Momentum is a fundamental vector quantity that describes how much motion an object has. In this chapter you will learn how forces change momentum over time (impulse), why colliding objects always conserve total momentum in an isolated system, and how to classify collisions as elastic or inelastic.
1.1 Defining Momentum
Definition
Momentum (p)
Momentum (p) is defined as the product of the mass of an object and its velocity: p = mv. It is a vector quantity — it has both magnitude and direction. The direction of momentum is the same as the direction of velocity. SI unit: kg·m·s⁻¹ (equivalent to N·s).
Because velocity is a vector, momentum is also a vector — it has both magnitude and direction. A heavy truck moving slowly can have the same momentum as a small car moving fast. The SI unit of momentum is kg·m·s⁻¹. When you calculate momentum always state the direction as part of your answer.
Formula
Momentum
p = momentum (kg·m·s⁻¹), m = mass (kg), v = velocity (m·s⁻¹)
SI unit: kg·m·s⁻¹
Definition
Newton's Second Law (momentum form)
The net (resultant) force acting on an object is equal to the rate of change of momentum of the object in the direction of the net force: F_net = Δp/Δt = mΔv/Δt. This is Newton's original formulation and is more general than F = ma (which applies only when mass is constant).
Formula
Newton's Second Law — momentum form
F_net = net force (N), Δp = change in momentum (kg·m·s⁻¹), Δt = time interval (s)
SI unit: N (newton)
Exam Tip
EXAM TIP: The momentum form of Newton's second law is tested most often in the context of impulse problems. Always define a positive direction at the start of your solution and keep it consistent throughout.
Worked Example
A 1 200 kg car accelerates from rest to 20 m·s⁻¹ east in 8 s. Calculate (a) the momentum at 20 m·s⁻¹ and (b) the net force required.
Given
- m = 1 200 kg
- v_i = 0 m·s⁻¹
- v_f = 20 m·s⁻¹ east
- Δt = 8 s
Find
(a) p at v_f (b) F_net
Solution
- 1(a) p = mv = 1 200 × 20 = 24 000 kg·m·s⁻¹
- 2(b) Δp = p_f − p_i = 24 000 − 0 = 24 000 kg·m·s⁻¹
- 3F_net = Δp/Δt = 24 000 / 8 = 3 000 N
Practice Question
A 0.15 kg cricket ball moves at 30 m·s⁻¹ west. Calculate its momentum.
(2 marks)
Practice Question
A 900 kg car's momentum changes from 18 000 kg·m·s⁻¹ north to 9 000 kg·m·s⁻¹ north in 3 s. Find the net force acting on the car.
(4 marks)
Watch Out
COMMON MISTAKE: Students often forget that momentum is a vector. Always include direction in your final answer, and take care with signs when an object reverses direction.
1.2 Impulse
Definition
Impulse
Impulse is the product of the net force acting on an object and the time interval (Δt) over which the force acts. Impulse = F_net·Δt = Δp (change in momentum). It is a vector quantity with the same direction as the net force. SI unit: N·s = kg·m·s⁻¹. A larger impulse produces a larger change in momentum.
Impulse is a vector measured in N·s (which equals kg·m·s⁻¹). A large force over a short time can deliver the same impulse as a small force over a long time. Engineers use this principle in car safety design — crumple zones increase the collision time Δt, which reduces the peak force F on occupants for the same change in momentum.
Formula
Impulse–Momentum Theorem
F_net = average net force (N), Δt = contact time (s), Δp = change in momentum (kg·m·s⁻¹)
SI unit: N·s = kg·m·s⁻¹
Exam Tip
EXAM TIP: When a force–time graph is given, find impulse by calculating the area under the graph (treat triangles and rectangles separately). The impulse equals the change in momentum.
Worked Example
A 0.06 kg tennis ball travelling at 25 m·s⁻¹ east is struck by a racket and returns at 30 m·s⁻¹ west. The racket is in contact with the ball for 0.004 s. Calculate (a) the impulse on the ball and (b) the average force exerted by the racket.
Given
- m = 0.06 kg
- v_i = +25 m·s⁻¹ (east = positive)
- v_f = −30 m·s⁻¹ (west = negative)
- Δt = 0.004 s
Find
(a) Impulse (b) Average F_net
Solution
- 1Δp = m(v_f − v_i) = 0.06 × (−30 − 25) = 0.06 × (−55) = −3.3 kg·m·s⁻¹
- 2Impulse = Δp = 3.3 N·s west
- 3F_net = Δp/Δt = −3.3 / 0.004 = −825 N
Practice Question
A 70 kg skateboarder is brought to rest from 6 m·s⁻¹ by a net force of 210 N. How long does it take to stop?
