Elastic & Inelastic Collisions
Key idea: Use momentum conservation and impulse to solve 1D collision problems, and distinguish elastic, inelastic, and completely inelastic collisions (A Level Physics).
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The core idea
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Learning objectives
- Use impulse and momentum conservation in one-dimensional elastic and inelastic collisions.
1. Definitions (Must Know)
A. Collision (impulsive interaction)
A collision is a short interaction where two bodies exert large forces on each other for a short time, so their momenta change significantly.
B. Elastic collision
An elastic collision is one where:
- total momentum is conserved, and
- total kinetic energy is conserved.
C. Inelastic collision
An inelastic collision is one where:
- total momentum is conserved, but
- total kinetic energy is not conserved (some is transferred to other energy stores, e.g. heating, sound, deformation).
D. Completely (perfectly) inelastic collision
A completely inelastic collision is an inelastic collision where the objects stick together after impact, so they have the same final velocity.
E. Impulse
Impulse is the change in momentum:
vector J = Δ vector p
See: Momentum & Impulse.
2. Key Ideas (What Earns Marks)
- Treat 1D collisions with a sign convention (e.g. right is +).
- Momentum conservation is a system law: ∑ p_before = ∑ p_after provided external impulse is negligible.
- Kinetic energy is conserved only in elastic collisions.
- Completely inelastic: “stick together” ⇒ v₁ = v₂ after collision.
- Perfectly elastic (1D): speed of approach = speed of separation: |u₁-u₂| = |v₂-v₁|
- Average collision force comes from impulse: F_avg = (Δ p)/(Δ t)
3. Detailed Explanations
A. When can you conserve momentum?
Momentum is conserved for a system when the resultant external force is zero, or when the external impulse is negligible during the short collision time.
This is why collision questions often say “smooth surface” or “no external forces”.
B. 1D collision setup (exam workflow)
- Choose a positive direction.
- Label initial velocities as u₁, u₂ and final velocities as v₁, v₂ (include signs).
- Apply momentum conservation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
- Use the collision type to get a second equation (if needed).
C. Completely inelastic collisions (stick together)
If the objects stick, they move together with a common velocity v:
m₁u₁ + m₂u₂ = (m₁ + m₂)v
Kinetic energy decreases because energy is transferred to other stores during deformation.
D. Perfectly elastic collisions (1D shortcut)
For a perfectly elastic collision in 1D, you use:
- momentum conservation, and
- speed of approach = speed of separation: |u₁-u₂| = |v₂-v₁|
This is equivalent to the “relative speed of approach equals relative speed of separation” statement in the syllabus.
E. Impulse and force in collisions
Collision forces vary with time, so questions often ask for an average force:
F_avg = (Δ p)/(Δ t)
For the two colliding objects, Newton’s 3rd law implies the impulses are equal in magnitude and opposite in direction.
4. Common Mistakes
- Forgetting momentum is a vector (dropping signs in 1D).
- Conserving momentum for one object instead of the system.
- Assuming “collision = elastic” (elastic must be stated or proven with KE/relative-speed rule).
- Using N = mg or friction assumptions without checking the question (external forces can break momentum conservation).
- Mixing up impulses: each object’s impulse is opposite, but total system impulse is zero (if isolated).
5. Exam Tips
- Write your sign convention first: “take rightwards as positive.”
- If the objects stick, replace v₁ and v₂ with one unknown v immediately.
- For elastic collisions, the relative-speed rule is often faster than using kinetic energy directly.
- After solving, do a quick sanity-check:
- does a heavier object change speed less?
- do directions (signs) make sense?
