Conservation of Momentum

Momentum brings mass and motion together into a single powerful idea: however complicated an interaction between objects — a collision, a coupling, an explosion — the total momentum of an isolated system never changes. In this workbook you will learn to define and calculate momentum, apply the principle of conservation of momentum to collisions and explosions, distinguish elastic from inelastic collisions using energy considerations, and see how impulse and a more general form of Newton's second law let us deal with forces that act for a short time, or on a changing mass such as a rocket burning fuel.

By the end of this workbook you should be able to:

define momentum, p = mv, and explain why it is a vector quantity
state and apply the principle of conservation of momentum to collisions in one dimension
distinguish between elastic and inelastic collisions, and analyse the kinetic energy transferred in each
apply conservation of momentum to explosions, including recoil
define impulse, J = FΔt, and relate it to the change of momentum of a system
explain why Newton's second law in the form F = Δp ÷ Δt is more general than F = ma, and apply it to situations where mass is changing

Name

1. Momentum and its conservation

The momentum of a moving object combines how much matter it has with how fast it is moving. A heavy lorry and a bicycle travelling at the same speed do not behave the same way in a collision — momentum is what captures that difference.

momentum, p = mv (SI unit: kg m s⁻¹)

Key idea. Momentum is a vector — it has both a size and a direction, the same direction as the object's velocity. This means that, just as with displacement, velocity and force in earlier workbooks, you must choose a positive direction before combining the momenta of objects moving in different directions.

Isolated system: a system that no matter or energy can flow into or out of — in particular, no resultant external force acts on it. Momentum is only guaranteed to be conserved for an isolated system.

The principle of conservation of momentum is one of the most powerful ideas in physics, one of the 5 conservation laws we use:

Principle of conservation of momentum. The total momentum of an isolated system is constant, provided no resultant external force acts on it.

This principle is a direct consequence of Newton's second and third laws, which you met in the Forces workbook. When two objects interact — for example, in a collision — Newton's third law tells us the force each one exerts on the other is equal and opposite, and they act for exactly the same time. Equal and opposite forces acting for the same time produce equal and opposite changes in momentum, so whatever momentum one object gains, the other loses exactly that much: the total is unchanged.

Worked example 1.1

A cyclist and her bicycle have a combined mass of 68 kg and travel at 6.5 m s⁻¹. Calculate her momentum.

Answer:
p = mv = 68 × 6.5

Worked example 1.2

A railway wagon of mass 12 000 kg moving at 3.5 m s⁻¹ couples onto a stationary wagon of mass 8000 kg. Calculate their common velocity immediately after coupling.

Answer:
momentum before = momentum after
(12 000 × 3.5) + (8000 × 0) = (12 000 + 8000) × v
v = [(12 000 × 3.5)] ÷ 20 000

Check your understanding

1Explain why momentum is described as a vector quantity, and why this matters when two objects are moving towards each other.

Momentum is a vector because it has both a size (magnitude) and a direction — the same direction as the object's velocity. This matters when two objects move towards each other because their momenta point in opposite directions: once a positive direction is chosen, one object's momentum must be written as a negative number, so the momenta partly (or fully) cancel when added, rather than simply adding as if both numbers were positive.

2A sprinter of mass 65 kg reaches a speed of 9.2 m s⁻¹. Calculate her momentum.

p = mv = 65 × 9.2

3A 1500 kg car moving at 4.0 m s⁻¹ collides with a stationary 1200 kg car, and their bumpers lock together. Calculate their common velocity immediately after the collision.

momentum before = momentum after
(1500 × 4.0) + (1200 × 0) = (1500 + 1200) × v
v = (1500 × 4.0) ÷ 2700

4State the principle of conservation of momentum, and explain what is meant by an "isolated system" in this context.

The principle of conservation of momentum states that the total momentum of an isolated system is constant, provided no resultant external force acts on it. An isolated system is one that no matter or energy can flow into or out of — in particular, one on which no resultant external force is acting, so only forces between the objects already inside the system (internal forces) are involved.

