Work, Energy, Power and Momentum
Scope of the domain
This domain links force and displacement to energy transfer, and force and time to momentum change. The exam asks you to calculate work, kinetic and potential energy, power and efficiency, and to apply conservation of momentum to collisions and explosions.
Core rules
- Work W = Fs cosθ: only the component of force along the displacement does work. Friction does negative work.
- Kinetic energy KE = ½mv²; gravitational potential change ΔGPE = mgΔh; elastic energy = ½kx².
- Power P = W/t; for constant speed against a steady force, P = Fv.
- Conservation of energy: total energy is constant, but mechanical energy falls when friction or drag converts it to thermal energy.
- Efficiency = useful output energy ÷ total input energy × 100%.
- Momentum p = mv is a vector. Impulse FΔt = Δp = m(v − u); the area under a force–time graph is impulse.
- Total momentum is conserved when the resultant external force is zero. Kinetic energy is conserved only in elastic collisions; in inelastic collisions it is reduced.
Worked example
A 1200 kg car accelerates uniformly from rest to 15 m/s in 8.0 s. Find the gain in kinetic energy, the average power and the average resultant force. KE = ½ × 1200 × 15² = 1.35 × 10⁵ J. Average power = 1.35 × 10⁵ ÷ 8.0 = 1.69 × 10⁴ W, about 17 kW. Momentum change = 1200 × 15 = 1.8 × 10⁴ kg m/s, so average force = 1.8 × 10⁴ ÷ 8.0 = 2.25 × 10³ N. A check with P = Fv gives 2250 × 7.5 = 16.9 kW at the average speed.
Traps that cost marks
- Omitting cosθ in work calculations, especially when the force is at an angle to the motion.
- Treating kinetic energy as conserved in every collision.
- Losing the vector nature of momentum: assign one direction positive and keep the signs.
- Mixing up impulse (force × time) with work (force × distance).
- Calculating efficiency with input and output reversed, producing values above 100%.
- Using P = Fv for an accelerating object when F is the resultant force and v is changing.