A-level Mechanics Key Concepts

Overview

Concise summary of A-level Mechanics topics covered in the lecture.
Focus on kinematics, variable acceleration, forces, equilibrium, friction, Newton's laws, projectiles, moments, and worked examples.
Emphasis on definitions, key formulas (SUVAT), problem strategies, and typical assumptions.

Kinematics: Basic Definitions

Distance: scalar quantity, total path length.
Displacement: vector quantity, straight-line distance from origin with direction.
Position: distance from a fixed origin; important to distinguish from distance.
Average speed = total distance / total time.
Average velocity = displacement / time (vector).

Position-Time and Velocity-Time Graphs

Displacement-time graph:
- Flat line: stationary.
- Positive slope: moving away from origin.
- Negative slope: moving toward origin.
Velocity-time graph:
- Gradient = acceleration.
- Flat line: constant velocity.
- Below x-axis: motion in opposite direction.
- Crossing x-axis: velocity = 0 (stop or change of direction).
Area under velocity-time graph = displacement.

SUVAT (Constant Acceleration) Summary

Symbols: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time).
Use these equations only for constant acceleration.
Typical g = 9.8 m/s^2 (use negative when upward positive).
Strategy: choose SUVAT equation that excludes the unknown you don't have.

Quantity	Symbol
Displacement	s
Initial velocity	u
Final velocity	v
Acceleration	a
Time	t

SUVAT Example Highlights

Example 1 (car): start from rest (u=0), v=15 m/s, s=25 m, solve v^2 = u^2 + 2as to find a = 4.5 m/s^2. Next 4 s distance: s = ut + 1/2 a t^2 = 96 m.
Example 2 (vertical projection): u = 12 m/s upward, initial height 1.5 m, at max v=0. Use v^2 = u^2 + 2as to get rise s ≈ 7.347 m; add 1.5 m → max height ≈ 8.85 m (3 s.f.).

Variable Acceleration (Differentiation / Integration Map)

If s(t) known → v(t) = ds/dt → a(t) = dv/dt.
If a(t) known → integrate to get v(t) (+ constant), integrate v(t) to get s(t) (+ constant).
Use initial conditions to find constants of integration.
Area under velocity-time graph = displacement.

Operation	Result
Differentiate s(t)	v(t)
Differentiate v(t)	a(t)
Integrate a(t)	v(t) (+ C)
Integrate v(t)	s(t) (+ C)

Variable Acceleration Examples

a(t) = 12 − 2t, initial v(0)=3:
- Integrate to v(t) = 12t − t^2 + C; C = 3 → v(t) = 12t − t^2 + 3.
- Integrate v(t) to x(t): x(t) = 6t^2 − (1/3)t^3 + 3t (C = 0 at origin).
Train speed v(t) = (1/5000) t (1200 − t):
- Expand, integrate to get x(t). Stops at t=0 and t=1200 s.
- Evaluate x(1200) gives distance ≈ 57 600 m (57.6 km).

Forces: Basics and Common Assumptions

Weight W = mg (mass × g). Weight depends on gravity, mass is invariant.
Reaction force (normal) = equal and opposite contact force.
Tension in light, inextensible strings is uniform.
Model assumptions:
- Objects as point masses (no rotation).
- Strings inextensible, rods light (ignore mass).
- Pulleys smooth (no friction).
Resolve forces along convenient axes (along slope and perpendicular for inclines).

Resolving Forces and Resultant Force

If system in equilibrium, resultant force = 0.
Resolve vectors into i (horizontal) and j (vertical) components and sum components.
Use F = ma for non-equilibrium cases.

Step	Action
Resolve components	Use sin/cos for angled forces
Sum horizontal components	Resultant in i
Sum vertical components	Resultant in j
Use F = ma	To find acceleration vector

Forces Examples

Example: three forces 12 N right, 10 N down, 35 N at 45°:
- Break 35 N into components: 35 cos45 and 35 sin45 (both = 35√2/2).
- Resultant components → use signs per chosen positive directions.
- For mass 0.8 kg, divide resultant force vector by mass to get acceleration vector, then magnitude via Pythagoras ≈ 24.4 m/s^2.
Example on slope (mass 5 kg, slope 30°):
- Weight = 49 N. Normal reaction R = 49 cos30 ≈ 42.4 N.
- Friction (parallel) = 49 sin30 = 24.5 N.

