Kinetic-energy change in collisions
Key idea: First use momentum conservation to find collision velocities, then compare total kinetic energies. In an inelastic collision, the difference transfers to deformation, heating and sound.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Add signed momenta and positive kinetic energies in one frame.
- Predict the total momentum and total kinetic energy.
- Follow the energy changes when bodies stick together.
- Explain how another energy store can increase kinetic energy.
- Find momentum, both kinetic energies and the percentage loss.
- Find common velocity, both kinetic energies and their difference.
- Find common velocity and the decimal fraction of energy lost.
- Find the cars’ common speed and percentage energy loss.
- Show the full calculation and explain the energy destinations.
- Use momentum and spring energy to find both signed velocities.
- Find momentum, kinetic energies, velocity and the energy destination.
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ContinueTwo cars collide and lock together. With negligible external impulse during contact, their total momentum is the same just after the crash as just before, but they are clearly moving much more slowly as a pair, and their crumpled, warm metal shows where the kinetic energy went. This lesson shows how to calculate the change in kinetic energy in a collision and explain where the energy goes.
Momentum is conserved; kinetic energy usually is not
When no resultant external force acts, the total momentum of the colliding bodies is always conserved. Their total kinetic energy usually changes. Only a perfectly elastic collision keeps it the same.
To compare, add the kinetic energies of all the bodies before and after:
Momenta in opposite directions can cancel, but kinetic energies cannot: each (1/2)mv² is nonnegative, with zero at rest and a positive value for motion in either direction. Measure every velocity relative to the same thing, usually the ground.
Check your understanding 1 Can momenta cancel but energies remain?
Two equal-mass carts have equal nonzero speeds in opposite directions. Predict their total momentum and total kinetic energy in the ground frame.
Show answer
The signed momenta cancel, giving zero total momentum. Each moving cart has positive kinetic energy, so their total kinetic energy is positive. Zero total momentum is compatible with motion.
Where the kinetic energy goes
In an inelastic collision, the kinetic energy that is lost is transferred mainly to internal energy of the bodies, which are deformed and warm up slightly, and to the surroundings as sound. Total energy is still conserved: the energy has been transferred, not destroyed.
For fixed initial masses and velocities, bodies that stick together lose the most kinetic energy that conservation of momentum allows. Assume negligible external impulse during the interactions below. Find the common velocity from momentum first, then compare kinetic energies.
Check your understanding 2 Can kinetic energy increase?
Can the total kinetic energy of the bodies be greater after an interaction than before?
Show answer
Yes, if energy is released from another store. In an explosion, chemical energy becomes kinetic energy: the pieces fly apart from rest with zero total momentum but plenty of kinetic energy. A compressed spring pushing two trolleys apart does the same.
Compare the source store, transfer into kinetic energy and cancellation of the pieces’ signed momenta. Momentum conservation alone does not fix the kinetic-energy total.
Worked example 1
Two carts that stick
Problem
On a level track, a 1.50 kg cart moving at + 4.00 m s⁻¹ collides with a 0.500 kg cart moving at -2.00 m s⁻¹. They stick together. Find their common velocity, the kinetic energy lost and the percentage of the initial kinetic energy lost.
Show full solution
Common velocity
Method
Conserve momentum, with one common final velocity.Reason
Momentum, not kinetic energy, is conserved in a collision where the bodies stick.Working
Answer
1.50(4.00) + 0.500(-2.00) = (1.50 + 0.500)v, so 5.00 = 2.00v and v = +2.50 m s⁻¹
Kinetic energy before and after
Reason
Before, both carts move; after, the joined carts move together.Working
Answer
Before: (1/2)(1.50)(4.00)² + (1/2)(0.500)(2.00)² = 12.0 + 1.00 = 13.0 J. After: (1/2)(2.00)(2.50)² = 6.25 J.
Loss
Working
Answer
Kinetic energy lost = 13.0-6.25 = 6.75 J, which is 6.75/13.0 × 100% = 51.9% of the initial kinetic energy.
Quick check
Momentum after is 2.00(2.50) = +5.00 N s, the same as before, even though half the kinetic energy has gone.
Guided practice 2
A cart sticks to an identical cart
Problem
A 2.00 kg cart moving at + 4.00 m s⁻¹ hits an identical stationary cart, and they stick together. Find the common velocity and the kinetic energy lost.
Find the velocity and the loss
Show the steps Optional working
Show the steps Optional working
Hints
Hint 1
combined mass
Which bodies move together after the collision?
