Physics I course sequence · Interactive lesson
Conservation of momentum
Before you write before = after, decide what sits inside the system and what the outside world does to it during the event. This lesson trains that decision, then spends it on recoil speeds, fragment velocities, throws measured relative to a moving thrower, and the energy an explosion releases.
UPI-08.03 · Lesson 3 of 8 in this unit. Position is not a completion score.Build the model
Internal forces trade momentum; only external impulse changes the total.
The key idea
A system's total momentum stays constant whenever the net external impulse on it is zero; forces between members of the system only move momentum from one member to another.
Example. A 0.020 kg dart leaves a 1.2 kg spring gun at 15 m s⁻¹ on a frictionless track, so the gun recoils at 0.020 × 15 ÷ 1.2 = 0.25 m s⁻¹ the other way: 0.30 − 0.30 = 0 kg m s⁻¹, exactly as before.
The complete physical model and assumptions
Momentum conservation is not an extra law to memorise; it is Newton's second and third laws summed over a system you choose. Every push one member exerts on another comes paired with an equal and opposite push lasting exactly as long, so internal impulses cancel in the total and only forces from outside the boundary can change it: ΔP = Jₑₓₜ. That turns conservation into two decisions. Draw the boundary around both partners of every strong interaction, and keep the time window short enough that gravity, supports and friction deliver little impulse compared with the momenta being traded. Do both and the total before equals the total after, however violent, brief or unknown the internal forces were, which is why recoil, explosions and chains of throws and catches can be solved without ever finding a force. Because impulse is a vector, the test runs one component at a time: a cannon and its shell keep their total horizontal momentum while the ground supplies the vertical impulse. The model fails when an external impulse is comparable to the momenta involved (a long push across rough ground, a collision with something anchored to the Earth, a boundary that leaves one partner outside), and it never promises anything about kinetic energy, which an explosion creates from a chemical or elastic store.
1Internal forces cancel; external ones do not
Two skaters on smooth ice, 50 kg and 75 kg, push palms and drift apart. For skater 1, dp₁/dt = F₁₂ + F₁, ext; for skater 2, dp₂/dt = F₂₁ + F₂, ext. Add the two equations. Newton's third law makes F₁₂ = −F₂₁ at every instant, so the pair drops out and dP/dt = ΣFₑₓₜ for the total P = p₁ + p₂. Integrate over the push and ΔP = Jₑₓₜ. On ice the weights are balanced by normal forces and friction is negligible, so Jₑₓₜ ≈ 0. If the push averages 300 N for 0.40 s, each skater receives 120 N s in opposite directions: 2.4 m s⁻¹ for the lighter one, 1.6 m s⁻¹ for the heavier, and the total is still zero. The total would be zero for any push; its size only decides how fast they separate.
2Draw the boundary before you write an equation
Conservation belongs to a system, so the first move is choosing one. A force is internal only when both of its third-law partners sit inside the boundary. Take one skater alone and the other's push is external: her momentum changes by 120 N s and nothing is conserved. Take both and the push is internal. The same test explains a bouncing ball. A 0.50 kg ball that strikes the floor at 6.0 m s⁻¹ and leaves at 5.0 m s⁻¹ changes its momentum by 5.5 N s upward, and the floor delivered it. Put the Earth inside the boundary and momentum is conserved again, but the Earth's recoil is 5.5 N s ÷ (6.0 × 10²⁴ kg) ≈ 9 × 10⁻²⁵ m s⁻¹, which is why the floor is normally left outside as an external agent.
3Isolated is a comparison, not a label
Almost nothing on Earth is truly isolated: gravity acts on everything, and surfaces push and rub. What licenses conservation is that the external impulse over the interaction, roughly |Fₑₓₜ| Δt, is small beside the momenta being exchanged. Short, violent events pass easily. A 2.4 kg firework shell bursting in 8.0 ms receives 2.4 × 9.81 × 0.0080 = 0.19 N s from gravity while its fragments leave with 24 to 40 N s each, so gravity's share is below 1%. Long or grinding events can fail. Two cars of 1400 kg and 1100 kg that lock their wheels and collide over 0.10 s can receive up to 0.70 × 2500 kg × 9.81 m s⁻² × 0.10 s ≈ 1.7 × 10³ N s from the road, a tenth of the 1.7 × 10⁴ N s the heavier car carries at 12 m s⁻¹. The same equation that is good to a fraction of a percent for the firework is only a ±10% estimate for the crash.