5The SI unit of momentum is given as kg m s⁻¹, but it is sometimes also written as N s. Show that these two units are equivalent.

From F = ma, the newton is defined as N = kg m s⁻². Multiplying both sides by a time in seconds: N s = kg m s⁻² × s = kg m s⁻¹. So N s and kg m s⁻¹ are the same unit, just written in two equivalent forms.

2. Collisions in one dimension

To apply conservation of momentum to a collision, every velocity must be measured in a consistent direction. Before starting any calculation, choose a positive direction and stick to it — any velocity in the opposite direction must then be written as a negative number.

Fig. 2.1 Before combining momenta, choose a positive direction. Here, object A moves in the positive direction (u_A is a positive number) while object B moves in the negative direction (u_B must be written as a negative number).

Worked example 2.1

Object A, of mass 3.0 kg, moves right at 2.4 m s⁻¹. Object B, of mass 2.0 kg, moves left at 1.6 m s⁻¹. They collide, and afterwards object A continues moving, now at 0.40 m s⁻¹ to the left. Determine the velocity of B immediately after the collision.

Answer:
Take "right" as positive, so B's velocity before is −1.6 m s⁻¹, and A's velocity after is −0.40 m s⁻¹.
momentum before = momentum after
(3.0 × 2.4) + (2.0 × −1.6) = (3.0 × −0.40) + (2.0 × v_B)
v_B = [(3.0 × 2.4) + (2.0 × −1.6) − (3.0 × −0.40)] ÷ 2.0

Live simulation: collision simulator

Set the mass and initial velocity of each trolley, choose whether the collision is elastic or perfectly inelastic, then press play. Watch the momentum and kinetic energy readouts — momentum is always conserved, whichever mode you choose.

Collision simulator

Mass of A, m_A = 1.5 kg

Initial velocity of A, u_A = 3.0 m s⁻¹

Mass of B, m_B = 1.0 kg

Initial velocity of B, u_B = 0.0 m s⁻¹

mode: elastic

Fig. 2.2 A 1D collision simulator. Momentum is conserved in both modes; kinetic energy is only conserved in the elastic mode.

Check your understanding

6Explain why a positive direction must be chosen before adding the momenta of two objects moving towards each other.

Momentum is a vector, so its direction is essential information. If two objects move towards each other, their velocities (and momenta) point in opposite directions. Unless a positive direction is chosen and one of the velocities is written as negative, adding the momenta as if both were positive would give the wrong total — it would add their sizes instead of correctly allowing them to partly (or fully) cancel.

7A 4.0 kg trolley moving right at 2.0 m s⁻¹ collides with a stationary 6.0 kg trolley. After the collision, the 4.0 kg trolley continues to the right at 0.40 m s⁻¹. Calculate the velocity of the 6.0 kg trolley after the collision.

momentum before = momentum after
(4.0 × 2.0) + (6.0 × 0) = (4.0 × 0.40) + (6.0 × v)
v = [(4.0 × 2.0) − (4.0 × 0.40)] ÷ 6.0

8Using the simulator, set A to 2.0 kg at 4.0 m s⁻¹ and B to 2.0 kg at 0 m s⁻¹, in elastic mode. Run the simulation, then describe what happens to each trolley and confirm that momentum is conserved.

With equal masses in an elastic collision, trolley A comes to rest and trolley B moves off with A's original velocity (4.0 m s⁻¹) — the moving trolley effectively hands all of its velocity to the one it strikes. The readout confirms momentum is conserved: momentum before = 2.0 × 4.0 = 8.0 kg m s⁻¹, and momentum after = 2.0 × 4.0 = 8.0 kg m s⁻¹ (now carried entirely by B).

9A 0.50 kg ball moving right at 6.0 m s⁻¹ strikes a stationary 1.5 kg ball. After the collision, the 0.50 kg ball continues to the right at 1.0 m s⁻¹. Calculate the velocity of the 1.5 kg ball after the collision.

momentum before = momentum after
(0.50 × 6.0) + (1.5 × 0) = (0.50 × 1.0) + (1.5 × v)
v = [(0.50 × 6.0) − (0.50 × 1.0)] ÷ 1.5

3. Elastic and inelastic collisions

Momentum is always conserved in an isolated system, whatever kind of collision occurs. What can change from one collision to another is whether kinetic energy is also conserved.

