C. Energy and Momentum · Momentum and CollisionsConservation of Momentum in One Dimension
C2.5 C2.6About 70 minutes
Learning goals- I can derive the law of conservation of momentum from Newton's third law and the impulse–momentum theorem.
- I can tell internal forces from external forces and decide when a system is isolated.
- I can solve one-dimensional collision problems, including objects that stick and objects that separate.
- I can explain why momentum is conserved during a brief collision even when friction acts.
Words to know
law of conservation of momentum
If the net external force on a system is zero, the total momentum of the system remains constant.
isolated system
A system on which the net external force is zero.
internal force
A force exerted by one object in the system on another object in the system.
external force
A force exerted on an object in the system by something outside the system.
collision
A brief interaction in which objects exert large forces on each other.
The great sharing rule
Watch two hockey players collide, two curling stones meet, or two railway cars couple together. Velocities change suddenly and in complicated ways. Yet one quantity never changes: the total momentum of the objects involved. This is the law of conservation of momentum, one of the most powerful laws in physics. It works for billiard balls, for colliding cars, and for subatomic particles, where Newton's other ideas need modification.
Why momentum is conserved
Think about two carts that collide on a track. During the collision:
- Cart 2 pushes on cart 1 with a force F₂₁.
- Cart 1 pushes on cart 2 with a force F₁₂.
- By Newton's third law, F₁₂ = −F₂₁ at every instant.
- The two forces act for exactly the same time Δt (the carts touch for the same time interval).
So the impulses are equal and opposite: F₂₁Δt = −F₁₂Δt. By the impulse–momentum theorem, that means Δp₁ = −Δp₂. Whatever momentum cart 1 loses, cart 2 gains. The total change is zero:
Δp₁ + Δp₂ = 0
The law
If the net external force on a system is zero, the total momentum of the system is constant.
For two objects in one dimension (primes ′ mean after):
m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′
- Internal forces are forces that objects in the system exert on each other (the collision forces). They always come in third-law pairs, so they cannot change the total momentum.
- External forces come from outside the system (friction from the ground, a push from your hand). Only an external net force can change a system's total momentum.
- A system with no net external force is called isolated.
What about friction and gravity?
In a real collision there are usually some external forces, such as road friction. But collisions are short and the collision forces are huge. The impulse of friction during those few hundredths of a second is tiny compared with the collision impulses. So we treat the system as isolated just before and just after the collision. Afterwards, friction slowly takes the momentum away, but that is a separate stage of the problem.
A method for momentum problems
- Choose the system (usually all the objects that collide).
- Draw before and after sketches with velocity arrows. Label masses and velocities, with unknowns shown.
- Choose a positive direction and give every velocity its sign.
- Write the conservation equation: total momentum before = total momentum after.
- Solve for the unknown and give its direction.
- Check: is the answer reasonable? (Can the objects pass through each other? Is a speed impossibly large?)
Worked exampleExample 1: Carts that stick
A 0.50 kg cart moving at 2.0 m/s [right] hits a 1.5 kg cart at rest. They stick together. Find their velocity after the collision.
- System: both carts. Right is positive.
- Before: p = (0.50)(2.0) + (1.5)(0) = 1.0 kg·m/s.
- After: the carts move together with velocity v′: p′ = (0.50 + 1.5)v′ = 2.0v′.
- Conservation: 2.0v′ = 1.0, so v′ = 0.50 m/s.
0.50 m/s [right]. Check: the pair is four times as massive as the moving cart, so it moves at a quarter of the speed. ✓
Worked exampleExample 2: Coupling railway cars
A 12 000 kg boxcar rolling at 3.0 m/s [E] catches up to an 8000 kg boxcar rolling at 1.0 m/s [E], and they couple. Find their common velocity.
- East positive. Before: (12 000)(3.0) + (8000)(1.0) = 36 000 + 8000 = 44 000 kg·m/s.
- After: (12 000 + 8000)v′ = 20 000v′.
- v′ = 44 000 ÷ 20 000 = 2.2 m/s.
2.2 m/s [E], between the two original velocities, as expected.
Worked exampleExample 3: A head-on check
An 85 kg hockey player skating at 6.0 m/s [E] collides head-on with a 95 kg player skating at 4.0 m/s [W]. They hold on to each other. Find their velocity just after the collision.
- East positive. p₁ = (85)(+6.0) = +510 kg·m/s. p₂ = (95)(−4.0) = −380 kg·m/s.
- Total before = 510 − 380 = +130 kg·m/s.
- After: (85 + 95)v′ = 180v′ = 130, so v′ = 130 ÷ 180 ≈ +0.72 m/s.
About 0.72 m/s [E]. The lighter but faster player had more momentum, so the pair drifts east.
Momentum bar chart for Example 3. Before the collision the players' momenta are +510 and −380 kg·m/s, a total of +130 kg·m/s. After, the pair has the same total, +130 kg·m/s. (Bars to scale.)Worked exampleExample 4: Objects that separate
A 0.20 kg puck moving at 5.0 m/s [E] hits a 0.30 kg puck at rest. Afterward the first puck moves at 1.0 m/s [E]. Find the velocity of the second puck.