6. Worked Examples
Modelled example 1
Completely inelastic: common velocity
Problem
Study the worked solution
Keep the stated velocity signs
Method
The initial momenta oppose each other.Reason
Momentum is a vector, so the cart moving in the negative direction contributes negative momentum.Working
pᵢₙᵢₜᵢₐₗ = 2.0(4.0) + 3.0(-2.0)Use the sticking condition
Method
Both carts share one final velocity v.Reason
A completely inelastic collision joins the objects into a combined 5.0 kg mass.Working
2.0(4.0) + 3.0(-2.0) = (2.0 + 3.0)vSolve and interpret
Method
v = +0.40 m s⁻¹.Reason
The initial positive momentum slightly exceeds the negative momentum.Working
8-6 = 5v ⇒ v = +0.40 m s⁻¹
Guided practice 2
Kinetic energy lost in a completely inelastic collision
Problem
Try this before viewing the solution
Hints
Hint 1: find the shared speed first
View solution step by step
Find the common velocity
Method
v = 5.00 m s⁻¹.Reason
Momentum is conserved for the two-object system during the collision.Working
5.00(15.0) + 10.0(0) = 15.0v ⇒ v = 5.00 m s⁻¹Calculate the initial kinetic energy
Method
E_(k,i) = 562.5 J.Reason
Only the 5.00 kg object is moving initially.Working
E_(k,i) = (1/2)(5.00)(15.0)² = 562.5 JCalculate the final kinetic energy
Method
E_(k,f) = 187.5 J.Reason
After sticking, the full 15.0 kg mass moves at the common speed.Working
E_(k,f) = (1/2)(15.0)(5.00)² = 187.5 JFind the kinetic-energy loss
Method
375 J is transferred from kinetic energy.Reason
An inelastic collision conserves total energy but not the system’s kinetic energy.Working
E_(k,i)-E_(k,f) = 562.5-187.5 = 375 J
Common misconception 3
Impulse and average force (collision time given)
Learner claim
Try this before viewing the solution
View solution step by step
Identify the sign error
Method
The velocity change is -4.0-6.0 = -10.0 m s⁻¹.Reason
The final velocity is negative, not a positive speed to subtract from the initial speed.Working
Δ v = v-u = -4.0-(+6.0) = -10.0 m s⁻¹Calculate the impulse
Method
Δ p = -2.0 N s.Reason
Impulse equals the ball’s signed momentum change.Working
Δ p = 0.20(-10.0) = -2.0 N sCalculate average-force magnitude
Method
|F_avg| = 200 N.Reason
Divide the impulse magnitude by the contact time.Working
|F_avg| = 2.0/0.010 = 200 N
Examiner practice 4
Concept check: impulses on the two vehicles
Examination question
Try this before viewing the solution
View solution step by step
Select statement 2
1 markMethod
The impulse magnitudes on the lorry and motorcycle are equal.Reason
Their interaction forces form a Newton’s-third-law pair.Working
|F_(L → M)| = |F_(M → L)|Connect force to impulse
1 markMethod
The impulses are equal in magnitude and opposite in direction.Reason
The equal and opposite forces act over the same contact time.Working
J_(L → M) = -J_(M → L)
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the selection and Newton's-third-law justification.
Challenge 5
Perfectly elastic collision (use the relative-speed rule)
Independent transfer
Try this before viewing the solution
Hints
Hint 1: identify the two constraints
View solution step by step
Write momentum conservation
Method
5 = v₁ + 2v₂.Reason
External impulse is negligible for the two-trolley system.Working
1.0(5.0) + 2.0(0) = 1.0v₁ + 2.0v₂Use the elastic relative-speed condition
Method
5 = v₂-v₁.Reason
For this one-dimensional perfectly elastic collision, speed of separation equals the initial speed of approach.Working
u₁-u₂ = v₂-v₁ ⇒ 5 = v₂-v₁Solve simultaneously
Method
v₁ = -1.67 m s⁻¹ and v₂ = +3.33 m s⁻¹.Reason
Substituting v₂ = v₁ + 5 into momentum conservation determines both signed velocities.Working
5 = v₁ + 2(v₁ + 5) ⇒ v₁ = -5/3, v₂ = 10/3 m s⁻¹
7. Mind Stretchers
Mind stretcher 1: When momentum conservation can failExtension
A ball hits the ground and rebounds. You model the ball alone as the system.
Is momentum conserved for this “system”? Explain.
Show Answer
No. The ball experiences a large external impulse from the ground during the collision, so momentum is not conserved for the ball alone.
If you treat “ball + Earth” as the system, momentum can be conserved (external impulse negligible).
Mind stretcher 2: Elastic vs inelastic: what changes?Extension
Two objects collide in 1D. Total momentum is conserved.
What additional condition must hold for the collision to be elastic?
Show Answer
Total kinetic energy must be conserved (no net transfer of kinetic energy to other energy stores).
Mind stretcher 3: Optional (Enrichment)Extension
A. Coefficient of restitution (not required)
Some courses define:
e = (speed of separation)/(speed of approach)
In this syllabus, you only need the special case for perfectly elastic collisions (relative speed approach = relative speed separation). You do not need to use e.
B. 2D collisions (beyond core scope)
For 2D collisions, you conserve momentum separately in perpendicular directions (e.g. x and y). These problems usually need extra information beyond momentum alone.
Continue with the next resource in this course.
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027