AQA GCSE Combined Science: Trilogy · 6.5

Forces

50 revision questions with answers for 6.5 Forces in AQA GCSE Combined Science: Trilogy (spec 8464). Practise with flashcards, a topic quiz, a random quiz across the course, 5 crosswords, or read every question and answer below. Not official AQA material.

All 50 questions and answers

Question 1 · Spec 6.5.1.1
How do scalar and vector quantities differ?
Answer:
Scalars have magnitude only; vectors have magnitude and direction. Speed is scalar; velocity and force are vectors.
Question 2 · Spec 6.5.1.2
What is a contact force? Give three examples.
Answer:
A force between touching objects; examples are friction, tension and normal contact force.
Question 3 · Spec 6.5.1.2
What is a non-contact force? Give three examples.
Answer:
A force between separated objects; examples are gravitational, magnetic and electrostatic forces.
Question 4 · Spec 6.5.1.3
How does weight differ from mass, and what equation links them?
Answer:
Mass is measured in kg; weight is gravitational force in N. W = mg, with g in N/kg.
Question 5 · Spec 6.5.1.3
Find the weight of a 60 kg person where g = 9.8 N/kg.
Answer:
W = 60 × 9.8 = 588 N.
Question 6 · Spec 6.5.1.3
Where can an object's weight be considered to act, and how is weight measured?
Answer:
At its centre of mass. Weight is measured with a calibrated spring balance or newtonmeter.
Question 7 · Spec 6.5.1.4
Forces of 12 N right and 7 N left act on an object. Find the resultant.
Answer:
12 − 7 = 5 N to the right.
Question 8 · Spec 6.5.1.4 · Higher only
What does a free-body diagram show?
Answer:
Forces acting on one chosen object, drawn as labelled arrows; arrow directions and lengths indicate force directions and magnitudes.
Question 9 · Spec 6.5.1.4 · Higher only
How can angled forces be analysed without changing their overall effect?
Answer:
Resolve a force into two perpendicular components, or use an accurate scale vector diagram to find the resultant's magnitude and direction.
Question 10 · Spec 6.5.2
When is mechanical work done, and what is its equation?
Answer:
When a force causes displacement. W = Fs, using distance moved along the force's line of action.
Question 11 · Spec 6.5.2
A 25 N force moves an object 4.0 m along its direction. Find work done.
Answer:
W = 25 × 4.0 = 100 J.
Question 12 · Spec 6.5.2
Where does energy transferred by work against friction go?
Answer:
It increases thermal energy stores of the object and its surroundings.
Question 13 · Spec 6.5.3
How do elastic and inelastic deformation differ?
Answer:
An elastic object returns to its original shape when force is removed; an inelastically deformed object does not fully return.
Question 14 · Spec 6.5.3
State Hooke's law and its condition.
Answer:
F = ke: force is proportional to extension, provided the limit of proportionality is not exceeded.
Question 15 · Spec 6.5.3
A spring has k = 80 N/m and extension 0.050 m. Find force.
Answer:
F = 80 × 0.050 = 4.0 N.
Question 16 · Spec 6.5.3 · Required practical
How can force and spring extension be investigated?
Answer:
Measure original length, add known forces in steps, measure new lengths and calculate extensions, staying within safe loading limits.
Question 17 · Spec 6.5.3 · Required practical
What does a straight force-extension graph through the origin show?
Answer:
Force is proportional to extension; its gradient gives the spring constant if force is on the vertical axis.
Question 18 · Spec 6.5.3
How is energy stored in an elastically stretched spring calculated?
Answer:
Ee = 0.5ke², within the elastic limit, using extension in metres.
Question 19 · Spec 6.5.3
A spring has length 15 cm unloaded and 19 cm loaded. What extension is used in calculations?
Answer:
19 − 15 = 4 cm = 0.040 m; use extension, not total length.
Question 20 · Spec 6.5.4.1.1
How does displacement differ from distance?
Answer:
Distance is total path length. Displacement is the straight-line change from start to finish, including direction.
Question 21 · Spec 6.5.4.1.2
Give typical walking, running, cycling and sound-in-air speeds.
Answer:
Walking about 1.5 m/s; running 3 m/s; cycling 6 m/s; sound in air about 330 m/s.
Question 22 · Spec 6.5.4.1.2
State the equation for distance travelled at constant speed.
Answer:
Distance = speed x time, or s = vt.
Question 23 · Spec 6.5.4.1.2
A journey covers 900 m in 60 s. Find average speed.
Answer:
Average speed = total distance / total time = 900 / 60 = 15 m/s.
Question 24 · Spec 6.5.4.1.3
What is velocity?
Answer:
Speed in a stated direction, so it is a vector quantity.
Question 25 · Spec 6.5.4.1.3 · Higher only
Why does travelling around a circle at constant speed involve acceleration?
Answer:
Velocity changes because direction changes; a change in velocity means acceleration.
