You're staring at a physics problem. Two carts on a track. Think about it: a ball bouncing off a wall. A car crash in slow motion. The question asks: which of these collisions demonstrate momentum conservation?
Here's the short answer: all of them. Every single one. But that's not what your teacher wants to hear, and it's not what the textbook means when it asks the question.
What Is Momentum Conservation Really
Momentum conservation isn't a suggestion. It's not a rule that applies sometimes. Always. This isn't negotiable. In any isolated system — no external forces, no friction messing things up, no rocket engines firing — the total momentum before a collision equals the total momentum after. It falls straight out of Newton's third law and the symmetry of space itself.
But here's where students get tripped up: isolated system is doing a lot of heavy lifting in that sentence.
When your textbook shows you three collision diagrams and asks which demonstrate momentum conservation, it's testing whether you can spot the hidden external forces. The wall attached to the Earth. Now, the friction between tires and asphalt. The air resistance nobody talks about but always exists.
The Math You Actually Need
Momentum is mass times velocity. That's -6 kg·m/s. A 2 kg cart moving right at 3 m/s has +6 kg·m/s momentum. Same cart moving left at 3 m/s? Vector quantity — direction matters. Total momentum adds up like vectors, not scalars.
Conservation means: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
The subscripts i and f mean initial and final. That's it. That's the entire equation. But applying it? That's where the thinking happens.
Why It Matters / Why People Care
You might wonder why physics professors obsess over this. It's not because they enjoy watching students suffer through vector addition.
Momentum conservation lets you solve problems you couldn't touch otherwise. In real terms, skid marks, debris, witness statements — investigators reconstruct the crash using momentum. Particle physicists smash protons together at near light speed and track the spray of debris. Same principle. Think about it: two cars crash at an intersection. Different scale.
It's also the principle behind rocket propulsion. The rocket pushes exhaust backward; the exhaust pushes the rocket forward. Here's the thing — no air to push against needed. Momentum conservation works in the vacuum of space because it's fundamental — not because of air pressure.
Real talk: if you understand momentum conservation deeply, you understand something about how the universe works at every scale. From billiard balls to galaxy collisions.
How It Works: The Collision Types That Show Up On Tests
Textbooks love categorizing collisions. This leads to you'll see these three categories constantly. They matter because they tell you what else* is conserved besides momentum.
Elastic Collisions
Both momentum and kinetic energy conserved. The textbook examples: billiard balls, gas molecules, those Newton's cradle toys executives keep on their desks.
In a perfectly elastic collision between two objects, they bounce apart with no energy lost to heat, sound, or deformation. The relative velocity of approach equals the relative velocity of separation. That's a useful shortcut: v₁ᵢ - v₂ᵢ = v₂f - v₁f
But here's the thing — perfectly elastic collisions don't exist in the macroscopic world. Every bounce makes sound. Sound is energy leaving the system. Practically speaking, every deformation generates heat. Because of that, billiard balls are close* to elastic. Consider this: gas molecules at room temperature are very* close. But perfectly elastic? That's an idealization.
Inelastic Collisions
Momentum conserved. Now, kinetic energy not conserved. Some kinetic energy transforms to internal energy — heat, sound, permanent deformation.
Most real collisions are inelastic. That's why a car crash. On top of that, a lump of clay hitting the floor. A baseball hitting a bat (the ball deforms, the bat vibrates, you hear the crack — all energy leaving the kinetic pool).
The extreme end: perfectly inelastic collisions. The objects stick together. But maximum kinetic energy loss consistent with momentum conservation. They move as one combined mass afterward.
The "Which of These" Trap
This is the section you actually came for. When a multiple-choice question shows you diagrams and asks which demonstrate momentum conservation, here's your checklist:
Is the system isolated?
- Two pucks on frictionless ice? Yes. Momentum conserved.
- Two pucks on a regular air hockey table? Close enough for intro physics. The air cushion minimizes friction.
- A ball bouncing off a wall? No — unless you include the Earth in your system. The wall is attached to the building, which is attached to the ground. The Earth recoils imperceptibly. If your system is just the ball, momentum is not conserved. External force from the wall.
- A car crashing into a barrier? Same problem. The barrier connects to Earth. External impulse.
- Two cars colliding on an icy road? If friction is truly negligible during the collision instant, yes. Momentum conserved during the collision*. Afterward, friction stops them — that's a different time interval.
- A rocket exploding mid-air? Yes, if you ignore air resistance and gravity during the explosion*. The explosion forces are internal. Total momentum of all fragments equals the rocket's momentum just before.
The key insight: momentum conservation applies to the system* during the collision interval*. External forces like gravity and friction act over time. If the collision is fast enough, their impulse (force × time) is negligible compared to the huge internal forces. That's why we can use momentum conservation for car crashes even though friction exists — the crash forces dwarf friction forces during the 0.1-second impact.
Common Mistakes / What Most People Get Wrong
I've graded a lot of physics exams. These errors show up every single semester.
For more on this topic, read our article on ethnic religion ap human geography definition or check out compare positive and negative feedback mechanisms..