(3 marks)
Practice Question
Explain, using the impulse–momentum theorem, why the padding inside a car's steering wheel reduces injury in a collision.
(3 marks)
Real World
REAL WORLD: Helmets, padding, and crumple zones all work on the same impulse principle — extending contact time reduces peak force. This is why mattresses are softer than concrete even if you land with the same momentum!
1.3 Conservation of Momentum and Collisions
Definition
Isolated system
A physical configuration that doesn't exchange any matter with its surroundings and is not subject to any force whose source is external to the system.
Definition
Conservation of Momentum
The total linear momentum of an isolated system (where no external net force acts) remains constant. This means: the total momentum before a collision = the total momentum after the collision. Mathematically: Σp_before = Σp_after. This law holds regardless of whether the collision is elastic or inelastic.
Formula
Law of Conservation of Momentum
Σp_before = total momentum before the event (kg·m·s⁻¹), Σp_after = total momentum after the event (kg·m·s⁻¹)
SI unit: kg·m·s⁻¹
Definition
Elastic Collisions
A collision where total momentum AND total kinetic energy are both conserved.
Definition
Inelastic Collisions
An inelastic collision is one in which the total momentum of the system is conserved, but the total kinetic energy is not conserved — some kinetic energy is converted to other forms of energy (heat, sound, deformation). In a perfectly inelastic (or totally inelastic) collision, the objects stick together and move with a common velocity after the collision.
Elastic vs Inelastic Collisions
| Property | Elastic | Inelastic |
|---|---|---|
| Momentum conserved? | Yes | Yes |
| Kinetic energy conserved? | Yes | No (some converted to heat/sound/deformation) |
| Objects after collision | Separate and bounce apart | May stick together (perfectly inelastic) or deform |
| Everyday example | Billiard balls, atomic collisions | Car crashes, clay balls sticking together |
Exam Tip
EXAM TIP: In all collision and explosion problems, first define a positive direction, then write Σp_before = Σp_after with signs. For explosions (e.g. guns, rockets), the total momentum before is usually zero.
Worked Example
Cart A (mass 2 kg, velocity 5 m·s⁻¹ east) collides with stationary Cart B (mass 3 kg). After the collision they stick together. Find the velocity of the combined carts after the collision.
Given
- m_A = 2 kg, v_Ai = 5 m·s⁻¹ east (positive)
- m_B = 3 kg, v_Bi = 0 m·s⁻¹
- They stick together after collision (perfectly inelastic)
Find
v_f (velocity of combined carts)
Solution
- 1Σp_before = m_A·v_Ai + m_B·v_Bi = 2×5 + 3×0 = 10 kg·m·s⁻¹
- 2Σp_after = (m_A + m_B)·v_f = 5·v_f
- 310 = 5·v_f → v_f = 2 m·s⁻¹
Worked Example
A 5 kg rifle fires a 0.01 kg bullet at 400 m·s⁻¹ east. The rifle was at rest before firing. Find the recoil velocity of the rifle.
Given
- m_rifle = 5 kg, m_bullet = 0.01 kg
- Σp_before = 0 (system at rest)
- v_bullet = +400 m·s⁻¹ (east = positive)
Find
v_rifle (recoil velocity)
Solution
- 1Σp_before = 0
- 2Σp_after = m_bullet·v_bullet + m_rifle·v_rifle = 0
- 30.01×400 + 5×v_rifle = 0
- 44 + 5·v_rifle = 0 → v_rifle = −0.8 m·s⁻¹
Practice Question
Trolley X (mass 4 kg, v = 3 m·s⁻¹ east) collides with Trolley Y (mass 2 kg, v = 1 m·s⁻¹ west). After the collision Trolley X moves at 1 m·s⁻¹ east. Find the velocity of Trolley Y after the collision.
(5 marks)
Practice Question
In the worked example of Cart A and Cart B above, determine whether the collision is elastic or inelastic by comparing kinetic energies before and after.
(4 marks)
Watch Out
COMMON MISTAKE: Students think that because momentum is conserved, kinetic energy must also be conserved. Remember — kinetic energy is only conserved in elastic collisions. In most real collisions (cars, balls of clay), kinetic energy is converted to sound, heat, and deformation energy.