9.1 Mechanics

9.1 Mechanics

Kinematics: Equations of Motion

Kinematics describes motion without asking what causes it. For motion along a straight line with constant acceleration (a), three equations link displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t):

• v = u + at
• s = ut + ½at²
• v² = u² + 2as

These equations are valid only when acceleration is uniform. If acceleration varies with time, you must integrate (or use an average) — a frequent trap on the GD Pilot Initial Test.

Units matter. Velocity is in m/s, acceleration in m/s², displacement in m, time in s. A common exam trick gives a speed in km/h; convert to m/s by dividing by 3.6 before substituting. A 720 km/h jet is 200 m/s.

Worked example — takeoff acceleration: An aircraft starts from rest (u = 0) and accelerates at 4 m/s² for 10 s. Find its final velocity and the runway distance covered.

• v = u + at = 0 + (4)(10) = 40 m/s
• s = ut + ½at² = 0 + ½(4)(10²) = 200 m

At 40 m/s the aircraft is doing 144 km/h — still well below rotation speed for most trainers, which is why runways are typically 1.5–2 km long.

Newton's Three Laws of Motion

Newton's laws are the backbone of mechanics and appear in some form on almost every PAF physics paper. The table below summarises them and ties each to an aviation situation.

Aviation link — the four forces: In steady, level, unaccelerated flight, lift balances weight (Newton's first law in the vertical direction) and thrust balances drag (first law horizontally). To climb, the pilot increases lift so it exceeds weight; to accelerate, thrust must exceed drag. Every manoeuvre is, at heart, an application of F = ma.

Momentum and Its Conservation

Momentum (p) is mass in motion: p = mv. Its SI unit is the kg·m/s, and it is a vector — direction matters just as much as magnitude.

In an isolated system (no external net force), total momentum is conserved before and after any collision or explosion:

• m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Worked example — inelastic collision: A 1500 kg car moving at 20 m/s rear-ends a stationary 1000 kg car. The two lock together. Find their common velocity.

• (1500)(20) + (1000)(0) = (1500 + 1000) v
• 30 000 = 2500 v ⇒ v = 12 m/s

Momentum conservation is also the principle behind jet and rocket propulsion: the engine throws mass backward at high speed, and the aircraft gains an equal amount of forward momentum. This is Newton's third law expressed through momentum.

Work, Energy, and the Work-Energy Theorem

Work (W) is done when a force moves its point of application through a displacement: W = Fd cos θ, where θ is the angle between force and displacement. The SI unit is the joule (J). Note that a force perpendicular to motion (like centripetal force in circular motion) does zero work.

Two forms of mechanical energy appear constantly on the exam:

• Kinetic energy: KE = ½mv²
• Gravitational potential energy: PE = mgh, with g = 9.8 m/s²

The work-energy theorem states that the net work done on a body equals the change in its kinetic energy:

• W_net = ΔKE = ½mv² − ½mu²

Worked example — acceleration in flight: A 1200 kg aircraft accelerates from 60 m/s to 80 m/s. What net work do the engines (minus drag) perform?

• W = ½(1200)(80² − 60²) = 600(6400 − 3600) = 600(2800) = 1 680 000 J ≈ 1.68 MJ

Projectile Motion

A projectile is any object launched into the air with only gravity acting on it (at FSc level we ignore air resistance). The crucial insight is that horizontal and vertical motions are independent and can be analysed separately.

• Horizontal: no acceleration (a_x = 0), so x = u_x · t, where u_x = u cos θ.
• Vertical: constant downward acceleration (a_y = −g = −9.8 m/s²), so v_y = u_y − gt and y = u_y t − ½gt², where u_y = u sin θ.

For a projectile launched on level ground at speed u and angle θ:

• Time of flight: T = 2u sin θ / g
• Range: R = u² sin(2θ) / g (maximum at θ = 45°)
• Maximum height: H = u² sin²θ / (2g)

Worked example — artillery range: A shell is launched at 30 m/s at 30°. Find its range (g = 9.8 m/s²).

• R = (30²) sin(60°) / 9.8 = 900(0.866) / 9.8 ≈ 79.6 m

Aviation link: A bomb released from a level, fast-moving aircraft is itself a projectile — it keeps the aircraft's horizontal velocity while accelerating downward under gravity. The pilot must release the bomb before reaching the target, not over it. This is why dive bombing is more accurate than level bombing: the steep dive angle shortens the fall time and reduces wind drift.