Kinetic-energy change in collisions
Key idea: A closed system can conserve momentum while its kinetic energy changes. The difference is transferred into deformation, internal energy and sound.
Continue where you stopped
The core idea
H1 Physics 8867 · Lesson 3 of 3
Check your understandingBy the end of this lesson, you should be able to
- Calculate kinetic energy before and after a collision.
- Explain why momentum conservation does not imply kinetic-energy conservation.
- Find the fraction or percentage of kinetic energy transferred in an inelastic collision.
Learn the idea
Big question: How can momentum remain unchanged while a collision transfers most of the kinetic energy?
Separate a vector rule from a scalar account
Momentum conservation follows from negligible external impulse and includes direction. Kinetic energy is a scalar store and can be transferred to deformation, internal energy and sound while total energy remains conserved.
For a sticking collision, first use momentum to find the common velocity. Only then calculate total kinetic energy before and after; trying to conserve kinetic energy would assume the conclusion.
Check your understanding: Is kinetic energy destroyed in an inelastic collision?
No. Some is transferred to other stores, while total energy remains conserved.
Use system totals for a percentage
Add the kinetic energies of every body before and after. The transferred fraction is (Ekinitial − Ekfinal)/Ekinitial.
Velocity signs matter in momentum, but kinetic energy uses v². A left-moving body therefore has positive kinetic energy.
Check your understanding: Can final total kinetic energy be larger in a collision?
Yes, if another store such as chemical or elastic energy is released; this is sometimes called a superelastic interaction.
Key ideas
- Calculate system totals, not percentage changes for one body alone.
- Kinetic energy uses speed squared and is never negative.
- A collision can transfer kinetic energy while conserving total energy.
Relationships to know
Ek = ½mv²fraction transferred = (Ekinitial − Ekfinal)/Ekinitial
Follow the reasoning
Worked example
Find the percentage transferred in a sticking collision
Question: A 1.5 kg cart at +4.0 m s⁻¹ sticks to a 0.50 kg cart at −2.0 m s⁻¹. Find their common velocity and the percentage of initial kinetic energy transferred to other stores.
Step 1: Conserve momentum
Why: The common speed comes from the vector conservation law.
Working: Initial p = 1.5(4.0) + 0.50(−2.0) = 5.0 kg m s⁻¹. Thus v = 5.0/2.0 = +2.5 m s⁻¹.
Step 2: Calculate both kinetic totals
Why: The percentage must refer to the complete system.
Working: Eki = ½(1.5)(4.0²) + ½(0.50)(2.0²) = 13.0 J. Ekf = ½(2.0)(2.5²) = 6.25 J.
Step 3: Find the transferred fraction
Why: The decrease is compared with the initial store.
Working: Fraction = (13.0 − 6.25)/13.0 = 0.519.
Answer: The carts move at +2.5 m s⁻¹ and about 51.9% of the initial kinetic energy is transferred to other stores.
Check: The final velocity lies between the two initial velocities, and both kinetic-energy totals are positive.
Now try it with support
Practise with support
A 2.0 kg cart at 3.0 m s⁻¹ sticks to a stationary 1.0 kg cart. Find the kinetic energy transferred.
Hints
- The common velocity is found from momentum first.
- Subtract final total kinetic energy from initial total kinetic energy.
View the guided answer
Common velocity = 2.0 m s⁻¹. Initial Ek = 9.0 J; final Ek = ½(3.0)(2.0²) = 6.0 J. Therefore 3.0 J is transferred internally.
Your turn
Practise independently
Two equal carts approach at 4.0 and 2.0 m s⁻¹ and stick. Determine their common velocity and the fraction of initial kinetic energy transferred to other stores.
Check your answer
Take the velocities as +4.0 and −2.0 m s⁻¹ for equal mass m. Their common velocity is (4m − 2m)/(2m) = +1.0 m s⁻¹. Initial Ek = ½m(4² + 2²) = 10m; final Ek = ½(2m)(1²) = m. The fraction transferred is 9m/10m = 0.90, or 90%.
Common mistakes and exam guidance
Watch out for
- Using momentum equations to claim kinetic energy must also be conserved.
- Calculating percentage loss from only one cart’s initial energy when both carts were moving.
In an exam
- Keep the momentum and energy calculations on separate lines because they follow different rules.
- Use ‘transferred to internal energy/deformation/sound’ for a complete explanation.
Put the ideas together
Exam-style practice [7 marks]
A 900 kg car at 20 m s⁻¹ hits a stationary 1100 kg car and they move together. Calculate their speed, the kinetic energy transferred, and the percentage transferred. Explain why momentum conservation does not conflict with this change.
Plan before you answer
- Use total momentum for the common speed.
- Compare system kinetic-energy totals.
- Name the different physical rules governing momentum and energy stores.
View the marking points and model answer
Marking points
- Finds common speed 9.0 m s⁻¹.
- Finds initial Ek = 1.80 × 10⁵ J.
- Finds final Ek = 8.10 × 10⁴ J.
- Finds transfer = 9.90 × 10⁴ J.
- Finds 55% transferred.
- Names deformation/internal energy/sound.
- Explains momentum remains conserved because net external impulse is negligible while energy changes store.
Model answer
Momentum gives 900(20) = 2000v, so v = 9.0 m s⁻¹. Initial Ek = ½(900)(20²) = 1.80 × 10⁵ J and final Ek = ½(2000)(9.0²) = 8.10 × 10⁴ J. Thus 9.90 × 10⁴ J, or 55%, is transferred to deformation, internal energy and sound. Momentum conservation concerns the vector total when external impulse is negligible; total energy is also conserved, but kinetic energy can change store.
Finish from memory
Three-question recap
What is a perfectly inelastic collision?
Check
A collision in which the bodies stick and share a final velocity.
Is kinetic energy ever negative?
Check
No, because it depends on speed squared.
Write the fraction of initial kinetic energy transferred.
Check
(Eki − Ekf)/Eki.
Continue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H1 Physics
- Edition
- GCE A-Level H1 Physics 2027