Mechanics: Kinematics, Dynamics, Momentum, Energy, Circular Motion
Mechanics carries the heaviest weight in the Pancyprian Physics paper. The exam tests whether you can describe the same motion three ways - kinematically, by forces, and through energy or momentum - and choose the fastest route. Expect a structured problem on accelerated or projectile motion, a force problem with friction, a lift or a pulley, and a conservation problem involving a collision or a vertical circle.
Kinematics
For constant acceleration only: v = u + at, s = ut + ½at², v² = u² + 2as. Treat projectile motion as two independent motions: horizontal at constant velocity, vertical with a = g. Time is the link between them. At the highest point the vertical velocity component is zero - the speed is not zero unless the launch was vertical.
Dynamics
Newton's second law is a vector equation, ΣF = ma. Draw the free-body diagram first and choose axes along the acceleration. Friction obeys f = μN, where N comes from the direction perpendicular to the surface; on an incline N = mg cos θ, not mg. In a lift, the reading of a scale is the normal force N, which equals mg only at rest or at constant velocity.
Momentum and energy
p = mv is conserved when the net external force on the system is zero, including during short impacts. Kinetic energy is conserved only in elastic collisions; in inelastic ones it is converted to heat, sound and deformation. The work–energy theorem states W_net = ΔK, with W = Fs cos θ, so friction does negative work. In circular motion the centripetal force mv²/r is the resultant of the real forces - never an additional arrow on the diagram.
Worked example
A 0.40 kg ball on a 0.90 m string is whirled in a vertical circle. Find the minimum speed at the top. At the top, tension T and weight mg both point toward the centre, so T + mg = mv²/r. The minimum speed occurs as T → 0, giving v = √(gr) = √(9.81 × 0.90) = 2.97 m/s ≈ 3.0 m/s. Any slower and the string slackens before the ball reaches the top.
Traps that cost marks
- Applying v² = u² + 2as across a stage where the acceleration changes (parachute opening, engine cut-off).
- Drawing a separate "centripetal force" arrow instead of identifying which real force provides it.
- Assuming momentum conservation implies energy conservation in a collision.
- Confusing μ (dimensionless) with a force, or using mg instead of N on an incline.
- Mixing g = 9.8 and 9.81 within one calculation, or leaving the calculator in degree mode when a phase angle in radians is required.