4Keep the components the outside world leaves alone
Impulse is a vector, so the isolation test runs axis by axis. A 400 kg cannon on smooth ground fires an 8.0 kg shell at 150 m s⁻¹, 30° above the horizontal. During the firing the shell's reaction drives the cannon down into the ground, and the ground pushes back up with whatever impulse keeps the cannon on the surface, so vertical momentum is not conserved. A frictionless surface cannot push sideways, so horizontal momentum is. Horizontally, 0 = 8.0 × 150 × cos 30° − 400V, giving V = 1039 ÷ 400 = 2.60 m s⁻¹. Vertically the ground supplies 8.0 × 150 × sin 30° = 600 N s. If the wheels skid with μ = 0.20, friction adds a forward impulse of about μ × 600 = 120 N s and the recoil drops to (1039 − 120) ÷ 400 = 2.30 m s⁻¹: the large vertical impulse drags a horizontal one with it.
5Recoil and explosions share momentum, not energy
When a system at rest splits into two pieces, m₁v₁ + m₂v₂ = 0: the momenta are equal and opposite, and the speeds are in inverse ratio to the masses. A 4.0 kg rifle firing a 0.010 kg bullet at 900 m s⁻¹ gives each 9.0 N s, so the rifle recoils at 9.0 ÷ 4.0 = 2.25 m s⁻¹. Kinetic energy splits the other way. With equal |p|, K = p²/2m makes each share inversely proportional to mass: the bullet carries ½ × 0.010 × 900² = 4050 J and the rifle ½ × 4.0 × 2.25² = 10.1 J, a ratio of exactly 400 = 4.0 ÷ 0.010. Momentum conservation fixes that split; the powder only sets the total. The 4060 J came from chemical energy, not from momentum, which is why an explosion conserves momentum while its kinetic energy rises from zero.
6Multi-object events: one ledger per stage
With several objects, split the event into stages and choose a system for each. A 60 kg skater at rest throws a 4.0 kg ball at 10 m s⁻¹ relative to herself, and a 50 kg friend catches it. Stage one, skater plus ball: the ball's ground velocity is V + 10, where V is the skater's velocity just after release, so 0 = 60V + 4.0(V + 10) and V = −0.625 m s⁻¹, with the ball at 9.375 m s⁻¹. Treating 10 m s⁻¹ as the ground speed would give −0.667 m s⁻¹, 7% too large. Stage two, ball plus friend: 4.0 × 9.375 = 54v, so v = 0.694 m s⁻¹. Then check all three together: 60 × (−0.625) + 54 × 0.694 ≈ −37.5 + 37.5 = 0. That whole-system sum costs one line and catches a sign error in either stage.
Key equations
Read the units, assumptions and limits before substituting.
External impulse changes the total
ΔP = Jₑₓₜ = ∫ ΣFₑₓₜ dt
The general statement. A system that is not isolated changes its total momentum by exactly the outside impulse, whatever happens inside it.
P = Σmᵢvᵢ in kg m s⁻¹; Jₑₓₜ in N s, the same unit; internal forces never appearIsolated system
Jₑₓₜ = 0 ⇒ Σmᵢvᵢ (before) = Σmᵢvᵢ (after)
The interaction forces never enter the sum, so a before-and-after table is the whole calculation and the event itself stays a black box.
signed velocity components in m s⁻¹, all measured in one inertial frameOne protected component
ΣFₓ(ext) = 0 ⇒ Pₓ constant, even while Py changes
Keeps the horizontal balance for a cannon, a leaping swimmer or a thrown ball while the ground or water supplies vertical impulse.
