Learning objectives
- I can explain and apply: Momentum conservation in an isolated system.
- I can explain and apply: Perfectly inelastic and central elastic collisions.
- I can explain and apply: Elastic rebound from a fixed wall.
Panhellenic Exams / Physics / Two-unit revision pilot
Collisions and relative motion
Collisions and relative motion: structured theory, worked examples, answered practice, and a mastery checklist for Panhellenic Exams.
CHAPTER PLAN
Learn, practise, and check your progress
Follow the steps in order or jump directly to the part you need.
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completed
Estimated active study time
156 minutes
1. Understand
Objectives, key ideas, and structured theory.
Open →2. Follow the method
Worked examples that explain every step.
Open →3. Practise
Graded tasks with hints and answer guidance.
Open →4. Consolidate
Assignments from core work to challenge.
Open →5. Check
A timed, marked test with solution guidance.
Open →
Progress is stored only in this browser and does not require AI credits.
Key ideas and checks
- p=mv and Δp=pafter−pbefore.
- Momentum has a sign along the chosen axis.
- Kinetic-energy conservation requires an elastic collision.
Unit
Core theory
The essential chapter ideas in a clear sequence before practice.
Understanding the concept
Choose a system of bodies and a common positive direction. If the net external impulse during collision is negligible, total system momentum is conserved. Each body's individual momentum is generally not conserved.
Method and conditions
After a perfectly inelastic collision the bodies move together, so final momentum is (m1+m2)V. Kinetic energy generally decreases. In a central elastic collision, both momentum and kinetic energy are conserved.
Checking and coverage boundaries
In an elastic normal rebound from a fixed wall, velocity changes sign but not magnitude. The wall exerts an impulse on the ball. This pilot focuses on collisions; it does not cover all the relative-motion content named in the existing route title.
Original pilot material — human educator review pending
Written specifically for YourFavTeacher with AI assistance, without copying past-paper prompts. Technical and numerical checks do not replace review by a human educator. It is not an official paper or endorsed by an examination body.
Physics
Worked examples
Follow the method step by step and check why every step is valid.
Worked example 1
In an isolated horizontal system, a 1 kg body at v=12 m/s sticks to a stationary 1 kg body. Find their common velocity.
- 1Total momentum is conserved; kinetic energy need not be.
- 21·12+1·0=(1+1)V ⇒ V=6 m/s
- 3The common velocity is in the initial positive direction.
6 m/s
Worked example 2
A 1 kg ball moves at v=6 m/s and rebounds elastically, normally from a fixed wall. Find its momentum change along the initially positive axis.
- 1Speed is unchanged, but the direction reverses.
- 2Δp=1·(−6)−1·6=−12 kg·m/s
- 3The ball alone does not conserve momentum: the wall exerts an external impulse.
−12 kg·m/s
Worked example 3
A 1 kg body's velocity changes from −5 to 6 m/s. Find its momentum change.
- 1Calculate final minus initial momentum with signs.
- 2Δp=1·[6−(−5)]=11 kg·m/s
- 3Subtracting a negative initial value becomes addition.
11 kg·m/s
Collisions and relative motion
Practice with answers
Eight graded tasks from core fluency to exam-style application. Work independently before opening a hint or answer.
Core fluency6 minutes
In an isolated horizontal system, a 1 kg body at v=6 m/s sticks to a stationary 1 kg body. Find their common velocity.
Hint
Total momentum is conserved; kinetic energy need not be.
Answer guide
Total momentum is conserved; kinetic energy need not be. 1·6+1·0=(1+1)V ⇒ V=3 m/s The common velocity is in the initial positive direction. 3 m/s
Core fluency6 minutes
In an isolated system two equal 1 kg masses stick together: the first has v=6 m/s and the second is stationary. Find the kinetic energy loss.
Hint
First use momentum to obtain common velocity V=v/2.
Answer guide
First use momentum to obtain common velocity V=v/2. K_i=0.5·6²=18 J; K_f=0.5·2·3²=9 J; K_i−K_f=9 J The positive loss is transferred to other forms of energy. 9 J
Core fluency6 minutes
Equal masses collide head-on elastically. Initially v₁=3 m/s and v₂=−2 m/s. Find final velocities.
Hint
Equal masses in a head-on elastic collision exchange velocities.
Answer guide
Equal masses in a head-on elastic collision exchange velocities. υ₁′=−2 m/s; υ₂′=3 m/s Both total momentum and kinetic energy are preserved. υ₁′=−2 m/s; υ₂′=3 m/s
Application6 minutes
A 2 kg body has v₁=3 m/s rightwards and a 1 kg body v₂=−2 m/s. Find total momentum.
Hint
Momentum is a vector; add signed values along the common axis.