Key idea. In an elastic collision, the total kinetic energy after the collision is the same as before. In an inelastic collision, the total kinetic energy after is less than before — some has been transferred to the surroundings, usually as heat and sound. If the colliding objects stick together, the collision is described as totally inelastic: this is the case in which the greatest possible amount of kinetic energy is transferred away.

Perfectly elastic collisions between everyday objects are rare — some kinetic energy is almost always lost as heat or sound. Elastic collisions are far more common between microscopic particles, such as gas molecules colliding with each other.

Fig. 3.1 Two trolleys of equal mass. Left: an elastic collision — the moving trolley stops and the stationary one moves off with its velocity, and kinetic energy is conserved. Right: a totally inelastic collision — the trolleys stick together and share the original momentum, but kinetic energy is not conserved.

Worked example 3.1

A 1.5 kg trolley moving at 0.80 m s⁻¹ collides with a stationary 1.5 kg trolley. After the collision, the first trolley is at rest and the second moves at 0.80 m s⁻¹. (a) Show that momentum is conserved. (b) Calculate the total kinetic energy before and after the collision, and state whether it is elastic.

Answer:
(a) momentum before = (1.5 × 0.80) + (1.5 × 0) = 1.2 kg m s⁻¹
momentum after = (1.5 × 0) + (1.5 × 0.80) = 1.2 kg m s⁻¹ — momentum is conserved.
(b) E_k before = ½ × 1.5 × 0.80²
E_k after = ½ × 1.5 × 0.80²
Since these are equal, the collision is elastic.

Worked example 3.2

A 2.0 kg trolley moving at 1.2 m s⁻¹ collides with a stationary 3.0 kg trolley, and they stick together. Calculate (a) their common velocity, (b) the total kinetic energy before and after the collision, and (c) the energy transferred to the surroundings.

Answer:
(a) momentum before = momentum after: (2.0 × 1.2) + (3.0 × 0) = (2.0 + 3.0) × v, so v = (2.0 × 1.2) ÷ 5.0
(b) E_k before = ½ × 2.0 × 1.2²
E_k after = ½ × 5.0 × v² (using your value of v from part (a))
(c) energy transferred = E_k before − E_k after

Set up two objects with different masses and velocities, choose "elastic" or "inelastic" in the simulation's controls, and run a collision. Compare the momentum and kinetic energy graphs before and after.

Interactive simulation — open the online version of this workbook to launch it.

Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC-BY 4.0.

Fig. 3.2 PhET Collision Lab simulation.

Check your understanding

10Explain the difference between an elastic and an inelastic collision, and state what happens to kinetic energy in each.

In an elastic collision, the total kinetic energy after the collision equals the total kinetic energy before — no kinetic energy is transferred away. In an inelastic collision, the total kinetic energy after is less than before, because some has been transferred to the surroundings, usually as heat and sound. In both types of collision, momentum is conserved.

11A 0.20 kg ball moving at 3.0 m s⁻¹ collides with a stationary 0.20 kg ball. After the collision, the first ball is at rest and the second moves at 3.0 m s⁻¹. Calculate the kinetic energy before and after the collision, and state whether it is elastic or inelastic.

E_k before = ½ × 0.20 × 3.0²
E_k after = ½ × 0.20 × 3.0²
These are equal, so the collision is elastic.

12A 4.0 kg trolley moving at 2.0 m s⁻¹ collides with a stationary 4.0 kg trolley and they stick together. Calculate (a) their common velocity, and (b) the kinetic energy transferred to the surroundings.

(a) (4.0 × 2.0) + (4.0 × 0) = (4.0 + 4.0) × v, so v = (4.0 × 2.0) ÷ 8.0
(b) E_k before = ½ × 4.0 × 2.0²
E_k after = ½ × 8.0 × v² (using your value of v from part (a))
energy transferred = E_k before − E_k after

13Using the PhET simulation, set up an elastic collision between two objects of equal mass, one stationary, then repeat as an inelastic collision with the same starting conditions. Describe one difference you observe between the two outcomes.