- East positive. Before: (0.20)(5.0) + 0 = 1.0 kg·m/s.
- After: (0.20)(1.0) + (0.30)v₂′ = 0.20 + 0.30v₂′.
- 0.20 + 0.30v₂′ = 1.0, so v₂′ = 0.80 ÷ 0.30 ≈ 2.67 m/s.
About 2.7 m/s [E]. Check: the second puck moves faster than the first afterward, so they do not pass through each other. ✓
Choosing the system matters
When a ball falls, its momentum increases. Is momentum not conserved? It depends on the system. If the system is just the ball, gravity is an external force and the ball's momentum changes. If the system is the ball and Earth, gravity is an internal force: the ball pulls up on Earth as hard as Earth pulls down on the ball, so Earth gains an equal upward momentum. Earth's mass is about 6.0 × 10²⁴ kg, so its velocity change is far too small to notice. When you solve a problem, choose a system for which the external forces are zero or negligible.
Just for funWhen you jump up at 3 m/s, Earth recoils the other way so that the total momentum stays zero. For a 60 kg jumper, Earth's speed is about 180 kg·m/s ÷ 6.0 × 10²⁴ kg = 3 × 10⁻²³ m/s. At that speed, Earth would need about 100 000 years to move the width of one atom.
Hands-on activityCoin collisions
You need: several loonies (or other identical coins), a toonie, and a smooth table or a hard floor.
- Place one loonie on the table. Flick a second loonie so it slides straight into the first, hitting it head-on. Watch closely: what does the moving coin do after the hit?
- Line up five loonies so they touch, in a straight row. Slide a sixth loonie straight into the end of the row. What happens at the far end?
- Slide two touching loonies into the row together. How many leave the far end?
- Slide a toonie (slightly heavier) into a resting loonie, and then a loonie into a resting toonie. Compare.
What to look for: in a head-on hit between equal coins, the moving coin nearly stops and the target moves off at almost the same speed: the momentum is handed over. With the row, one coin flies off the far end, and two coins in gives two coins out. When the heavier toonie hits the loonie, the toonie keeps moving forward slowly; when the loonie hits the toonie, the loonie stops or bounces back slightly. You will explain these results with energy as well as momentum in later lessons of this unit.
Your turn
1A 2.0 kg cart moving at 3.0 m/s [right] hits a 1.0 kg cart at rest and they stick. Find their velocity.
(2.0)(3.0) = (3.0)v′, so v′ = 2.0 m/s [right].
2A 0.25 kg puck moving at 4.0 m/s [E] hits an identical puck at rest head-on. The first puck stops. Find the second puck's velocity.
(0.25)(4.0) = (0.25)v₂′, so v₂′ = 4.0 m/s [E]. The momentum is handed over completely.
3A 3000 kg truck moving at 10 m/s [N] collides head-on with a 1000 kg car moving at 20 m/s [S]. They lock together. Find their velocity just after.
North positive: (3000)(10) + (1000)(−20) = 30 000 − 20 000 = 10 000 kg·m/s. v′ = 10 000 ÷ 4000 = 2.5 m/s [N].
4A 90 kg hockey player skating at 5.0 m/s [E] collides with a 70 kg player skating at 4.0 m/s [W], and they hold on. Find their velocity.
(90)(5.0) + (70)(−4.0) = 450 − 280 = 170. v′ = 170 ÷ 160 ≈ 1.1 m/s [E].
5A 1.2 kg cart moving at 2.5 m/s [right] hits a 0.80 kg cart at rest. Afterward the 1.2 kg cart moves at 0.50 m/s [right]. Find the velocity of the 0.80 kg cart.
Before: (1.2)(2.5) = 3.0. After: (1.2)(0.50) + 0.80v = 0.60 + 0.80v. So 0.80v = 2.4 and v = 3.0 m/s [right].
6A 15 000 kg railway car moving at 2.0 m/s couples with a 10 000 kg car at rest. Find their common speed.
(15 000)(2.0) = (25 000)v′, so v′ = 1.2 m/s.
7Road friction acts on two cars during a collision. Explain why we can still use conservation of momentum for the instant just before and just after the collision.
The collision lasts a very short time and the collision forces are very large. The impulse of friction (force × short time) is tiny compared with the collision impulses, so the total momentum is very nearly unchanged across the collision.
More practice: warm-up
1State the law of conservation of momentum.
If the net external force on a system is zero, the total momentum of the system stays constant.
2Give an example of an internal force in a collision between two carts.
The force of one cart on the other (either member of the third-law pair).
3Before a collision, the total momentum of two pucks is 12 kg·m/s [E]. What is it after, if the system is isolated?
4A 1.0 kg cart at 2.0 m/s hits an identical cart at rest and they stick. Find their speed.
(1.0)(2.0) = (2.0)v′, so v′ = 1.0 m/s.