Answer guide
Momentum is a vector; add signed values along the common axis. p=2·3+1·(−2)=4 kg·m/s Keep SI units and check the sign. 4 kg·m/s →
Application6 minutes
A 1 kg ball moves at v=3 m/s and rebounds elastically, normally from a fixed wall. Find its momentum change along the initially positive axis.
Hint
Speed is unchanged, but the direction reverses.
Answer guide
Speed is unchanged, but the direction reverses. Δp=1·(−3)−1·3=−6 kg·m/s The ball alone does not conserve momentum: the wall exerts an external impulse. −6 kg·m/s
Application6 minutes
Can kinetic energy conservation be used for every collision?
Hint
Momentum is conserved when external impulse is negligible.
Answer guide
Momentum is conserved when external impulse is negligible. Kinetic energy is conserved only in an elastic collision. In a sticking collision there is deformation and other energy transfer. No; an elastic collision must be stated or justified.
Reasoning6 minutes
In an isolated system, a 2 kg body at v=6 m/s sticks to a stationary 1 kg body. Find common velocity.
Hint
Add the masses for the motion after sticking.
Answer guide
Add the masses for the motion after sticking. 2·6=3V ⇒ V=4 m/s Common velocity lies between the initial velocities. 4 m/s
Reasoning6 minutes
A 1 kg body's velocity changes from −2 to 3 m/s. Find its momentum change.
Hint
Calculate final minus initial momentum with signs.
Answer guide
Calculate final minus initial momentum with signs. Δp=1·[3−(−2)]=5 kg·m/s Subtracting a negative initial value becomes addition. 5 kg·m/s
Common mistakes
- Adding momentum magnitudes instead of signed values.
- Conserving kinetic energy in a perfectly inelastic collision.
- Forgetting the combined mass after sticking together.
Mastery check
- I can solve independently and check: Momentum conservation in an isolated system.
- I can solve independently and check: Perfectly inelastic and central elastic collisions.
- I can solve independently and check: Elastic rebound from a fixed wall.
Collisions and relative motion
Chapter homework
Six distinct assignments from core fluency to challenge, each with an estimated time, hint, and answer guide.
Core8 minutes
In an isolated horizontal system, a 1 kg body at v=8 m/s sticks to a stationary 1 kg body. Find their common velocity.
Hint
Total momentum is conserved; kinetic energy need not be.
Answer guide
Total momentum is conserved; kinetic energy need not be. 1·8+1·0=(1+1)V ⇒ V=4 m/s The common velocity is in the initial positive direction. 4 m/s
Core8 minutes
In an isolated system two equal 1 kg masses stick together: the first has v=8 m/s and the second is stationary. Find the kinetic energy loss.
Hint
First use momentum to obtain common velocity V=v/2.
Answer guide
First use momentum to obtain common velocity V=v/2. K_i=0.5·8²=32 J; K_f=0.5·2·4²=16 J; K_i−K_f=16 J The positive loss is transferred to other forms of energy. 16 J
Stretch8 minutes
Equal masses collide head-on elastically. Initially v₁=4 m/s and v₂=−3 m/s. Find final velocities.
Hint
Equal masses in a head-on elastic collision exchange velocities.
Answer guide
Equal masses in a head-on elastic collision exchange velocities. υ₁′=−3 m/s; υ₂′=4 m/s Both total momentum and kinetic energy are preserved. υ₁′=−3 m/s; υ₂′=4 m/s
Stretch8 minutes
A 2 kg body has v₁=4 m/s rightwards and a 1 kg body v₂=−3 m/s. Find total momentum.
Hint
Momentum is a vector; add signed values along the common axis.
Answer guide
Momentum is a vector; add signed values along the common axis. p=2·4+1·(−3)=5 kg·m/s Keep SI units and check the sign. 5 kg·m/s →
Challenge8 minutes
A 1 kg ball moves at v=4 m/s and rebounds elastically, normally from a fixed wall. Find its momentum change along the initially positive axis.
Hint
Speed is unchanged, but the direction reverses.
Answer guide
Speed is unchanged, but the direction reverses. Δp=1·(−4)−1·4=−8 kg·m/s The ball alone does not conserve momentum: the wall exerts an external impulse. −8 kg·m/s
Challenge8 minutes
Can kinetic energy conservation be used for every collision?
Hint
Momentum is conserved when external impulse is negligible.
Answer guide
Momentum is conserved when external impulse is negligible. Kinetic energy is conserved only in an elastic collision. In a sticking collision there is deformation and other energy transfer. No; an elastic collision must be stated or justified.
45 minutes / 40 marks
Full chapter test
A timed, full-mark self-assessment with model-answer guidance.
Show working, units and a final check. Equivalent correct methods are accepted. Timing is a revision guide, not official examination conditions.
1. In an isolated horizontal system, a 1 kg body at v=10 m/s sticks to a stationary 1 kg body. Find their common velocity.