In the elastic collision, the moving object comes to rest and the stationary object moves off with (close to) the first object's original velocity, and the total kinetic energy shown stays the same before and after. In the inelastic collision, the two objects move off together at a shared, lower velocity, and the total kinetic energy shown after the collision is noticeably less than before — some of it has been transferred away.

4. Explosions

In physics, the word explosion describes any event in which internal forces cause a system to separate into two or more parts that move apart from each other — this could be a firework, but it could equally be a spring pushing two trolleys apart, or a rifle firing a bullet.

Key idea. Momentum before an explosion = momentum after. If the system starts at rest, the total momentum before is zero, so the total momentum immediately afterwards must also be zero — the separating parts fly apart with equal and opposite momenta.

Fig. 4.1 Two trolleys held together at rest by a compressed spring. When released, they push each other apart with equal and opposite momenta, so the total momentum stays zero throughout.

Recoil: when part of a system (for example a bullet) is propelled forwards, the rest of the system (the gun) must gain equal and opposite momentum, so it moves backwards. This "kickback" motion is called recoil.

Worked example 4.1

A rifle of mass 3.2 kg fires a bullet of mass 15 g (0.015 kg) at 320 m s⁻¹. Calculate the recoil velocity of the rifle.

Answer:
total momentum before = total momentum after
0 = (3.2 × v) + (0.015 × 320)
v = −(0.015 × 320) ÷ 3.2

Worked example 4.2

Two identical 1.0 kg trolleys are held together at rest by a compressed spring, as in Fig. 4.1. When released, each trolley moves apart at 0.60 m s⁻¹. Calculate (a) the total momentum before and after (confirming it is conserved), and (b) the total kinetic energy before and after the explosion.

Answer:
(a) momentum before = 0. Taking one trolley's direction as positive: momentum after = (1.0 × 0.60) + (1.0 × −0.60) = 0 — confirmed.
(b) E_k before = 0 (nothing is moving)
E_k after = (½ × 1.0 × 0.60²) + (½ × 1.0 × 0.60²)

Worth rememberingKinetic energy is not conserved in an explosion. Unlike a collision, an explosion always increases the total kinetic energy of the system — the extra energy comes from somewhere else, such as the chemical energy stored in gunpowder, or the elastic potential energy stored in a compressed spring.

Check your understanding

14Explain why, if a system starts at rest, the parts must fly apart with equal and opposite momenta after an explosion.

Momentum is conserved, and the system started at rest, so the total momentum before the explosion was zero. Since momentum before must equal momentum after, the total momentum immediately after the explosion must also be zero. The only way for two (or more) separating parts to have a total momentum of zero is for their individual momenta to be equal in size and opposite in direction, so that they cancel when added.

15An astronaut of mass 70 kg is stationary in space next to her spacecraft. She pushes a 4.0 kg tool away from herself at a speed of 2.5 m s⁻¹. Calculate the speed at which she recoils.

total momentum before = total momentum after
0 = (70 × v) + (4.0 × 2.5)
v = −(4.0 × 2.5) ÷ 70

16Explain where the kinetic energy gained by the parts of an exploding system comes from, given that energy cannot be created from nothing.

The extra kinetic energy is not created from nothing — it is converted from another store of energy that was already present in the system before the explosion, such as chemical energy stored in an explosive or fuel, or elastic potential energy stored in a compressed or stretched spring.

17A cannon of mass 450 kg fires a 6.0 kg cannonball at 42 m s⁻¹. Calculate the recoil speed of the cannon, and suggest one reason the cannon's recoil is far less noticeable than the cannonball's launch speed.

0 = (450 × v) + (6.0 × 42)
v = −(6.0 × 42) ÷ 450
The cannon's recoil speed is much smaller than the cannonball's speed because the cannon's mass is far greater — for equal and opposite momenta, a much larger mass only needs a much smaller speed.

