PHYSICS NOTES 4U / DETAILED TOPIC SERIES
Mechanics / force, energy and
momentum
AQA A-level Physics 7408 • Section 3.4.1 • Review edition
Detailed topic pack / AQA 7408 section 3.4.1
Start with the object and model, not the first equation you recognise. Mechanics links vector
directions, changes in motion, energy transfers and momentum. Choose the right description
and keep signs and assumptions consistent.
Resolved forces, moments, motion graphs, constant acceleration, projectiles, drag,
Newton's laws, collisions, impulse, work, efficiency and free fall. Includes 16 original tasks
and worked solutions.
Find it Page
Scalars, vectors and equilibrium 2
Moments, couples and centre of mass 3
Motion graphs and signs 4
Constant acceleration and free fall 5
Projectiles and drag 6
Newton's laws and force models 7
Momentum, collisions and explosions 8
Impulse and transport design 9
Work, energy, power and efficiency 10
Required practical 3 and analysis 11
Practice 1--16
Solutions 1--16
Checklist, scope and sources 16
Independent original material. Not produced, approved or endorsed by AQA. Practice has no official mark
allocation. Review edition. Materials and circular motion/SHM are separate packs.
Independent revision notes • Edition 1.0 for review • October 2026 1
,PHYSICS NOTES 4U / DETAILED TOPIC SERIES
Vectors / choose directions first
AQA A-level Physics 7408 • Section 3.4.1 • Review edition
Scalar: magnitude only Vector: magnitude and direction
Distance / speed Displacement / velocity
Mass / energy / time Force / weight / acceleration / momentum
Define axes, then resolve. For force F at angle θ above horizontal: horizontal component F cosθ;
vertical component F sinθ. Cosine belongs to the component adjacent to the stated angle, not
automatically to horizontal.
Perpendicular resultant: 6.0 N east and 8.0 N north give √(6²+8²)=10 N, at
tan−1(8/6)=53.1° north of east. Use head-to-tail scale drawings for other angles. Direct
addition calculations in this section are limited to two perpendicular vectors.
A slope is a useful choice of axes
On a slope at angle θ above horizontal, weight resolves into mg sinθ down the slope and mg
cosθ into it. The normal force is perpendicular to the surface. It equals mg cosθ only with no
other perpendicular force components and no perpendicular acceleration.
Equilibrium: a 2.0 kg block rests on a smooth 30° slope held by a string parallel to the
slope. T=mg sin30°=9.81 N; N=mg cos30°=17.0 N. On a rough slope, friction could share
the balancing role, so tension would not follow this simple equation.
Equilibrium does not mean motionless
Zero resultant force means constant velocity, including rest. Two balancing forces are equal and
opposite. Three coplanar forces at a point balance when their head-to-tail triangle closes, or
when components sum to zero along both axes.
For an extended rigid object, zero resultant force alone is insufficient: a non-zero resultant moment can change its
rotation.
Independent revision notes • Edition 1.0 for review • October 2026 2
, PHYSICS NOTES 4U / DETAILED TOPIC SERIES
Moments / perpendicular distance
AQA A-level Physics 7408 • Section 3.4.1 • Review edition
Moment = force × perpendicular distance
Use distance from pivot to the force's line of action, not necessarily to its application point. State
clockwise or anticlockwise. Moment has unit N m, not joule: it is not energy.
Angled force: 20 N acts at the end of a 0.30 m handle at 40° to the handle. Moment=20 ×
0.30 × sin40°=3.86 N m. Using 20 × 0.30 assumes a perpendicular force.
Principle of moments
For rotational equilibrium, total clockwise moment equals total anticlockwise moment about any
point. Choosing a support as pivot removes its unknown force from that moment equation. Force
balance is still needed.
Beam: a uniform 3.0 m beam weighs 120 N, carrying 180 N at 2.0 m from its left end.
Supports act upward at both ends. About left: 3RR=120(1.5)+180(2)=540. Thus RR=180 N.
Vertical balance gives RL=300−180=120 N.
Couples and centre of mass
A couple is two equal, opposite, parallel forces along different lines of action. Net force is zero;
moment is either force × perpendicular separation. Total couple moment is independent of the
reference point.
Centre of mass is where mass can be considered concentrated for translational motion. In a
uniform gravitational field, weight acts through it. A uniform regular solid has its centre of mass
at its geometrical centre. A freely suspended object settles with its centre of mass below the
suspension point.
If the vertical line through the centre of mass moves outside the support base, weight can cause overturning. Do
not put every non-uniform object's weight at its geometric centre.
