Learning Objectives
4 objectivesBy the end of this note, you should be able to:
- Understand that momentum is defined as p = mv.
- Know the principle of conservation of linear momentum.
- Relate conservation of momentum to Newton’s laws of motion.
- Apply conservation of momentum to one-dimensional problems.
Defining Linear Momentum
Key Equations
Linear momentum:
$$p=mv$$
Variables:
- $p$ = linear momentum
- $m$ = mass
- $v$ = velocity
SI unit: kg m s⁻¹
Rearrangements:
$$m=\frac{p}{v}$$
$$v=\frac{p}{m}$$
ProportionalityAt constant mass, momentum is directly proportional to velocity. At constant velocity, momentum is directly proportional to mass.
Linear momentum is the product of an object’s mass and velocity, giving a vector quantity that describes the motion the object carries.
Because momentum depends on velocity, it is a vector. Its direction is always the same as the direction of the velocity. The SI unit kg m s⁻¹ has no special name in physics.
When solving any momentum problem, a positive direction must be chosen first. A negative momentum value then means the object moves opposite to that chosen positive direction. This sign convention is essential for one-dimensional calculations involving collisions or recoil.
A heavy slow lorry can carry the same momentum as a light fast car. This is why doubling either the mass or the velocity doubles the momentum. Doubling both quantities together quadruples the momentum.
Worked Example: Momentum of a Bullet
A bullet of mass 12 g leaves a rifle barrel travelling at 540 km h⁻¹. Calculate the magnitude of its momentum.
Step 1 — Convert the mass to kilograms.
$$m=\frac{12}{1000}$$
$$m=0.012 \text{kg}$$
Step 2 — Convert the velocity to m s⁻¹.
$$v=\frac{540\times 1000}{3600}$$
$$v=150 {\text{m s}}^{-1}$$
Step 3 — Apply the momentum equation.
$$p=mv$$
$$p=0.012\times 150$$
$$p=1.8 {\text{kg m s}}^{-1}$$
The bullet carries 1.8 kg m s⁻¹ of momentum directed along its line of flight. Direction must always be stated for a vector quantity.
Examiner InsightMomentum is a vector. A correct numerical answer without direction loses the vector mark in many Edexcel question types. Always state direction or sign with the magnitude.
Exam TipState both magnitude and direction in every momentum answer.
Conservation of Linear Momentum
Key Equations
Conservation of momentum (two bodies, one dimension):
$${m}_{1}{u}_{1}+{m}_{2}{u}_{2}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$$
Variables:
- ${m}_{1},{m}_{2}$ = masses of the two bodies
- ${u}_{1},{u}_{2}$ = velocities before interaction
- ${v}_{1},{v}_{2}$ = velocities after interaction
SI unit: kg m s⁻¹
Newton’s second law in momentum form:
$$F=\frac{\Delta p}{t}$$
Variables:
- $F$ = resultant force
- $\Delta p$ = change in momentum
- $t$ = time interval over which the force acts
SI unit: N
The principle of conservation of linear momentum states that, for a closed system with no external resultant force, the total momentum remains constant.
A closed system is one that no external resultant force acts on. The principle applies to all interactions inside such a system, including collisions, explosions, and recoil. The total momentum measured before the interaction equals the total momentum measured after.
The principle is a direct consequence of Newton’s laws. Newton’s second law, written as $F=\frac{\Delta p}{t}$, tells us that a force produces a change of momentum. Newton’s third law states that forces between two interacting bodies are equal in magnitude and opposite in direction. Combining these two laws shows that any momentum gained by one body is exactly cancelled by an equal and opposite momentum loss in the other.
Derivation: Conservation of Momentum from Newton’s Laws
Starting point: Newton’s second law and Newton’s third law applied to two interacting bodies A and B.
$$F=\frac{\Delta p}{t}$$
$${F}_{AB}=-{F}_{BA}$$
Step 1 — Apply Newton’s second law to each body. The force on A from B produces a change in A’s momentum, and similarly for B.
$${F}_{AB}=\frac{\Delta {p}_{A}}{t}$$
$${F}_{BA}=\frac{\Delta {p}_{B}}{t}$$
Step 2 — Substitute Newton’s third law. The two contact forces act over the same interaction time $t$.
$$\frac{\Delta {p}_{A}}{t}=-\frac{\Delta {p}_{B}}{t}$$
Step 3 — Multiply through by $t$.
$$\Delta {p}_{A}=-\Delta {p}_{B}$$
Step 4 — Add the two changes of momentum.
$$\Delta {p}_{A}+\Delta {p}_{B}=0$$
$$\Delta {p}_{\text{total}}=0$$
The total momentum of the two-body system does not change during the interaction. This is the principle of conservation of linear momentum.
Worked Example: Collision of Two Trolleys
A 0.50 kg trolley moving right at 4.0 m s⁻¹ collides with a stationary 0.30 kg trolley. After the collision the 0.50 kg trolley continues right at 1.5 m s⁻¹. Calculate the velocity of the 0.30 kg trolley after the collision.
Taking right as positive.
Step 1 — Write the conservation equation.
$${m}_{1}{u}_{1}+{m}_{2}{u}_{2}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$$
Step 2 — Substitute the given values.
$$(0.50)(4.0)+(0.30)(0)=(0.50)(1.5)+(0.30)({v}_{2})$$
$$2.0=0.75+0.30{v}_{2}$$
Step 3 — Rearrange for ${v}_{2}$.
$${v}_{2}=\frac{2.0-0.75}{0.30}$$
$${v}_{2}=\frac{1.25}{0.30}$$
$${v}_{2}=4.17 {\text{m s}}^{-1}\approx 4.2 {\text{m s}}^{-1}$$
The 0.30 kg trolley moves to the right at 4.2 m s⁻¹. The positive sign confirms it travels in the same direction as the original moving trolley.
MisconceptionMomentum and kinetic energy are not the same. Momentum is always conserved in a closed system. Kinetic energy is only conserved in elastic collisions. In inelastic collisions, momentum is conserved but some kinetic energy transfers to thermal or sound stores.
Exam TipState which quantity is conserved before claiming it.
QUICK RECAP
Key Points
- Momentum is defined by $p=mv$.
- Momentum is a vector with SI unit kg m s⁻¹.
- Direction of momentum is the same as direction of velocity.
- Total momentum is conserved in a closed system.
- A closed system has no external resultant force.
- Newton’s second law: $F=\frac{\Delta p}{t}$.
- Newton’s third law: action and reaction forces are equal and opposite.
- Equal and opposite forces over equal time give equal and opposite impulses.
- Equal and opposite impulses give equal and opposite momentum changes.
- These changes cancel, so total momentum is unchanged.
- Always choose a positive direction before substituting values.
- A negative answer means motion in the opposite direction.
- Conservation applies to collisions, explosions, and recoil.
- Momentum is conserved in both elastic and inelastic collisions.
CAN I…? PROGRESS CHECK
Self-Assessment
- Define linear momentum and state its SI unit?
- Calculate momentum from given mass and velocity values, including unit conversions?
- Rearrange $p=mv$ to find mass or velocity?
- State the principle of conservation of linear momentum, including the closed-system condition?
- Derive conservation of momentum from Newton’s second and third laws?
- Apply conservation of momentum to a one-dimensional collision problem?
- Apply conservation of momentum to a recoil or explosion problem?
- Use a positive-direction sign convention correctly throughout a calculation?
- Predict how momentum changes when mass or velocity is doubled?
- Distinguish between momentum and kinetic energy in collision problems?