Question 26 · Spec 6.5.4.1.4
How can speed be found from a distance-time graph?
Answer:
Calculate its gradient: change in distance / change in time. A horizontal section represents no movement.
Question 27 · Spec 6.5.4.1.4 · Higher only
How is instantaneous speed found from a curved distance-time graph?
Answer:
Draw a tangent at the required time and calculate the tangent's gradient.
Question 28 · Spec 6.5.4.1.5
State the acceleration equation and units.
Answer:
a = (v - u)/t; acceleration in m/s², velocities in m/s and time in s.
Question 29 · Spec 6.5.4.1.5
Velocity rises from 5.0 to 17.0 m/s in 4.0 s. Find acceleration.
Answer:
a = (17.0 − 5.0) / 4.0 = 3.0 m/s².
Question 30 · Spec 6.5.4.1.5
What does the gradient of a velocity-time graph give?
Answer:
Acceleration; a negative gradient indicates acceleration in the negative direction, often deceleration.
Question 31 · Spec 6.5.4.1.5 · Higher only
A velocity-time graph stays at 8.0 m/s for 5.0 s. Find displacement.
Answer:
Area under the graph = 8.0 × 5.0 = 40 m in the positive direction.
Question 32 · Spec 6.5.4.1.5
State the equation linking uniform acceleration, distance and initial and final speeds.
Answer:
v² - u² = 2as; it applies to motion with constant acceleration.
Question 33 · Spec 6.5.4.1.5
What is free-fall acceleration close to Earth's surface?
Answer:
About 9.8 m/s² downward when air resistance is negligible.
Question 34 · Spec 6.5.4.1.5
How does a falling object reach terminal velocity?
Answer:
As speed rises, drag increases until it balances weight. Resultant force and acceleration become zero, so speed stays constant.
Question 35 · Spec 6.5.4.2.1
State Newton's First Law.
Answer:
If resultant force is zero, an object stays at rest or continues moving at constant velocity.
Question 36 · Spec 6.5.4.2.2
State Newton's Second Law.
Answer:
Resultant force = mass x acceleration, F = ma. At fixed mass, acceleration is proportional to resultant force.
Question 37 · Spec 6.5.4.2.2
A resultant force of 1,200 N acts on a 600 kg object. Find acceleration.
Answer:
a = F/m = 1,200 / 600 = 2.0 m/s².
Question 38 · Spec 6.5.4.2.1 · Higher only
What is inertia?
Answer:
The tendency of an object to remain at rest or continue at constant velocity unless a resultant force acts.
Question 39 · Spec 6.5.4.2.2 · Higher only
What is inertial mass?
Answer:
A measure of resistance to a change in velocity, defined by the ratio of resultant force to acceleration: m = F/a.
Question 40 · Spec 6.5.4.2.2 · Required practical
How can force's effect on acceleration be investigated?
Answer:
Vary the pulling force on a trolley while keeping total moving mass constant; measure acceleration using suitable motion sensors or light gates.
Question 41 · Spec 6.5.4.2.2 · Required practical
How can mass's effect on acceleration be investigated?
Answer:
Keep pulling force constant, vary total moving mass and measure acceleration; control track slope and minimise friction.
Question 42 · Spec 6.5.4.2.3
State Newton's Third Law and distinguish its force pair from balanced forces on one object.
Answer:
Interacting objects exert equal opposite forces on each other. The pair acts on different objects, whereas balanced forces act on the same object.
Question 43 · Spec 6.5.4.3.1
What is stopping distance?
Answer:
Thinking distance plus braking distance.
Question 44 · Spec 6.5.4.3.2
Which factors can increase reaction time and thinking distance?
Answer:
Tiredness, alcohol, drugs and distraction; at the same reaction time, greater speed also increases thinking distance.
Question 45 · Spec 6.5.4.3.3
Which factors can increase braking distance?
Answer:
Wet or icy roads, worn tyres or brakes, and greater initial speed.
Question 46 · Spec 6.5.4.3.4
Why does a higher initial speed greatly increase braking distance for the same braking force?
Answer:
Kinetic energy is proportional to speed squared; more work is needed to stop, requiring a greater braking distance.
Question 47 · Spec 6.5.4.3.4
Why can a very large deceleration cause injury?
Answer:
F = ma means a large deceleration requires a large force on the body.
Question 48 · Spec 6.5.5.1 · Higher only
State the momentum equation and unit.
Answer:
p = mv; momentum in kg m/s, mass in kg and velocity in m/s. Momentum has direction.
Question 49 · Spec 6.5.5.2 · Higher only
State conservation of momentum for a closed system.
Answer:
Total momentum before an event equals total momentum after it, provided no external resultant force acts on the system.
Question 50 · Spec 6.5.5.2 · Higher only
A 2 kg trolley at 3 m/s joins a stationary 1 kg trolley. Find their shared velocity, with no external force.
Answer:
Initial momentum = 2 × 3 = 6 kg m/s. Shared mass = 3 kg; velocity = 6 / 3 = 2 m/s.

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