Momentum is always conserved in a collision, but kinetic energy is not. That single distinction divides every collision into two types: elastic (kinetic energy preserved) and inelastic (some kinetic energy lost to heat and deformation). In this final lesson of the focus area you will classify collisions, solve them with conservation of momentum, and design a practical to test the law for yourself.
A moving trolley collides with an identical stationary trolley. In one trial they stick together; in another, the moving one stops dead and the other moves off at the original speed.
Before reading: in which trial is kinetic energy preserved, and is momentum conserved in both? Write your prediction and one sentence of reasoning.
Momentum always survives; kinetic energy may not
Every collision keeps the same total momentum, but whether the total kinetic energy survives is what tells the two collision types apart.
In any collision of an isolated system, momentum is always conserved. The difference between collision types lies in the kinetic energy:
To classify a collision, calculate the total kinetic energy before and after: if it is unchanged, the collision is elastic; if it has decreased, it is inelastic.
Momentum is conserved in every collision of an isolated system. In an elastic collision total kinetic energy is also conserved; in an inelastic collision some kinetic energy is converted to heat, sound and deformation (maximum loss when objects stick together). Classify by comparing total $KE$ before and after.
Pause, copy the highlighted distinction into your book before moving on.
Did you get this? True or false: momentum is conserved in both elastic and inelastic collisions.
Total momentum before equals total momentum after
We just classified collisions by kinetic energy. That raises a question: how do we actually find the velocity after a collision? This card answers it → apply conservation of momentum, $\sum p_{before} = \sum p_{after}$, keeping directions consistent.
Because total momentum is unchanged, you can write the momentum before equal to the momentum after and solve for the unknown velocity.
Conservation of momentum, on the NESA formula sheet, gives:
$$\sum p_{\text{before}} = \sum p_{\text{after}} \qquad m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$
Choose a positive direction and keep signs consistent (momentum is a vector). For a perfectly inelastic collision, the objects share a common final velocity, so the right side becomes $(m_1 + m_2)v$.
A $1.0\ \text{kg}$ trolley moving at $6.0\ \text{m s}^{-1}$ collides with and sticks to a stationary $2.0\ \text{kg}$ trolley. Find their common velocity after the collision, and state whether the collision is elastic.
Solve collisions with $\sum p_{before} = \sum p_{after}$. For sticking objects, $m_1u_1 + m_2u_2 = (m_1 + m_2)v$. Here $v = 6.0/3.0 = 2.0\ \text{m s}^{-1}$; momentum is conserved but kinetic energy falls from $18\ \text{J}$ to $6.0\ \text{J}$, so the collision is inelastic.
Add the highlighted method to your notes before the check below.
Quick check: A $2.0\ \text{kg}$ ball at $3.0\ \text{m s}^{-1}$ strikes and sticks to a stationary $1.0\ \text{kg}$ ball. Their common velocity is…
Pattern C, Design, measure, analyse
Design a practical to test whether momentum is conserved when two dynamics trolleys collide. In your book: (1) describe the equipment and how you would measure the velocity of each trolley before and after a collision (e.g. motion sensors, light gates, or ticker tape); (2) state two variables you would control (e.g. a level, low-friction track; known trolley masses); (3) describe how you would calculate the total momentum before and after and compare them; (4) explain why your measured "after" momentum might be slightly less than "before", and how repeating trials improves the reliability of your conclusion.
Odd one out: three of these are true of a PERFECTLY INELASTIC collision. Which one is not?
Dot point: Classify collisions as elastic or inelastic by calculating and comparing the total kinetic energy before and after the collision
A vector that always balances, and a scalar that can be spent
We just solved an inelastic collision where energy was lost. That raises a question: how can momentum be conserved while kinetic energy is not? This card answers it → because the two quantities are governed by different rules.
Momentum and kinetic energy are different kinds of bookkeeping, so one can be perfectly preserved while the other is partly spent.
Momentum is conserved because of Newton's third law: the colliding objects exert equal and opposite forces for the same time, so the impulses cancel and the total momentum cannot change, regardless of what happens to the objects. Kinetic energy, however, can be converted into other forms (heat, sound, the energy of permanent deformation). In an inelastic collision, some kinetic energy is "spent" this way, so $KE$ decreases even though momentum is unchanged. Total energy is still conserved overall, it has just left the mechanical (kinetic) form. This is why a real car crash conserves momentum exactly while crumpling metal and generating heat.
Momentum is conserved because colliding objects exert equal and opposite impulses (Newton's third law), so the total cannot change. Kinetic energy can be converted into heat, sound and deformation, so it decreases in inelastic collisions. Total energy is still conserved, it has simply left the kinetic form.
Pause, copy the highlighted reasoning into your book before the check below.
Last check: Decide if the statement is true or false.
In a perfectly inelastic collision, both momentum and kinetic energy are conserved.
Match each collision type with what is conserved:
Dot point: Solve problems involving the conservation of momentum in elastic collisions using conservation of p and KE
Momentum is conserved in every collision; kinetic energy only in elastic ones
Use the simulation. Set $m_1 = 2.0\ \text{kg}$ at $u_1 = +3.0\ \text{m s}^{-1}$ striking $m_2 = 4.0\ \text{kg}$ at rest, then drag the restitution slider to $e = 0.50$ and read the ledger. What does it show?
Use the simulation. Press Light hits heavy: a $0.5\ \text{kg}$ body at $+4.0\ \text{m s}^{-1}$ strikes a stationary $10.0\ \text{kg}$ body perfectly elastically, and finishes at $-3.62\ \text{m s}^{-1}$. True or false: because body 1 reversed, the total momentum after the collision is smaller than before.