5. Impulse

Many forces act only for a short time — a bat striking a ball, a car crumpling in a crash, a rocket engine firing briefly. The impulse of a force describes the combined effect of its size and how long it acts for.

impulse, J = FΔt (SI unit: N s, equivalent to kg m s⁻¹)

Key idea. Impulse is equal to the change in momentum it produces: J = Δp. If a force varies while it acts, an average force can be used to calculate the impulse.

Because J = FΔt = Δp, for a given change in momentum, a longer contact time, Δt, means a smaller average force, F, is needed — and a shorter contact time means a larger force. This single idea explains a huge range of everyday safety features and techniques.

Worth rememberingWhy "give" reduces force. A car's crumple zone, a gymnast's landing mat, bending your knees when you jump down, or catching an egg with hands that move backwards as it lands — all of these work by increasing the time over which a momentum change happens, which reduces the average force needed to produce that same change in momentum.

Fig. 5.1 A force–time graph. The area under the graph — whatever its shape — equals the impulse, FΔt, which is also the change in momentum produced.

Worked example 5.1

A 0.16 kg hockey puck is struck from rest. The force–time graph is a triangle, rising to a peak of 60 N at t = 0.008 s and returning to zero at t = 0.016 s. Calculate (a) the impulse given to the puck, and (b) its resulting speed.

Answer:
(a) impulse = area of triangle = ½ × base × height = ½ × 0.016 × 60
(b) J = Δp = mv, so v = J ÷ 0.16 (using your value of J from part (a))

Live simulation: impulse and force–time graphs

Adjust the peak force and contact time to see how the shape of the force–time graph changes, and how the impulse (the shaded area) and resulting change in velocity respond.

Impulse explorer

Peak force, F_max = 80 N

Contact time, Δt = 0.20 s

Mass = 0.50 kg

impulse = 0 N s

Fig. 5.2 Exploring how peak force and contact time combine to determine impulse.

Check your understanding

18A constant force of 15 N acts on a stationary 0.30 kg ball for 0.12 s. Calculate (a) the impulse given to the ball, and (b) its resulting velocity.

(a) J = FΔt = 15 × 0.12
(b) J = mΔv, so Δv = J ÷ 0.30 (using your value of J from part (a))

19A force–time graph for a bat striking a ball is a triangle, rising to a peak of 900 N at t = 4.0 ms and returning to zero at t = 9.0 ms. Estimate the impulse given to the ball.

impulse = area of triangle = ½ × base × height = ½ × (9.0 × 10⁻³) × 900

20Using Newton's second law in the form F = Δp ÷ Δt, explain why bending your knees when you land after jumping down from a wall reduces the force on your legs.

Landing always produces the same change in momentum, Δp, since your speed changes from your landing speed to zero either way. Bending your knees increases the time, Δt, over which this change in momentum happens. Since F = Δp ÷ Δt, increasing Δt for the same Δp means the average force, F, on your legs is smaller.

21Two force–time graphs for two different collisions are both triangles with the same area, but one is tall and narrow while the other is short and wide. Explain what this tells you about the two collisions.

Since the area under each graph is the same, both collisions produce the same impulse — the same change in momentum. However, the tall, narrow graph represents a collision with a large peak force acting for a short time, while the short, wide graph represents a smaller peak force acting for a longer time. The two collisions have the same overall effect on momentum, but very different peak forces.

6. When mass changes: Newton's second law revisited

In the Forces workbook, Newton's second law was written as F = ma. This form assumes the object's mass stays constant throughout — which is true for most everyday situations, but not all of them.

Key idea. The more general form of Newton's second law is F = Δp ÷ Δt, the resultant force equals the rate of change of momentum. When mass is constant, Δp = mΔv, so F = Δp ÷ Δt = mΔv ÷ Δt = ma — the familiar equation is just a special case. When mass is not constant, F = ma no longer works correctly, but F = Δp ÷ Δt still does.

