4
What does area under a force-extension graph represent?
Answer:
Work done in changing the extension, equal to energy transferred to the spring in an ideal elastic stretch.
Question 22 · Spec P15.6 · Core practical
Outline the spring core practical.
Answer:
Measure unloaded length, add known loads in steps, record length and extension, then plot force against extension.
Question 23 · Spec P15.6 · Core practical
Why measure unloaded length before adding masses?
Answer:
It provides the reference needed to calculate extension.
Question 24 · Spec P15.6 · Core practical
How is force found from an added mass?
Answer:
Use weight = mg; for example, 0.10 kg gives 1.0 N with g = 10 N/kg.
Question 25 · Spec P15.6 · Core practical
Why fix a ruler close to the spring?
Answer:
It provides a consistent scale and reduces reading uncertainty.
Question 26 · Spec P15.6 · Core practical
Why read the spring's marker at eye level?
Answer:
To reduce parallax error.
Question 27 · Spec P15.6 · Core practical
Why wait for oscillations to stop before reading length?
Answer:
Movement makes the position uncertain; a stationary reading is more consistent.
Question 28 · Spec P15.6 · Core practical
Why secure the stand and avoid excessive loading?
Answer:
It prevents apparatus toppling, spring damage or falling masses.
Question 29 · Spec P15.6 · Core practical
What precaution can reduce injury from a falling mass or recoiling spring?
Answer:
Keep feet clear, use an appropriate catch arrangement and wear suitable eye protection.
Question 30 · Spec P15.6 · Core practical
Why repeat readings for each load?
Answer:
To assess variation and calculate a more reliable mean extension.
Question 31 · Spec P15.6 · Core practical
Why take readings while unloading as well as loading?
Answer:
It checks whether the spring returns along the same relationship and regains its original length.
Question 32 · Spec P15.6 · Core practical
Why choose loads within the spring's safe working range?
Answer:
Excess force can cause permanent distortion or failure, invalidating the intended elastic investigation.
Question 33 · Spec P15.6 · Core practical
Why use several well-spaced loads instead of only two?
Answer:
More points reveal whether the relationship is linear and make its gradient more dependable.
Question 34 · Spec P15.6 · Core practical
What should stay the same when testing load against extension?
Answer:
The spring, support arrangement, ruler reference and measurement method.
Question 35 · Spec P15.6 · Core practical
Why can very small extensions have large percentage uncertainty?
Answer:
A fixed ruler uncertainty is a large fraction of a small extension.
Question 36 · Spec P15.4
State the energy equation for a linearly elastic spring.
Answer:
E = 0.5kx².
Question 37 · Spec P15.4
Find energy for k = 200 N/m and x = 0.10 m.
Answer:
E = 0.5 × 200 × 0.10² = 1.0 J.
Question 38 · Spec P15.4
How does doubling extension change stored energy in the linear region?
Answer:
Energy becomes four times larger.
Question 39 · Spec P15.4
At the same extension, how does doubling k change stored energy?
Answer:
It doubles the stored energy.
Question 40 · Spec P15.4
A spring stores 0.40 J at 0.080 m extension. Find k.
Answer:
k = 2E/x² = 0.80/0.0064 = 125 N/m.
Question 41 · Spec P15.4
A 100 N/m spring stores 0.50 J. Find extension.
Answer:
x² = 2E/k = 0.010; x = 0.10 m.
Question 42 · Spec P15.4
A linear graph reaches 10 N at 0.20 m. Find work from its area.
Answer:
Triangular area = 0.5 × 10 × 0.20 = 1.0 J.
Question 43 · Spec P15.4
Why does work for a linear stretch equal half final force times extension?
Answer:
Force rises from zero to its final value, so average force is half the final force.
Question 44 · Spec P15.4
Can 0.5 x final force x extension always be used for a curved graph?
Answer:
No. It assumes a straight-line relationship; use the actual area for a non-linear graph.
Question 45 · Spec P15.4
Find energy for k = 50 N/m and extension 4 cm.
Answer:
4 cm = 0.040 m; E = 0.5 × 50 × 0.040² = 0.040 J.
Question 46 · Spec P15.2
What can stored elastic energy do when the force is removed?
Answer:
It can transfer to kinetic energy as the spring returns towards its original shape.
Question 47 · Spec P15.2
Why may an unloading cycle return less energy than was supplied?
Answer:
Some energy is dissipated to thermal stores during deformation and recovery.
Question 48 · Spec P15.3
Two springs have k values 80 and 160 N/m. Which extends farther under the same force?
Answer:
The 80 N/m spring; its extension is twice as large in their linear regions.
Question 49 · Spec P15.6 · Core practical
Why should a force-extension plot use extension rather than total length?
Answer:
Total length includes the unloaded length, which would give an incorrect origin and interpretation.
Question 50 · Spec P15.6 · Core practical
How should a spring practical conclusion be justified?
Answer:
Use the graph's shape, gradient and unloading evidence to state proportionality, spring constant and whether deformation remained elastic.

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