5Give an example of an external force on a system of two colliding cars.
Friction from the road (or the normal force or gravity from Earth).
More practice: on your own
1A 0.20 kg puck moving at 3.0 m/s [E] hits a 0.40 kg puck moving at 1.0 m/s [W]. Afterward the 0.20 kg puck moves at 1.0 m/s [W]. Find the velocity of the 0.40 kg puck.
East positive. Before: (0.20)(3.0) + (0.40)(−1.0) = 0.60 − 0.40 = 0.20. After: (0.20)(−1.0) + 0.40v = −0.20 + 0.40v. So 0.40v = 0.40, v = 1.0 m/s [E]. Both pucks reversed direction.
2A 20 000 kg railway car at 4.0 m/s catches up to a 30 000 kg car moving at 2.0 m/s in the same direction, and they couple. Find their common speed.
(20 000)(4.0) + (30 000)(2.0) = 80 000 + 60 000 = 140 000. v′ = 140 000 ÷ 50 000 = 2.8 m/s.
3A 0.60 kg cart moving at 2.0 m/s hits a cart of unknown mass at rest. They stick and move at 0.80 m/s. Find the unknown mass.
(0.60)(2.0) = (0.60 + m)(0.80). 1.2 ÷ 0.80 = 1.5 = 0.60 + m, so m = 0.90 kg.
4A 1500 kg car rear-ends a stopped 1000 kg car. Locked together, they move at 9.0 m/s just after the impact. How fast was the first car moving?
(1500)v = (2500)(9.0) = 22 500, so v = 15 m/s (54 km/h).
5A ball falls toward the ground, speeding up. Explain how momentum can be conserved.
If the system is the ball and Earth, gravity is an internal force. As the ball gains downward momentum, Earth gains an equal upward momentum. Earth's huge mass means its velocity change is unmeasurably small, but the total momentum stays constant.
Challenge
1Show that in any collision between two objects in an isolated system, Δv₁ ÷ Δv₂ = −m₂ ÷ m₁. What does this mean for a small car colliding with a large truck?
Δp₁ = −Δp₂ gives m₁Δv₁ = −m₂Δv₂, so Δv₁ ÷ Δv₂ = −m₂ ÷ m₁. The velocity changes are in opposite directions and inversely proportional to the masses: the lighter vehicle has the much larger change in velocity, which is why occupants of smaller vehicles are at greater risk.
2A 60 kg teen runs at 4.0 m/s and jumps onto a 15 kg toboggan at rest on level snow. (a) Find their common speed. (b) They slide 12 m before stopping. Find the coefficient of kinetic friction. Use g = 9.8 m/s².
(a) (60)(4.0) = (75)v′, so v′ = 3.2 m/s. (b) Friction gives a = μg, and v² = 2ad: (3.2)² = 2μ(9.8)(12), so μ = 10.24 ÷ 235.2 ≈ 0.044.
3A 1.0 kg cart moving at 6.0 m/s hits and sticks to an identical cart at rest. The pair then hits and sticks to a third identical cart at rest. Find the final speed, and the total kinetic energy at each stage.
Stage 1: (1.0)(6.0) = (2.0)v, v = 3.0 m/s. Stage 2: (2.0)(3.0) = (3.0)v, v = 2.0 m/s. Kinetic energy: ½(1.0)(6.0)² = 18 J, then ½(2.0)(3.0)² = 9.0 J, then ½(3.0)(2.0)² = 6.0 J. Momentum (6.0 kg·m/s) is conserved at every stage, but kinetic energy is not. The next lessons explain why.
Quiz
1. The total momentum of a system is conserved when:
2. A 3.0 kg cart moving at 4.0 m/s hits a 1.0 kg cart at rest, and they stick together. What is their speed?
3. During a collision, cart A pushes on cart B with a force of 500 N [E]. The force of cart B on cart A is:
4. Why can the momentum of two colliding cars be treated as conserved even though road friction acts on them?
5. Two skaters stand at rest, facing each other, and push apart. What is their total momentum just after the push?
6. A 1500 kg car moving at 20 m/s rear-ends a 1000 kg car moving at 10 m/s in the same direction. They lock together. What is their speed just after?
Big idea (C2.5, C2.6): because of Newton's third law, objects that interact give each other equal and opposite impulses. So the total momentum of an isolated system never changes: m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′.
How to help: Do the coin activity together. Ask your teen to predict what happens before each flick and then explain the result using "momentum before = momentum after".
Common difficulties:
- Signs. Objects moving in opposite directions must have opposite signs. Most errors in this lesson are sign errors.
- Forgetting the target's mass after sticking (using m₁ instead of m₁ + m₂).
- Confusing conservation of momentum with conservation of kinetic energy. Momentum is conserved in every collision of an isolated system; kinetic energy usually is not (see Challenge 3 and the next lessons).
- Choosing the system. If an outside force acts for a long time (braking, sliding to a stop), momentum is not conserved for that stage.