Momentum

CORE VS EXTENDED GUIDE

• Core students study only the unlabelled sections.
• Extended students must study everything, including Extended Extended points.
• Extended = Core + Supplement.

Extended Momentum — Definition and Equation

Key Equations

Momentum:

$$p=mv$$

Variables:

• $p$ = momentum, in kg m s⁻¹ (or N s)
• $m$ = mass, in kg
• $v$ = velocity, in m s⁻¹

Rearrangements:

$$m=\frac{p}{v} v=\frac{p}{m}$$

ProportionalityMomentum is directly proportional to mass when velocity is constant. Momentum is directly proportional to velocity when mass is constant. Therefore, doubling the velocity of an object doubles its momentum.

Momentum is defined as mass × velocity.

Momentum measures how difficult a moving object is to stop. Because velocity is a vector quantity, momentum is also a vector — it has both magnitude and direction.

A 2 kg ball travelling at 3 m s⁻¹ to the right has a different momentum from the same ball travelling at 3 m s⁻¹ to the left, because the directions differ.

The SI unit of momentum is kg m s⁻¹, which is equivalent to N s.

MisconceptionStudents sometimes treat momentum as a scalar and ignore direction. In one-dimensional problems, objects moving in opposite directions must be assigned opposite signs. Forgetting this causes sign errors in conservation calculations. Exam cue: Always state your positive direction before any momentum calculation.

Worked Example

A cyclist and bicycle have a combined mass of 75 kg and travel at 8.0 m s⁻¹ due east. Calculate the momentum.

Finding the momentum

Equation used

$$p=mv$$

Given

$$m=75\text{ kg}$$

$$v=8.0{\text{ m s}}^{-1}$$

Working — substitution

$$p=75\times 8.0$$

$$p=600{\text{ kg m s}}^{-1}$$

Answer

$$p=600{\text{ kg m s}}^{-1}\text{ due east}$$

Exam TipAlways label the momentum arrow with both magnitude and direction; never omit the unit.

Extended Impulse — Definition and Equation

Key Equations

Impulse:

$$\text{impulse}=F\Delta t=\Delta (mv)$$

Variables:

• $F$ = force, in N
• $\Delta t$ = time for which the force acts, in s
• $\Delta (mv)$ = change in momentum, in kg m s⁻¹ (or N s)

Rearrangements:

$$F=\frac{\Delta (mv)}{\Delta t} \Delta t=\frac{\Delta (mv)}{F}$$

ProportionalityImpulse is directly proportional to force when the time of action is constant. Impulse is directly proportional to time when force is constant. Doubling the force while keeping the time unchanged doubles the impulse and therefore doubles the change in momentum.

Impulse is defined as force × time for which the force acts.

Impulse equals the change in momentum of the object. This relationship connects force and time to the effect on an object's motion. A large force acting for a short time can produce the same impulse — and therefore the same change in momentum — as a smaller force acting for a longer time.

This principle explains why car crumple zones and airbags increase the collision time: a longer Δt means a smaller force is needed for the same change in momentum, so the force on the occupant decreases.

The SI unit of impulse is N s, which is numerically identical to kg m s⁻¹.

Examiner InsightCIE frequently asks why safety features reduce injury. The full causal chain is: crumple zone increases the time of collision → for the same change in momentum, the force is reduced (because F = Δp / Δt) → smaller force causes less injury. Exam cue: Always link increased time to reduced force via the impulse equation.

Worked Example

A tennis ball of mass 0.060 kg is struck by a racket. The ball was initially stationary and leaves the racket at 45 m s⁻¹. The racket is in contact with the ball for 5.0 ms. Calculate the average force exerted on the ball.

Step 1: Finding the change in momentum

Equation used

$$\Delta (mv)=m{v}_{\text{final}}-m{v}_{\text{initial}}$$

Given

$$m=0.060\text{ kg}$$

$${v}_{\text{initial}}=0{\text{ m s}}^{-1}$$

$${v}_{\text{final}}=45{\text{ m s}}^{-1}$$

Working — substitution

$$\Delta (mv)=(0.060\times 45)-(0.060\times 0)$$

$$\Delta (mv)=2.7-0$$

$$\Delta (mv)=2.7{\text{ kg m s}}^{-1}$$

Step 2: Converting the contact time to SI units

$$\Delta t=\frac{5.0}{1000}$$

$$\Delta t=0.0050\text{ s}$$

Step 3: Finding the average force

Equation used

$$F=\frac{\Delta (mv)}{\Delta t}$$

Working — substitution

$$F=\frac{2.7}{0.0050}$$

$$F=540\text{ N}$$

Answer

$$F=540\text{ N}$$

Extended Conservation of Momentum in One Dimension

The principle of the conservation of momentum states that the total momentum of a system of objects remains constant provided no resultant external force acts on the system.