Independent revision notes • Edition 1.0 for review • October 2026 3
Mechanics / force, energy and
momentum
AQA A-level Physics 7408 • Section 3.4.1 • Review edition
Detailed topic pack / AQA 7408 section 3.4.1
Start with the object and model, not the first equation you recognise. Mechanics links vector
directions, changes in motion, energy transfers and momentum. Choose the right description
and keep signs and assumptions consistent.
Resolved forces, moments, motion graphs, constant acceleration, projectiles, drag,
Newton's laws, collisions, impulse, work, efficiency and free fall. Includes 16 original tasks
and worked solutions.
Find it Page
Scalars, vectors and equilibrium 2
Moments, couples and centre of mass 3
Motion graphs and signs 4
Constant acceleration and free fall 5
Projectiles and drag 6
Newton's laws and force models 7
Momentum, collisions and explosions 8
Impulse and transport design 9
Work, energy, power and efficiency 10
Required practical 3 and analysis 11
Practice 1--16
Solutions 1--16
Checklist, scope and sources 16
Independent original material. Not produced, approved or endorsed by AQA. Practice has no official mark
allocation. Review edition. Materials and circular motion/SHM are separate packs.
Independent revision notes • Edition 1.0 for review • October 2026 1
,PHYSICS NOTES 4U / DETAILED TOPIC SERIES
Vectors / choose directions first
AQA A-level Physics 7408 • Section 3.4.1 • Review edition
Scalar: magnitude only Vector: magnitude and direction
Distance / speed Displacement / velocity
Mass / energy / time Force / weight / acceleration / momentum
Define axes, then resolve. For force F at angle θ above horizontal: horizontal component F cosθ;
vertical component F sinθ. Cosine belongs to the component adjacent to the stated angle, not
automatically to horizontal.
Perpendicular resultant: 6.0 N east and 8.0 N north give √(6²+8²)=10 N, at
tan−1(8/6)=53.1° north of east. Use head-to-tail scale drawings for other angles. Direct
addition calculations in this section are limited to two perpendicular vectors.
A slope is a useful choice of axes
On a slope at angle θ above horizontal, weight resolves into mg sinθ down the slope and mg
cosθ into it. The normal force is perpendicular to the surface. It equals mg cosθ only with no
other perpendicular force components and no perpendicular acceleration.
Equilibrium: a 2.0 kg block rests on a smooth 30° slope held by a string parallel to the
slope. T=mg sin30°=9.81 N; N=mg cos30°=17.0 N. On a rough slope, friction could share
the balancing role, so tension would not follow this simple equation.
Equilibrium does not mean motionless
Zero resultant force means constant velocity, including rest. Two balancing forces are equal and
opposite. Three coplanar forces at a point balance when their head-to-tail triangle closes, or
when components sum to zero along both axes.
For an extended rigid object, zero resultant force alone is insufficient: a non-zero resultant moment can change its
rotation.
Independent revision notes • Edition 1.0 for review • October 2026 2
, PHYSICS NOTES 4U / DETAILED TOPIC SERIES
Moments / perpendicular distance
AQA A-level Physics 7408 • Section 3.4.1 • Review edition
Moment = force × perpendicular distance
Use distance from pivot to the force's line of action, not necessarily to its application point. State
clockwise or anticlockwise. Moment has unit N m, not joule: it is not energy.
Angled force: 20 N acts at the end of a 0.30 m handle at 40° to the handle. Moment=20 ×
0.30 × sin40°=3.86 N m. Using 20 × 0.30 assumes a perpendicular force.
Principle of moments
For rotational equilibrium, total clockwise moment equals total anticlockwise moment about any
point. Choosing a support as pivot removes its unknown force from that moment equation. Force
balance is still needed.
Beam: a uniform 3.0 m beam weighs 120 N, carrying 180 N at 2.0 m from its left end.
Supports act upward at both ends. About left: 3RR=120(1.5)+180(2)=540. Thus RR=180 N.
Vertical balance gives RL=300−180=120 N.
Couples and centre of mass
A couple is two equal, opposite, parallel forces along different lines of action. Net force is zero;
moment is either force × perpendicular separation. Total couple moment is independent of the
reference point.
Centre of mass is where mass can be considered concentrated for translational motion. In a
uniform gravitational field, weight acts through it. A uniform regular solid has its centre of mass
at its geometrical centre. A freely suspended object settles with its centre of mass below the
suspension point.
If the vertical line through the centre of mass moves outside the support base, weight can cause overturning. Do
not put every non-uniform object's weight at its geometric centre.
Independent revision notes • Edition 1.0 for review • October 2026 3