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
AnalyseBand 6(5 marks) 1. A $3.0\ \text{kg}$ trolley moving at $4.0\ \text{m s}^{-1}$ collides with a stationary $1.0\ \text{kg}$ trolley and they stick together. (a) Find their common velocity. (b) Calculate the total kinetic energy before and after. (c) Classify the collision and explain.
1 mark: applies $m_1u_1 = (m_1+m_2)v$ · 1 mark: $v = 12/4.0 = 3.0\ \text{m s}^{-1}$ · 1 mark: $KE_{before} = \tfrac{1}{2}(3.0)(4.0)^2 = 24\ \text{J}$ · 1 mark: $KE_{after} = \tfrac{1}{2}(4.0)(3.0)^2 = 18\ \text{J}$ · 1 mark: inelastic because KE decreased ($6\ \text{J}$ lost) while momentum was conserved
ApplyBand 6(5 marks) 2. A $0.20\ \text{kg}$ ball moving at $5.0\ \text{m s}^{-1}$ collides head-on and elastically with a stationary $0.20\ \text{kg}$ ball. Using conservation of momentum AND conservation of kinetic energy, determine the velocity of each ball after the collision, and explain why momentum alone is not enough to solve this.
1 mark: writes momentum conservation · 1 mark: writes kinetic energy conservation · 1 mark: recognises two unknowns require two equations · 1 mark: $v_1 = 0$, $v_2 = 5.0\ \text{m s}^{-1}$ · 1 mark: explains momentum alone gives one equation in two unknowns
Dot point: Solve problems involving the conservation of momentum in elastic collisions using conservation of p and KE
EvaluateBand 6(4 marks) 3. Describe how a practical investigation using two trolleys on a track could test the conservation of momentum in a collision, identifying the measurements required and one significant source of error.
1 mark: measure both masses · 1 mark: measure velocities before and after, e.g. with light gates or motion sensors · 1 mark: compare total $\sum p$ before with $\sum p$ after · 1 mark: a valid error source, such as friction or track misalignment, with its effect
Dot point: Conduct a practical investigation to analyse the conservation of momentum in collisions
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (5 marks): (a) By conservation of momentum, $m_1u_1 + m_2u_2 = (m_1 + m_2)v$, so $3.0 \times 4.0 + 0 = (3.0 + 1.0)v$, giving $v = \dfrac{12}{4.0} = 3.0\ \text{m s}^{-1}$. (b) $KE_{before} = \tfrac{1}{2}(3.0)(4.0)^2 = 24\ \text{J}$; $KE_{after} = \tfrac{1}{2}(4.0)(3.0)^2 = 18\ \text{J}$. (c) The collision is inelastic: the total kinetic energy has decreased from $24\ \text{J}$ to $18\ \text{J}$ (a loss of $6\ \text{J}$ to heat, sound and deformation), even though momentum is conserved ($12\ \text{kg m s}^{-1}$ before and after). Because the trolleys stick together, it is a perfectly inelastic collision.
Q2 (5 marks): Let the final velocities be $v_1$ and $v_2$, taking the initial direction as positive. Conservation of momentum gives $$0.20(5.0) + 0 = 0.20v_1 + 0.20v_2 ;\Rightarrow; v_1 + v_2 = 5.0$$ Because the collision is elastic, kinetic energy is also conserved: $$\tfrac{1}{2}(0.20)(5.0)^2 = \tfrac{1}{2}(0.20)v_1^2 + \tfrac{1}{2}(0.20)v_2^2 ;\Rightarrow; v_1^2 + v_2^2 = 25$$ Substituting $v_2 = 5.0 - v_1$ into the second equation gives $v_1^2 + (5.0-v_1)^2 = 25$, so $2v_1^2 - 10v_1 = 0$ and $v_1(v_1 - 5.0) = 0$. The root $v_1 = 5.0$ is the pre-collision state, so the physical solution is $v_1 = 0$ and $v_2 = 5.0\ \text{m s}^{-1}$: the first ball stops dead and the second moves off at the original speed. Momentum alone is not enough because it supplies a single equation containing two unknown velocities; the kinetic energy condition supplies the second equation the system needs.
Q3 (4 marks): Measure the mass of each trolley on a balance. Set the track as level as possible and use light gates or a motion sensor to record the velocity of each trolley immediately before and immediately after the collision, repeating each run several times to average random error. Compute $\sum p$ before as $m_1u_1 + m_2u_2$ and $\sum p$ after as $m_1v_1 + m_2v_2$, keeping a consistent sign convention because momentum is a vector, and compare the two totals. Within experimental uncertainty they should agree. The most significant source of error is friction between the trolleys and the track, together with any slight tilt in the track: both act as external forces on the system, so the system is not perfectly isolated and the measured momentum after the collision comes out slightly lower than before. Using a low-friction air track, or compensating the tilt so a trolley coasts at constant velocity, reduces this.
You can now classify collisions as elastic or inelastic, solve them with conservation of momentum, and design a practical to test the law. That completes Fundamentals of mechanics, from vectors and forces through energy and momentum. These tools underpin all of Year 12 dynamics.
↩ Back to focus-area hubAt the start you predicted which trolley trial preserves kinetic energy and whether momentum is conserved in both. The answer: momentum is conserved in both trials; kinetic energy is preserved only in the second (where the moving trolley stops and the other moves off at the original speed, an elastic collision), while sticking together loses kinetic energy (inelastic). If you thought momentum was lost when they stick, revisit Card 1.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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