Confusing Momentum Conservation with Energy Conservation
Students see "collision" and start writing ½mv² equations. Stop. Ask yourself: is kinetic energy conserved? This leads to only if the problem says "elastic" or "perfectly elastic. Because of that, " If it says "inelastic" or "stick together" or gives you a coefficient of restitution, kinetic energy is not conserved. But momentum always* is (in an isolated system).
Forgetting Vector Directions
A 3 kg object moving right at 4 m/s collides with a 2 kg object moving left at 6 m/s. After collision, total momentum must still be zero. Total initial momentum: (3)(+4) + (2)(-6) = 12 - 12 = 0. So naturally, it stops dead. Also, if they stick together, the combined 5 kg mass has velocity zero. Students who treat momentum as scalar get this wrong every time.
Most people don't realize how important this is.
Including External Forces in the System
The ball-wall example again. If the system is ball + wall + Earth, momentum is conserved. Even so, the Earth gains momentum equal and opposite to the ball's change. But the Earth's mass is 6×10²⁴ kg. Its velocity change is immeasurably small. That's why we usually say "momentum isn't conserved for the ball" — because we're not including the Earth. Be precise about what your system includes.
Applying Conservation to the Wrong Time Interval
Momentum is conserved during* the collision. Not before, not after. Now, before the collision, external forces might be acting. In practice, after, friction and air resistance change momentum. The conservation equation only connects the instant before impact to the instant after.
Practical Tips / What Actually Works
Every time you sit down to solve a collision problem, follow this sequence. It works every time.
1. Define your system explicitly. Write it down. "System: cart A + cart B
… + cart B (no external forces during the impact).**
2. Sketch a before‑and‑after diagram. Label each object’s mass, velocity vector, and direction. Arrow conventions (right = positive, left = negative) keep the vector bookkeeping honest.
3. Write the momentum‑conservation equation for the chosen interval.
[
\sum \mathbf{p}{\text{before}} = \sum \mathbf{p}{\text{after}}
]
Insert the known quantities; leave the unknown(s) as symbols.
4. Solve algebraically. If the collision is perfectly inelastic, set the final velocities equal ( (v_{1f}=v_{2f}=V) ). If it’s elastic, you may also use the relative‑speed condition (v_{1i}-v_{2i}=-(v_{1f}-v_{2f})) as a second equation, but remember that this step is optional—momentum alone already gives you one relation; the extra condition comes from energy conservation, not from momentum.
5. Check your answer.
- Does the total momentum before equal the total after (within rounding)?
- Are the directions sensible? A negative result simply means motion opposite to your chosen positive axis.
- If the problem gave a coefficient of restitution (e), verify that (|v_{2f}-v_{1f}| = e|v_{2i}-v_{1i}|).
Quick Worked Example
A 0.15 kg hockey puck sliding east at 8 m/s strikes a stationary 0.Also, 30 kg puck. After the impact the first puck moves north at 4 m/s. Find the velocity of the second puck.
- System: puck 1 + puck 2 (impact is brief, ice friction negligible).
- Diagram:
- Before: ( \vec{v}{1i}=+8\hat{i}) m/s, ( \vec{v}{2i}=0).
- After: ( \vec{v}{1f}=+4\hat{j}) m/s, ( \vec{v}{2f}=v_{2x}\hat{i}+v_{2y}\hat{j}) (unknown).
- Momentum conservation:
[ m_1\vec{v}{1i}+m_2\vec{v}{2i}=m_1\vec{v}{1f}+m_2\vec{v}{2f} ]
[ 0.15(8\hat{i})+0.30(0)=0.15(4\hat{j})+0.30(v_{2x}\hat{i}+v_{2y}\hat{j}) ] - Component‑wise solve:
- (x): (1.2 = 0.30v_{2x}) → (v_{2x}=4.0) m/s.
- (y): (0 = 0.15(4)+0.30v_{2y}) → (v_{2y}= -2.0) m/s.
- Result: ( \vec{v}_{2f}=4.0\hat{i}-2.0\hat{j}) m/s (i.e., 4 m/s east, 2 m/s south).
Check: total (x)‑momentum before = 1.2 kg·m/s, after = 0.15·0 + 0.30·4 = 1.2 kg·m/s; total (y)‑momentum before = 0, after = 0.15·4 + 0.30·(‑2) = 0.6 ‑ 0.6 = 0.
The method works for any number of bodies, any dimensionality, and any type of collision—as long as you isolate the system and treat the impact interval as instantaneous relative to external impulses.
Bottom Line
Momentum conservation is a powerful, universally valid tool provided you:
- Clearly delineate the system,
- Restrict the analysis to the short collision interval,
- Treat momentum as a vector, and
- Resist the temptation to mix in energy concepts unless the problem explicitly states an elastic interaction.
Follow the five‑step routine, keep your diagrams tidy, and you’ll avoid the most common pitfalls that trip up even seasoned students. When the internal forces dwarf gravity, friction, or air resistance during the impact, the simple equation (\sum \mathbf{p}{\text{before}} = \sum \mathbf{p}{\text{after}}) becomes your reliable shortcut to the post‑collision velocities—no matter how messy the real‑world surroundings might look.
