Calculate kinetic energy and changes of speed
Use ½mv² with positive mass and speed magnitude, convert mass units and distinguish kinetic energy from directed momentum.
Section 4.5 · Energy, work and power
Exam scope and assessment notes
Mechanics (Paper 4, AS & A Level) · Mathematics 9709, 2026–2027, section 4.5. Use g = 10 m s⁻² throughout these problems. Masses are positive and constant. Work by a constant force is Fd cos θ, with θ between force and displacement; scalar-product notation is not required. Kinetic energy is ½mv² and gravitational potential-energy change is mgΔh. State the system and count gravitational work or potential-energy change once, not both. Mechanical energy is conserved only under the stated loss-free model; friction and supplied resistance transfer mechanical energy to other forms. Smooth curved-slide examples use overall energy changes and assume contact throughout, rather than requiring a normal-reaction calculation. Average power is work divided by elapsed time; P = Fv applies to a force along the motion. Constant driving power generally gives variable force and acceleration; instantaneous calculations do not justify constant-acceleration formulae over a whole journey. Check string, contact and motion-state changes before continuing a model. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Square signed numbers and rearrange a quadratic relation.
- Convert grams to kilograms and distinguish velocity from speed.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
A particle doubles speed without changing mass. What happens to its kinetic energy?
What you will learn
- Calculate kinetic energy and its change with compatible units.
- Explain why reversing velocity sign alone does not change kinetic energy.
Kinetic energy depends on the square of speed
A particle of mass m moving at speed v has kinetic energy K = ½mv². With m in kilograms and v in m s⁻¹, K is in joules. A 3 kg particle at 4 m s⁻¹ has K = 24 J; at 6 m s⁻¹ it has K = 54 J. Its increase is 30 J. Doubling speed multiplies kinetic energy by four for the same mass, rather than doubling it.
A 4 kg particle with signed velocity −5 m s⁻¹ has K = ½ × 4 × 25 = 50 J. The negative velocity specifies direction; kinetic energy remains non-negative and is the same at +5. A change of direction can change momentum without changing speed or kinetic energy. Do not use a signed kinetic energy to describe leftward motion.
Changes use final energy minus initial energy
For a 0.5 kg particle slowing from speed 10 to 6 m s⁻¹, initial K is 25 J and final K is 9 J. Thus ΔK = −16 J and the amount lost is positive 16 J. A loss of kinetic energy need not mean energy ceases to exist; it can be transferred to heating, deformation or other forms under the model’s stated interactions.
To recover speed from a known positive energy, use v = √(2K/m) and report the non-negative speed. Energy alone cannot choose a velocity direction. A 250 g particle at 20 m s⁻¹ has mass 0.25 kg and K = 50 J, not 50 000 J. Check both conversion and the squared-speed dependence before using the result in a larger energy balance.
Pause and explain
A 4 kg particle has velocity −5 m s⁻¹. What is its kinetic energy?
Worked example
A 3 kg particle increases speed from 4 to 6 m s⁻¹. Find its two kinetic energies and change.
Show the worked solution
- Initial K = ½ × 3 × 4² = 24 J.
- Final K = ½ × 3 × 6² = 54 J.
- The change is 54 − 24 = +30 J.
Answer Initial 24 J; final 54 J; increase 30 J.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Find the kinetic energy of a 2 kg particle moving at speed 3 m s⁻¹.
Give me a hint
Use ½mv² and square the numerical speed first.
Compare my reasoning
- The squared speed is 9 m² s⁻².
- K = ½ × 2 × 9 = 9 J.
- The result is a non-negative scalar energy.
K = 9 J.
Look for these in your work
- Retain the factor ½.
- Report joules rather than momentum units.
A 4 kg particle has velocity −5 m s⁻¹. Find its kinetic energy and compare it with the energy at velocity +5 m s⁻¹.
Give me a hint
Both signed velocities have the same square.
Compare my reasoning
- At −5, K = ½ × 4 × (−5)² = 50 J.
- At +5, K = ½ × 4 × 5² = 50 J.
- The energies are equal even though the momenta have opposite directions.
K = 50 J at either velocity; the direction cannot be inferred from K.
Look for these in your work
- Keep mass positive and square the velocity.
- Separate energy magnitude from velocity direction.
A 500 g particle slows from speed 10 to 6 m s⁻¹. Find initial energy, final energy, signed change and amount lost.
Give me a hint
Convert 500 g to kilograms, then subtract initial energy from final energy.
Compare my reasoning
- Mass is 0.5 kg, so initial K = ½ × 0.5 × 100 = 25 J.
- Final K = ½ × 0.5 × 36 = 9 J.
- The signed change is −16 J and the positive amount lost is 16 J.
Initial 25 J; final 9 J; ΔK = −16 J; loss 16 J.
Look for these in your work
- Convert the mass before calculating SI energy.
- Distinguish a negative change from a positive loss amount.
Common mistakes
- Square the speed; do not omit the factor ½.
- Kinetic energy is non-negative even for negative velocity.
- Distinguish signed change from the positive amount lost.
What can you explain now?
Can kinetic energy alone tell you whether a particle moves left or right?
Compare with the explanation
No; opposite velocities with equal speed have equal kinetic energy.
The squared velocity loses the direction information carried by momentum.
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Make it stick
Use positive mass and squared speed, then calculate energy change as final minus initial.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.