This means that in any collision or explosion, the total momentum before the event equals the total momentum after the event:

$$\text{total }{p}_{\text{before}}=\text{total }{p}_{\text{after}}$$

$${m}_{1}{u}_{1}+{m}_{2}{u}_{2}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$$

where $u$ represents velocity before the event and $v$ represents velocity after. All velocities must include their correct sign based on the chosen positive direction.

Conservation of momentum applies to three main situations:

• Collisions — two objects meet and may stick together or bounce apart.
• Explosions — one object separates into two or more parts. Before the explosion the total momentum is zero (objects at rest), so the total momentum after must also be zero. The parts move in opposite directions with momenta that cancel.
• Recoil — a specific type of explosion, such as a gun firing a bullet. The gun recoils in the opposite direction to the bullet so that total momentum remains zero.

MisconceptionStudents sometimes forget that momentum is a vector and add all magnitudes together regardless of direction. Objects moving in opposite directions have momenta of opposite sign. If object A moves right (+) and object B moves left (−), their momenta partially cancel. Exam cue: Assign a positive direction at the start, then apply signs consistently throughout.

Worked Example

A trolley of mass 2.0 kg moves to the right at 3.0 m s⁻¹ and collides with a stationary trolley of mass 1.0 kg. After the collision, the trolleys stick together. Calculate the velocity of the combined trolleys after the collision.

Taking rightward as positive:

Step 1: Finding the total momentum before the collision

$${p}_{\text{before}}={m}_{1}{u}_{1}+{m}_{2}{u}_{2}$$

Given

$${m}_{1}=2.0\text{ kg}, {u}_{1}=+3.0{\text{ m s}}^{-1}$$

$${m}_{2}=1.0\text{ kg}, {u}_{2}=0{\text{ m s}}^{-1}$$

Working — substitution

$${p}_{\text{before}}=(2.0\times 3.0)+(1.0\times 0)$$

$${p}_{\text{before}}=6.0{\text{ kg m s}}^{-1}$$

Step 2: Applying conservation of momentum

$${p}_{\text{after}}={p}_{\text{before}}=6.0{\text{ kg m s}}^{-1}$$

The combined mass after collision:

$${m}_{\text{total}}=2.0+1.0=3.0\text{ kg}$$

Rearranging for velocity:

$$v=\frac{p}{m}$$

$$v=\frac{6.0}{3.0}$$

$$v=2.0{\text{ m s}}^{-1}$$

Answer

$$v=2.0{\text{ m s}}^{-1}\text{ to the right}$$

Extended Resultant Force and Rate of Change of Momentum

Key Equations

Resultant force (Newton's second law in terms of momentum):

$$F=\frac{\Delta p}{\Delta t}$$

Variables:

• $F$ = resultant force, in N
• $\Delta p$ = change in momentum, in kg m s⁻¹
• $\Delta t$ = time interval, in s

Rearrangements:

$$\Delta p=F\Delta t \Delta t=\frac{\Delta p}{F}$$

ProportionalityResultant force is directly proportional to the change in momentum when time is constant. Resultant force is inversely proportional to time when the change in momentum is constant. Therefore, halving the time interval doubles the resultant force for the same change in momentum.

The resultant force acting on an object is defined as the change in momentum per unit time.

This equation is the most general form of Newton's second law. When mass remains constant, Δp = mΔv, so F = mΔv / Δt, which simplifies to F = ma. However, the momentum form is essential when mass changes (e.g. a rocket expelling fuel) or when only momentum data is provided.

The direction of the resultant force is the same as the direction of the change in momentum. A positive Δp means the force acts in the positive direction; a negative Δp means the force acts in the negative direction.

Examiner InsightCIE may give a speed–time graph and ask for the force. The strategy is: read off the change in velocity, calculate Δp = mΔv, then divide by the time interval read from the graph. Exam cue: Extract Δv from the graph before substituting into F = Δp / Δt.

Worked Example

A car of mass 1200 kg accelerates from 10 m s⁻¹ to 25 m s⁻¹ in 6.0 s. Calculate the resultant force acting on the car.

Taking the direction of motion as positive:

Step 1: Finding the change in momentum

$$\Delta p=m{v}_{\text{final}}-m{v}_{\text{initial}}$$

Given

$$m=1200\text{ kg}$$

$${v}_{\text{initial}}=10{\text{ m s}}^{-1}$$

$${v}_{\text{final}}=25{\text{ m s}}^{-1}$$

Working — substitution

$$\Delta p=(1200\times 25)-(1200\times 10)$$

$$\Delta p=30 000-12 000$$

$$\Delta p=18 000{\text{ kg m s}}^{-1}$$

Step 2: Finding the resultant force

Equation used

$$F=\frac{\Delta p}{\Delta t}$$

Given

$$\Delta t=6.0\text{ s}$$

Working — substitution

$$F=\frac{18 000}{6.0}$$

$$F=3000\text{ N}$$

Answer

$$F=3000\text{ N in the direction of motion}$$