choose x perpendicular to the external forces; weight and normal forces are usually verticalRecoil from rest
m₁v₁ + m₂v₂ = 0 ⇒ v₂ = −(m₁/m₂) v₁
Equal and opposite momenta, unequal speeds: a 4.0 kg rifle recoils 400 times more slowly than its 0.010 kg bullet.
m in kg; v in m s⁻¹, measured in the frame where both bodies began at restEnergy shared in a two-body explosion
K = p²/2m ⇒ K₁/K₂ = m₂/m₁, so K₁ = E m₂/(m₁ + m₂)
Momentum fixes the split and the energy store only sets its size: the lighter fragment always carries the larger share.
two bodies starting from rest; E is the total kinetic energy released, in JImpulse approximation
|Fₑₓₜ| Δt ≪ |Δpᵢ| ⇒ treat the event as isolated
Short and violent passes the test; slow and grinding fails it. A millisecond burst leaves gravity no time to act, while a push lasting a second across rough ground gives friction all the time it needs.
Δt is the interaction time in s; Fₑₓₜ is the largest outside force in N; Δpᵢ are the momenta tradedInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Sweep θ with μ = 0 and the dashed total stays vertical while the cannon's arrow shrinks as cos θ; raise μ to 0.40 and the total tilts forward as friction cancels some backward momentum; change M and only the recoil speed V moves, never the momentum diagram.
Live results
- RECOIL SPEED V
- 2.60 m s⁻¹
- CANNON MOMENTUM MV
- 1039 kg m s⁻¹
- GROUND IMPULSE, UP
- 600 N s
- FRICTION IMPULSE, FORWARD
- 0 N s
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Launch angle θ: 30 °; Ground friction μ: 0.00; Cannon mass M: 400 kg
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The sentence “momentum is always conserved” causes most of the wrong answers here, because it attaches conservation to objects or events instead of to a system. A ball striking the floor at 6.0 m s⁻¹ and leaving at 5.0 m s⁻¹ breaks no law: the floor sits outside the ball and supplied the 5.5 N s. A cannon fired upward at an angle does not recoil down into the ground, so writing m v sin θ = M Vy conserves a component the ground is actively changing. A block sliding to rest on a rough floor has not hidden its momentum anywhere; friction removed it. Third-law pairs cancel only when both partners are inside the boundary, so before = after is licensed by one check: the external impulse on the chosen system, in that component, over that interval, is negligible.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
An astronaut whose mass with her suit is 92 kg, not counting the 4.0 kg tool bag she holds, floats at rest 6.0 m from a space-station hatch. To get back, she throws the bag directly away from the hatch at 2.3 m s⁻¹ relative to the station. Find her velocity, how long she takes to reach the hatch, and the kinetic energy her arm supplied.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- System: astronaut plus bag. The throw is internal, and in the station's freely falling frame nothing outside pushes on the pair, so the total momentum is zero before and after.
- Take +x away from the hatch. 0 = (4.0 kg)(+2.3 m s⁻¹) + (92 kg)V, so V = −9.2 kg m s⁻¹ ÷ 92 kg = −0.100 m s⁻¹: she moves towards the hatch.
- Nothing slows her after the throw, so t = 6.0 m ÷ 0.100 m s⁻¹ = 60 s.
- Kinetic energy: bag ½ × 4.0 × 2.3² = 10.58 J; astronaut ½ × 92 × 0.100² = 0.46 J; total 11.04 J, all supplied by her muscles.
- Check the share rule: 10.58 ÷ 0.46 = 23 = 92 ÷ 4.0. The light bag takes 96% of the energy even though the two momenta are equal in size.
Answer. She drifts towards the hatch at 0.10 m s⁻¹ and reaches it after 60 s. The throw supplied 11 J of kinetic energy, 96% of it to the bag.
MediumExample 2
Two 150 kg rafts float at rest, one behind the other. A 50 kg swimmer standing on raft A leaps horizontally onto raft B, leaving A at 4.0 m s⁻¹ relative to raft A, and lands and stays on B. Ignoring water drag during the brief push and landing, find both rafts' final velocities and the kinetic energy dissipated in the landing.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Two interactions, two systems: push-off is swimmer plus raft A, landing is swimmer plus raft B. Gravity and buoyancy are vertical and drag is negligible over each brief event, so horizontal momentum is conserved in each stage.