The clearest example of changing mass is a rocket. As a rocket burns fuel, hot gas is ejected out of the back at high speed, and the rocket's own mass steadily decreases. By the same reasoning used for explosions in Section 4, the backward momentum gained by the ejected gas is matched by an equal and opposite forward momentum gained by the rocket — this is what pushes the rocket forwards, with no need for anything outside the rocket to push against.

Fig. 6.1 A rocket ejects hot exhaust gas backwards. If the rocket and its fuel started at rest, the forward momentum gained by the rocket is always equal and opposite to the backward momentum gained by the ejected gas.

Worked example 6.1

A model rocket has a total mass of 0.40 kg (including fuel) and is at rest on the launch pad. Its motor burns and ejects 0.050 kg of gas at a speed of 60 m s⁻¹ relative to the ground, in the downward direction. Calculate the speed gained by the remaining 0.35 kg of the rocket body.

Answer:
total momentum before = total momentum after
0 = (0.35 × v) + (0.050 × −60)
v = −(0.050 × −60) ÷ 0.35

Check your understanding

22Explain why F = ma cannot be correctly applied to a rocket, considered on its own, as it burns fuel.

F = ma assumes that the mass, m, stays constant while the force acts. A rocket's mass is continuously decreasing as it burns and ejects fuel, so this assumption does not hold, and F = ma cannot be correctly applied to the rocket on its own. The general form, F = Δp ÷ Δt, must be used instead, since it correctly accounts for momentum carried away by the ejected exhaust gas as well as any change in the rocket's speed.

23Starting from F = Δp ÷ Δt, show algebraically that this reduces to F = ma when the mass of the object is constant.

F = Δp ÷ Δt. Since p = mv, Δp = Δ(mv). If the mass, m, is constant, it can be taken outside the change: Δp = mΔv. Substituting this in: F = mΔv ÷ Δt. Since Δv ÷ Δt is exactly the definition of acceleration, a, this becomes F = ma — showing that F = ma is simply the constant-mass special case of the more general F = Δp ÷ Δt.

24A firework of total mass 0.80 kg is at rest in the air when it bursts, ejecting a 0.10 kg burning fragment downwards at 45 m s⁻¹. Calculate the velocity gained by the remaining 0.70 kg of the firework.

0 = (0.70 × v) + (0.10 × −45)
v = −(0.10 × −45) ÷ 0.70

25A garden hose sprays water forwards out of its nozzle at high speed. Explain, in terms of momentum, why the hose pushes back against the person holding it.

The water leaving the nozzle gains momentum in the forward direction. Since the water and hose system started with no net momentum change from any external force, this forward momentum must be balanced by an equal and opposite (backward) momentum given to the hose — the same principle as a rocket ejecting exhaust gas, or a gun recoiling as it fires a bullet. This backward push is what the person holding the hose feels.

Glossary

Momentum: The product of an object's mass and velocity, p = mv; a vector, measured in kg m s⁻¹ (equivalent to N s).
Principle of conservation of momentum: The total momentum of an isolated system is constant, provided no resultant external force acts on it.
Isolated system: A system that no matter or energy can flow into or out of; in particular, one on which no resultant external force acts.
Elastic collision: A collision in which the total kinetic energy after equals the total kinetic energy before.
Inelastic collision: A collision in which the total kinetic energy after is less than before. If the objects stick together, it is described as totally inelastic — the greatest possible amount of kinetic energy has been transferred away.
Explosion: An event in which internal forces cause a system to separate into parts that move apart; momentum is conserved, but kinetic energy increases, drawn from another energy store (such as chemical or elastic potential energy).
Recoil: The backward motion gained by the remaining part of a system (such as a gun) when another part (such as a bullet) is propelled forwards, so that total momentum is conserved.
Impulse: The product of a force and the time for which it acts, J = FΔt; equal to the change of momentum it produces, J = Δp, measured in N s (equivalent to kg m s⁻¹).
Newton's second law (general form): F = Δp ÷ Δt is the general form of Newton's second law, valid even when mass is changing. F = ma is the special case that only applies when mass is constant.
Kinetic energy, E_k: The energy an object has because of its motion, E_k = ½mv² (covered in full detail in a later workbook).