Show solution step by step
Common velocity
Reason
Momentum is conserved.Working
Answer
2.00(4.00) = 4.00v, so v = +2.00 m s⁻¹
Kinetic energy lost
Working
Answer
Before: (1/2)(2.00)(4.00)² = 16.0 J. After: (1/2)(4.00)(2.00)² = 8.00 J. Lost: 8.00 J, half of the initial kinetic energy.
Check your understanding 3
Two carts of equal mass m move towards each other at + 4.00 m s⁻¹ and -2.00 m s⁻¹, and stick together. Find their common velocity and the fraction of the initial kinetic energy that is lost. Enter the fraction as a decimal, rather than a percentage.
Show the steps Optional working
Show solution
Momentum: 4.00m-2.00m = 2mv, so v = +1.00 m s⁻¹. Kinetic energy before: (1/2)m(4.00² + 2.00²) = 10.0m. After: (1/2)(2m)(1.00)² = 1.00m. The fraction lost is (10.0m-1.00m)/10.0m = 0.900, or 90%. The mass cancels, so you do not need its value.
Check your understanding 4 A car crash
A 900 kg car moving at 20.0 m s⁻¹ hits a stationary 1100 kg car, and they move off together. Ignoring friction from the road during the brief collision, find their common speed and the percentage of the kinetic energy lost. Where does that energy go?
Show the steps Optional working
Show solution
Momentum: 900(20.0) = 2000v, so v = 9.00 m s⁻¹. Kinetic energy before: (1/2)(900)(20.0)² = 1.80 × 10⁵ J. After: (1/2)(2000)(9.00)² = 8.10 × 10⁴ J. The loss is 9.90 × 10⁴ J, or 55.0%. It goes mainly into deforming and heating the cars, and into sound.
Check your understanding 5 Explain the crash energy account
Show your full momentum and kinetic-energy calculation for the cars, including the 55.0% loss. Explain where the energy is transferred; paper working is welcome.
Show answer
Momentum gives 900(20.0) = 2000v, so v = 9.00 m s⁻¹. Before: (1/2)(900)(20.0)² = 180000 J. After: (1/2)(2000)(9.00)² = 81000 J. Loss: 99000 J, so 99000/180000 × 100% = 55.0%. The energy is transferred mainly to deformation and heating of the cars, and to the surroundings as sound. Compare the combined mass after contact, percentage denominator and physical destinations.
Check your understanding 6 The reverse energy change
Two identical 0.20 kg carts start at rest, separated by a compressed spring of negligible mass storing 1.8 J. All this energy becomes translational kinetic energy, with negligible external impulse. Cart A moves in the positive direction. Find the signed velocity of each cart.
Show the steps Optional working
Show answer
Momentum gives 0 = 0.20vA + 0.20vB, so vB = -vA. Energy gives 1.8 = (1/2)(0.20)vA² + (1/2)(0.20)vB² = 0.20vA². Thus vA = +3.0 m s⁻¹ and vB = -3.0 m s⁻¹. Each cart gains 0.90 J; the spring loses 1.8 J. A shared-velocity equation would be inappropriate because these carts separate rather than stick.
Common mistakes
- Using conservation of kinetic energy to find the velocity after an inelastic collision.
- Thinking that because momentum is conserved, kinetic energy must be too.
- Letting a body moving in the negative direction have negative kinetic energy.
- Using the mass of one cart instead of the combined mass for the kinetic energy after they stick.
- Saying kinetic energy “disappears”; it is transferred to internal energy and sound.
Before you move on
Check your understanding 7
Two 0.50 kg carts approach each other at + 3.0 m s⁻¹ and -3.0 m s⁻¹ and stick together. No external impulse acts. Find the total momentum and kinetic energy before contact, then their common velocity and the kinetic energy transferred during the collision.
Show the steps Optional working
Show answer
The total momentum is 0.50(3.0) + 0.50(-3.0) = 0 kg m s⁻¹. The initial kinetic energy is 2[(1/2)(0.50)(3.0)²] = 4.5 J. Conservation of momentum gives (1.0)v = 0, so the joined carts are at rest and have zero kinetic energy. All 4.5 J is transferred, mainly to internal energy and sound. Zero total momentum did not imply zero initial kinetic energy.
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Syllabus and review details
- GCE A-Level H2 Physics 2027 · 2027
Content Overview, PDF pages 9–10; Subject Content, PDF pages 11–30
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