- Push-off, with +x towards B. The 4.0 m s⁻¹ is measured relative to A, so the swimmer's ground velocity is u = vA + 4.0. Then 0 = 50(vA + 4.0) + 150vA, so 200vA = −200 and vA = −1.00 m s⁻¹, and the swimmer leaves at u = 3.00 m s⁻¹.
- Landing: 50 × 3.00 = (50 + 150)vB, so vB = 150 ÷ 200 = 0.75 m s⁻¹.
- Whole-system check: 150 × (−1.00) + 200 × 0.75 = −150 + 150 = 0 kg m s⁻¹, as it must be for the swimmer and both rafts together.
- Energy: the leap created ½ × 50 × 3.00² + ½ × 150 × 1.00² = 225 + 75 = 300 J. After landing there is ½ × 200 × 0.75² + 75 = 56.25 + 75 = 131.25 J, so 168.75 J is dissipated as the swimmer lands. Momentum survives the landing; kinetic energy does not.
- The trap: reading 4.0 m s⁻¹ as a ground speed gives vA = −1.33 m s⁻¹ and vB = 1.00 m s⁻¹, each a third too large, because raft A was already moving backward when the swimmer's feet left it.
Answer. Raft A moves at 1.0 m s⁻¹ away from B; raft B, carrying the swimmer, moves at 0.75 m s⁻¹ away from A. About 1.7 × 10² J of kinetic energy is dissipated in the landing.
HardExample 3
A 2.4 kg firework shell is moving straight up at 18 m s⁻¹ when it bursts, over 8.0 ms, into three fragments. Just afterwards a 0.80 kg fragment moves horizontally east at 30 m s⁻¹ and a 1.0 kg fragment moves straight up at 40 m s⁻¹. Find the velocity of the third fragment, test whether gravity may be ignored during the burst, and find the kinetic energy released.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- System: all three fragments. The burst forces are internal and gravity is the only external force. Take x east and y up. Before the burst, P = (0, 2.4 × 18) = (0, 43.2) kg m s⁻¹, and the third fragment has mass 2.4 − 0.80 − 1.0 = 0.60 kg.
- Treat the burst as isolated and conserve each component for the third fragment. x: 0 = 0.80 × 30 + pₓ, so pₓ = −24.0 kg m s⁻¹. y: 43.2 = 1.0 × 40 + py, so py = +3.2 kg m s⁻¹.
- |p| = √(24.0² + 3.2²) = √586.24 = 24.21 kg m s⁻¹, so its speed is 24.21 ÷ 0.60 = 40.4 m s⁻¹, directed atan(3.2/24.0) = 7.6° above the horizontal, towards the west.
- Gravity during the burst: 2.4 × 9.81 × 0.0080 = 0.19 N s downward, 0.4% of the 43.2 kg m s⁻¹ total. Including it makes py = 3.01 kg m s⁻¹, which moves the speed to 40.3 m s⁻¹ and the angle to 7.2°. The approximation is excellent for the speed and only fair for the small vertical part, which is the difference of two large numbers.
- Kinetic energy before: ½ × 2.4 × 18² = 388.8 J. After: ½ × 0.80 × 30² + ½ × 1.0 × 40² + |p|²/(2 × 0.60) = 360 + 800 + 488.5 = 1648.5 J. Released: 1648.5 − 388.8 = 1259.7 J.
- Conservation described the 8.0 ms burst, not the flight: one second later gravity has changed the fragments' total momentum by 2.4 × 9.81 × 1.0 = 23.5 N s, more than half of the original 43.2 kg m s⁻¹.
Answer. The third fragment moves at 40 m s⁻¹, 7.6° above the horizontal towards the west (7.2° once gravity's 0.19 N s is included). The burst releases about 1.3 kJ of kinetic energy.
Quick check
Select an answer and explain it before checking.
Review & continue
A system's total momentum stays constant whenever the net external impulse on it is zero; forces between members of the system only move momentum from one member to another.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics I completion tasks.