Kinematics (SUVAT equations), Newton's laws, forces, friction, connected particles, and momentum for CIE A Level Mathematics 9709 Mechanics.
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For constant acceleration motion, five variables: s (displacement), u (initial velocity), v (final velocity), a (acceleration), t (time).
| Equation | Variables used |
|---|---|
| v = u + at | v, u, a, t |
| s = ut + ½at² | s, u, a, t |
| s = vt − ½at² | s, v, a, t |
| v² = u² + 2as | v, u, a, s |
| s = ½(u + v)t | s, u, v, t |
Method: Identify three known variables, choose the equation containing the fourth (unknown).
Always define a positive direction first. Objects moving in the negative direction have negative velocity/acceleration.
Example: A ball is thrown vertically upward at 15 m/s. How long before it hits the ground (from the same level)?
Take up as positive: u = 15, a = −9.8, s = 0.
s = ut + ½at²: 0 = 15t − 4.9t² = t(15 − 4.9t) → t = 0 or t = 3.06 s
Maximum height: v = 0. v² = u² + 2as: 0 = 225 − 19.6s → s = **11.5 m**
First Law: An object remains at rest or moves at constant velocity unless acted on by a resultant force.
Second Law: **F = ma** (resultant force = mass × acceleration; in Newtons when m in kg, a in m/s²)
Third Law: For every action there is an equal and opposite reaction.
Weight: W = mg (g ≈ 9.8 m/s² or 10 m/s² as stated in question).
Resultant force: Vector sum of all forces. F_net = ma.
Equilibrium: F_net = 0 (no acceleration). Resolve horizontally and vertically → two equations.
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Friction: F = μN (maximum static friction = μₛN; kinetic friction = μₖN)
Example: Block of mass 5 kg on a 30° smooth slope. Find acceleration.
F_net along slope = mg sin 30° = 5 × 9.8 × 0.5 = 24.5 N
a = F_net/m = 24.5/5 = 4.9 m/s² (down slope)
Example with friction: Same slope, μ = 0.2, rough.
N = mg cos 30° = 5 × 9.8 × (√3/2) = 42.4 N
Friction (up slope, opposing motion) = 0.2 × 42.4 = 8.48 N
F_net = 24.5 − 8.48 = 16.0 N
a = 16.0/5 = 3.20 m/s²
Masses m₁ and m₂ (m₁ > m₂), connected by light inextensible string over smooth pulley.
Example: m₁ = 6 kg, m₂ = 4 kg.
a = (6−4)(9.8)/(6+4) = 2×9.8/10 = 1.96 m/s²
T = 2(6)(4)(9.8)/10 = 47.0 N
Key insight: Tension is the same throughout a light inextensible string over a smooth pulley.
Momentum: p = mv (kg m/s — a vector)
Newton's Second Law in momentum form: F = Δp/Δt (rate of change of momentum)
Impulse: J = FΔt = Δp = m(v − u)
Conservation of momentum: For an isolated system (no external forces):
Example: Car of mass 1200 kg at 20 m/s collides with stationary van of 2000 kg. They couple together. Find final velocity.
1200 × 20 = (1200 + 2000) × v
24000 = 3200v → v = 7.5 m/s
The Karakoram Highway (KKH) — the world's highest paved road — provides rich mechanics problems. A truck of mass 8000 kg descends a 15° slope. Find the braking force needed to maintain constant velocity (a=0). Along slope: weight component down = 8000×9.8×sin15°=20,327 N. Normal reaction: N=8000×9.8×cos15°=75,705 N. For constant velocity, braking force + friction = 20,327 N. On steep KKH descents, brake failure is catastrophic — SUVAT gives the stopping distance: if brakes fail at v=20 m/s on this slope with additional rolling resistance providing only 5000 N deceleration force, F_net=20327−5000=15327 N down the slope, a=15327/8000=1.916 m/s². Using v²=u²+2as with v=0: 0=400+2(1.916)s → wait, the truck accelerates! This is why runaway truck ramps (sand traps) exist on KKH — vital mechanics engineering.
Mathematics — Quick Revision
UVAT: v=u+at; s=ut+½at²; v²=u²+2as; s=½(u+v)t — identify 3 known vars, choose equationarea under=displacementF=ma (resultant force); weight=mg; resolve forces on inclined planesF=μN; always opposes motion; N=mg cosθ on slopeMechanics in Pakistan: Karakoram Highway Physics
The Karakoram Highway (KKH) — the world's highest paved road — provides rich mechanics problems. A truck of mass 8000 kg descends a 15° slope. Find the braking force needed to maintain constant velocity (a=0). Along slope: weight component down = 8000×9.8×sin15°=20,327 N. Normal reaction: N=8000×9.8×cos15°=75,705 N. For constant velocity, braking force + friction = 20,327 N. On steep KKH descents, brake failure is catastrophic — SUVAT gives the stopping distance: if brakes fail at v=20 m/s on this slope with additional rolling resistance providing only 5000 N deceleration force, F_net=20327−5000=15327 N down the slope, a=15327/8000=1.916 m/s². Using v²=u²+2as with v=0: 0=400+2(1.916)s → wait, the truck accelerates! This is why runaway truck ramps (sand traps) exist on KKH — vital mechanics engineering.
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