Friction

Limiting friction F_max = μ R (μ = coefficient of friction).
Static friction ≤ μR; when at verge of motion, friction = μR.
Kinetic friction often taken as constant (opposes motion).
Determine direction of friction by imagining motion tendency; friction opposes that direction.

Newton's Laws

1st Law: Body remains at rest or in uniform motion unless acted on by resultant external force.
2nd Law: Resultant force = mass × acceleration (F = ma). Use to relate net force and acceleration.
3rd Law: Action and reaction equal in magnitude and opposite in direction.

Newton's Laws Examples

Driving force 12 000 N, resistances 6 000 N, mass 8 000 kg → resultant 6 000 N so a = 6000/8000 = 0.75 m/s^2.
Man in lift (m=80 kg):
- Moving down at constant velocity → equilibrium, R = mg = 784 N.
- Lift accelerating upward a = 2 m/s^2 → resultant F = ma upward = 160 N. R − mg = 160 → R = 944 N.

Connected Particles and Pulleys

Tension uniform in light, inextensible string over smooth pulley.
Write equations for each mass: T − weight = m a or weight − T = m a depending on direction.
Solve simultaneous equations to find a and T.
Example: 5 kg and 4 kg connected: derive simultaneous equations, solve for a ≈ 5.22 m/s^2 (direction indicated by sign).

Projectiles (2D SUVAT Approach)

Resolve initial velocity into horizontal (u_x = u cosθ) and vertical (u_y = u sinθ).
Horizontal motion: acceleration ≈ 0 → constant speed; s_x = u_x t.
Vertical motion: acceleration = −g → use SUVAT vertically.
Max height: v_vertical = 0 at top.
Range: solve vertical motion for time when vertical displacement returns to launch height, then use horizontal motion to get range.

Component	Equation / Note
Horizontal velocity	u cosθ (constant)
Vertical velocity	u sinθ (affected by gravity)
Time to land/height	Solve vertical SUVAT for t or v

Projectile Example

Shot put: release height 2.5 m, speed 10 m/s at 50°.
- u_x = 10 cos50 ≈ 6.43 m/s, u_y = 10 sin50 ≈ 7.66 m/s.
- Use vertical SUVAT with s = −2.5 m to solve for t; positive root t ≈ 1.84 s.
- Horizontal distance = u_x × t ≈ 6.43 × 1.84 ≈ 11.83 m.

Moments (Turning Effects)

Moment = force × perpendicular distance from pivot.
For equilibrium: sum of clockwise moments = sum of anticlockwise moments.
Take moments about a point to eliminate unknown forces passing through that point.
Uniform lamina/rod: weight acts at center (midpoint) of a uniform object.

Rule	Use
Moment = F × d_perp	Compute turning effect
Sum clockwise = Sum anticlockwise	Condition for rotational equilibrium
Take moments about point	Eliminate forces through that point

Moments Examples

Uniform bridge (5 m, mass 100 kg) with child (45 kg) at C:
- Reaction at A is 3/4 of reaction at B → R_A = (3/4)R_B.
- Resolve vertical forces to find R_B and R_A (R_B ≈ 812 N, R_A ≈ 609 N).
- Take moments about A or B to find child's distance from A → x ≈ 3.65 m.
Uniform ladder (3 m, 20 kg), leaning at 60° against smooth wall:
- Smooth wall → no horizontal reaction at top; floor provides friction and normal reaction.
- Vertical equilibrium gives normal at wall R2 = weight = 196 N.
- Take moments about base to find frictional force R1 using perpendicular distances (R1 ≈ 56.6 N).
- Horizontal equilibrium: friction = R1.

Key Terms and Definitions

Displacement: vector from origin to position.
Velocity: rate of change of displacement (vector).
Acceleration: rate of change of velocity.
Reaction (normal) force: perpendicular contact force.
Tension: force transmitted through a string.
Moment: turning effect = F × perpendicular distance.
Center of mass: point where weight acts for uniform bodies.

Action Items / Problem-Solving Tips

Always draw a clear labelled diagram before solving.
Choose coordinate axes to simplify (e.g., along and perpendicular to slope).
For SUVA(T) problems: select equation that excludes unknown variable.
For variable acceleration: integrate or differentiate using initial conditions.
For forces: resolve into components; sum components separately.
For equilibrium problems: use both force sum = 0 and moment sum = 0.
For projectiles: separate horizontal and vertical analysis; use common time t.
For friction questions: decide likely direction of impending motion first.