Motion, Forces, Energy & Graphs: Exam-focused Summary

Displacement: the vector distance and direction from a reference point; SI unit: m.
Velocity: rate of change of displacement; a vector; SI unit: m s^{-1}; average velocity = total displacement ÷ total time.
Speed: rate of change of distance; magnitude of velocity; average speed may differ from average velocity if motion reverses direction.

Constant-acceleration equations (u = initial, v = final, a = acceleration, s = displacement, t = time):
- v = u + a t
- v^2 = u^2 + 2 a s
- s = ut + \tfrac{1}{2} a t^2
- s = \tfrac{1}{2} (u + v) t
Signs can be positive or negative for displacement (s), velocity (u, v), and acceleration (a).
Approach to problems (summary):
1) verify constant acceleration; 2) note knowns (u, v, a, s, t); 3) decide which quantity to find; 4) choose the equation containing knowns and required quantity; 5) substitute and solve; 6) check reasonableness.

Gravity acceleration: g = 9.81\ \mathrm{m\,s^{-2}} downward.
Sign convention: choose upwards as positive; acceleration is -g when up is positive.
Key results: moving upward slows until velocity zero; then falls.
Useful equations (vertical):
- v^2 = u^2 + 2 a s with a = -g for upward-positive convention.
- When starting from rest, s = \tfrac{1}{2} a t^2.
Maximum height (v = 0): H = \dfrac{u^2}{2g} (for vertical launch).

Start with speed U at angle (\theta) to horizontal.
Components:
- Horizontal: Ux = U\cos\theta, acceleration ax = 0 (constant velocity)
- Vertical: Uy = U\sin\theta, acceleration ay = -g
Key times and heights:
- Time to reach max height: t{\text{up}} = \dfrac{Uy}{g} = \dfrac{U\sin\theta}{g}
- Maximum height: H = \dfrac{U_y^2}{2g} = \dfrac{(U\sin\theta)^2}{2g}
- Horizontal range (landing at same height): R = \dfrac{U^2 \sin(2\theta)}{g}
Velocity at any time is the vector with components vx = U\cos\theta and vy = U\sin\theta - g t.
Resultant velocity at impact: |\mathbf{v}| = \sqrt{vx^2 + vy^2}.

Displacement-time (d-t) graph:
- Slope = velocity; a straight line => constant velocity; curved => changing velocity.
Velocity-time (v-t) graph:
- Gradient = acceleration; straight line with constant gradient => constant acceleration; horizontal line => zero acceleration (constant velocity).
- Area under v-t graph (for the time interval) = displacement.
Instantaneous velocity from a curved d-t: use the gradient of the tangent at the desired point.
Exam tips:
- When calculating gradients, use long segments for accuracy.
- Use tangents carefully to find instantaneous velocity.

Velocity is a vector: magnitude and direction.
In one-dimensional motion, forward vs backward is represented by positive/negative sign.
Displacement and velocity signs indicate direction relative to chosen axis.

Resultant of two vectors: draw vectors head-to-tail and connect tail to final head; or use components.
Components: for a vector (V) making angle (\phi) with a chosen axis,
- Horizontal: V_x = V \cos\phi
- Vertical: V_y = V \sin\phi
For two vectors, parallelogram rule or triangle rule can be used; for three or more, use a vector polygon or sum components.

Free-body diagram shows all forces acting on a body:
- Weight: W = mg (acts downward through the centre of mass)
- Normal reaction: N (perpendicular to contact surface)
- Friction: F_f (along contact surface; direction opposes motion or tendency to move)
- Tension/Thrust: along strings or springs
- Air resistance: opposite to motion
Newton's Laws:
- First Law (equilibrium): a body at rest or moving with constant velocity has net force zero.
- Second Law: vector form, \mathbf{F}_{\text{net}} = m \mathbf{a}. For components along a given direction: sum of forces in that direction = m a along that direction.
- Third Law: action-reaction pairs are equal in magnitude and opposite in direction; act on different bodies.
Problem-solving approach: draw FBD, resolve forces along the direction of acceleration, apply Newton's laws, and solve.

Resolve forces parallel and perpendicular to the incline:
- Parallel: sum = 0 gives friction if in limiting equilibrium or motion.
- Perpendicular: normal force balances weight component perpendicular to plane.
Friction: static friction up to a maximum value; direction depends on whether tendency is to move up or down the plane.

Work: W = F s \cos\theta for constant forces; for a non-constant force, use integral or area under a force-distance graph.
Kinetic Energy: KE = \tfrac{1}{2} m v^2
Gravitational Potential Energy: GPE = m g h
Work-Energy Principle (mechanical energy): if no non-conservative forces, total mechanical energy is conserved:
- Initial: KEi + GPEi; Final: KEf + GPEf; set equal to solve for unknowns.
Work-Energy Theorem: when non-conservative forces are present, work done by these forces equals the change in kinetic energy.
Power: P = \frac{dW}{dt} = F v (instantaneous: (\mathbf{P} = \mathbf{F} \cdot \mathbf{v})).

Energy lost to friction manifests as heat (and sometimes sound).
In vertical motion with air resistance, energy transfer includes work done against air; total mechanical energy not strictly conserved.
Ferris wheel example: energy exchange between kinetic and gravitational potential energy; some energy dissipated as heat.

Stress: \sigma = \dfrac{F}{A} (force per unit cross-sectional area).
Strain: \varepsilon = \dfrac{\Delta L}{L_0} (extension per original length).
Young's Modulus: E = \dfrac{\sigma}{\varepsilon} = \dfrac{F L_0}{A \Delta L}
Elastic region obeys Hooke's law; yield point, UTS (ultimate tensile stress), and breaking point are key material properties.
Ductile vs brittle: ability to undergo plastic deformation before breaking vs sudden fracture.

Always define quantities clearly and use vector directions (positive/negative) consistently.
For graphs, identify whether you are dealing with d-t or v-t graphs and apply the correct interpretation (slope vs area vs gradient).
When asked for a resultant of multiple forces, draw free-body diagrams and resolve into two perpendicular directions to sum components.
In energy problems, decide whether conservation of energy applies (no non-conservative forces) or whether work-energy must be used (non-conservative forces present).

Kinematics (uniform acceleration):
- v = u + a t
- v^2 = u^2 + 2 a s
- s = ut + \tfrac{1}{2} a t^2
- s = \tfrac{1}{2} (u + v) t
Projectile components: Ux = U\cos\theta, { } Uy = U\sin\theta; R = \dfrac{U^2 \sin 2\theta}{g}
Gravity: GPE = m g h,\ ext{KE} = \tfrac{1}{2} m v^2
Work: W = F s \cos\theta; Power: P = F v = \dfrac{W}{t}
Stress/Strain: \sigma = \dfrac{F}{A},\ \varepsilon = \dfrac{\Delta L}{L_0},\ E = \dfrac{\sigma}{\varepsilon}
Gravity constant: g \approx 9.81\ \mathrm{m\,s^{-2}}
Sign conventions: always state which direction is positive for displacements, velocities, and accelerations.

Title: Motion and Energy – Exam-focused Summary for Quick Review