Calculate kinetic energy from mass and speed
Use one half of mass times speed squared and convert to kilograms and metres per second.
Exam scope and assessment notes
Extended only (Supplement) · Physics 0972, 2026–2028, section 1.7.1. Energy stores, transfers and simple conservation are Core; kinetic and gravitational potential energy equations, multistage conservation and Sankey diagrams are Supplement. State the system, interval and model assumptions. Written practice is self-checked. Supervised practical experience, cumulative assessment and human subject-teacher review remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Square numbers and convert grams to kilograms.
- Distinguish speed from signed velocity.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is 6²?
What you will learn
- Recall and use Ek = ½mv².
- Explain the square dependence without treating energy as signed momentum.
Square the speed, then multiply
Kinetic energy is Ek = ½mv², with mass in kilograms and speed in metres per second, giving joules. A 2 kg cart at 3 m/s has Ek = ½ × 2 × 3² = 9 J. At 6 m/s it has 36 J. Doubling speed at fixed mass gives four times the energy because the whole speed value is squared.
A 500 g object at 4 m/s has mass 0.500 kg and Ek = ½ × 0.500 × 16 = 4.0 J. Using 500 as kilograms or forgetting the half gives a different physical answer. The formula concerns macroscopic kinetic energy in the specified reference frame, not every store of the object.
Energy is not momentum with another unit
Kinetic energy does not acquire a leftward negative sign. A fixed-mass object at equal speeds left and right has equal kinetic energy because the squared velocity magnitude is the same, even though its momenta have opposite signs. A signed kinetic-energy change can be negative during slowing, but the kinetic energy amount itself is non-negative.
To find a change, subtract the initial kinetic energy from the final kinetic energy. The 2 kg cart slowing from 6 to 3 m/s changes by 9 − 36 = −27 J. Squaring the speed change as ½m(3 − 6)² would give the wrong energy change: a difference of squares is not the square of a difference.
Pause and explain
Equal speeds left and right at the same mass give which kinetic energies?
Worked example
Calculate kinetic energies of a 2 kg cart at 3 m/s and 6 m/s.
Show the worked solution
- At 3 m/s, Ek = ½ × 2 × 3² = 9 J.
- At 6 m/s, Ek = ½ × 2 × 6² = 36 J.
- The energy increases by a factor of four for doubled speed.
Answer 9 J and 36 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 kinetic energies of 2 kg at 3 m/s and 6 m/s.
Give me a hint
Use the same mass with each speed squared.
Compare my reasoning
- ½ × 2 × 3² = 9 J.
- ½ × 2 × 6² = 36 J.
- The ratio is 36/9 = 4.
9 J; 36 J.
Look for these in your work
- Square each speed.
- Retain consistent mass and energy units.
Find kinetic energy of 500 g at 4 m/s, then the change for 2 kg slowing from 6 to 3 m/s.
Give me a hint
Convert the first mass and subtract two full energies in the second task.
Compare my reasoning
- 0.500 kg at 4 m/s has 4.0 J.
- The 2 kg object's energies are 36 J initially and 9 J finally.
- Change = 9 − 36 = −27 J.
4.0 J; ΔEk = −27 J.
Look for these in your work
- Convert grams.
- Distinguish an energy amount from its signed change.
A 2 kg cart slows from 6 to 3 m/s. A student calculates ½ × 2 × (3 − 6)² = 9 J as its energy change. Correct the method.
Give me a hint
Subtract final and initial squared-speed energies separately.
Compare my reasoning
- Final kinetic energy is 9 J.
- Initial kinetic energy is 36 J.
- Their difference is −27 J, not the positive square of the speed difference.
Use ½m(v² − u²); the change is −27 J.
Look for these in your work
- Keep the difference-of-squares structure.
- Explain the decreasing-store sign.
Common mistakes
- Keep the one half.
- Square speed before completing the product.
- Find an energy change by subtracting energies, not squaring a speed difference.
What can you explain now?
What is kinetic energy of 0.500 kg at 4 m/s?
Compare with the explanation
4.0 J.
½ × 0.500 × 4² = 4.0 J.
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Make it stick
Recall the kinetic-energy formula, explain doubled speed and distinguish kinetic energy from signed momentum.
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.