5 marksAnswer guide
Total momentum is conserved; kinetic energy need not be. 1·10+1·0=(1+1)V ⇒ V=5 m/s The common velocity is in the initial positive direction. 5 m/s
2. In an isolated system two equal 1 kg masses stick together: the first has v=10 m/s and the second is stationary. Find the kinetic energy loss.
5 marksAnswer guide
First use momentum to obtain common velocity V=v/2. K_i=0.5·10²=50 J; K_f=0.5·2·5²=25 J; K_i−K_f=25 J The positive loss is transferred to other forms of energy. 25 J
3. Equal masses collide head-on elastically. Initially v₁=5 m/s and v₂=−4 m/s. Find final velocities.
5 marksAnswer guide
Equal masses in a head-on elastic collision exchange velocities. υ₁′=−4 m/s; υ₂′=5 m/s Both total momentum and kinetic energy are preserved. υ₁′=−4 m/s; υ₂′=5 m/s
4. A 2 kg body has v₁=5 m/s rightwards and a 1 kg body v₂=−4 m/s. Find total momentum.
5 marksAnswer guide
Momentum is a vector; add signed values along the common axis. p=2·5+1·(−4)=6 kg·m/s Keep SI units and check the sign. 6 kg·m/s →
5. A 1 kg ball moves at v=5 m/s and rebounds elastically, normally from a fixed wall. Find its momentum change along the initially positive axis.
5 marksAnswer guide
Speed is unchanged, but the direction reverses. Δp=1·(−5)−1·5=−10 kg·m/s The ball alone does not conserve momentum: the wall exerts an external impulse. −10 kg·m/s
6. Can kinetic energy conservation be used for every collision?
5 marksAnswer guide
Momentum is conserved when external impulse is negligible. Kinetic energy is conserved only in an elastic collision. In a sticking collision there is deformation and other energy transfer. No; an elastic collision must be stated or justified.
7. In an isolated system, a 2 kg body at v=12 m/s sticks to a stationary 1 kg body. Find common velocity.
5 marksAnswer guide
Add the masses for the motion after sticking. 2·12=3V ⇒ V=8 m/s Common velocity lies between the initial velocities. 8 m/s
8. A 1 kg body's velocity changes from −4 to 5 m/s. Find its momentum change.
5 marksAnswer guide
Calculate final minus initial momentum with signs. Δp=1·[5−(−4)]=9 kg·m/s Subtracting a negative initial value becomes addition. 9 kg·m/s
Unit
Official sources and verification
Curriculum reference sources. Always confirm the teaching sequence with the school and tutor.
Physics
Strict methodology by question: diagram, principles/laws, equations, numerical accuracy, and qualitative explanation.
What this chapter covers
The structure follows the official textbook layout and is used to organise study.
Momentum conservation in an isolated system
Perfectly inelastic and central elastic collisions
Elastic rebound from a fixed wall
Where to focus
The areas that usually create mistakes or need extra revision.
I can solve independently and check: Momentum conservation in an isolated system.
I can solve independently and check: Perfectly inelastic and central elastic collisions.
I can solve independently and check: Elastic rebound from a fixed wall.
Sources, daily material, and resources
Where to start: textbook, daily material, PDFs, videos, and worked examples.
Start from the official textbook or specification referenced on the subject page.
Use the notes and examples as support, not as a replacement for the official syllabus.
Check the current syllabus version before exam preparation.
Practice by subtopic
Targeted practice before full tests so coverage is clear.
In an isolated horizontal system, a 1 kg body at v=6 m/s sticks to a stationary 1 kg body. Find their common velocity.
In an isolated system two equal 1 kg masses stick together: the first has v=6 m/s and the second is stationary. Find the kinetic energy loss.
Equal masses collide head-on elastically. Initially v₁=3 m/s and v₂=−2 m/s. Find final velocities.
A 2 kg body has v₁=3 m/s rightwards and a 1 kg body v₂=−2 m/s. Find total momentum.
A 1 kg ball moves at v=3 m/s and rebounds elastically, normally from a fixed wall. Find its momentum change along the initially positive axis.
Can kinetic energy conservation be used for every collision?
In an isolated system, a 2 kg body at v=6 m/s sticks to a stationary 1 kg body. Find common velocity.
A 1 kg body's velocity changes from −2 to 3 m/s. Find its momentum change.
Mocks and progress checks
How to measure progress in this chapter and when it enters a cumulative mock.
Start with untimed practice by subtopic.
Move to a short timed checkpoint only after completing the mastery checklist.
Record each error with the correct method and revisit it after 48 hours.
Next step
What to do after finishing the chapter and how it connects to the next unit.
Complete the practice without support.
Explain the core method aloud in under two minutes.
Continue to the next chapter or request targeted tutor support.
Note: for the official examinable syllabus of each school year, always confirm with the school, tutor, and current Ministry/IEP announcements.