Momentum and Newton's laws of motion

1. Overview

Dynamics is the branch of mechanics that explains why objects move. The fundamental principle of this topic is that the motion of an object is governed by the resultant force acting upon it and its mass, which is the intrinsic property that resists change in motion (inertia). By defining linear momentum as the product of mass and velocity, we establish Newton’s Second Law as the most universal description of force: the rate of change of momentum. These laws, combined with the concept of weight as a gravitational force, allow for the precise calculation of acceleration and the analysis of complex interactions between multiple bodies.

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Key Definitions

• Mass: The property of an object that resists change in motion (a measure of its inertia).
• Linear Momentum ($p$): The product of an object's mass and its velocity.
• Force ($F$): The rate of change of momentum of an object.
• Newton’s First Law of Motion: An object will remain at rest or continue to move with constant velocity unless acted upon by a resultant force.
• Newton’s Second Law of Motion: The resultant force acting on an object is directly proportional to the rate of change of its momentum, and this change takes place in the direction of the force.
• Newton’s Third Law of Motion: If body A exerts a force on body B, then body B exerts an equal and opposite force of the same type on body A.
• Weight: The force acting on an object due to the effect of a gravitational field on its mass.
• Inertia: The tendency of an object to resist acceleration (changes in its state of rest or uniform motion).

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Content

3.1 Mass and Inertia

In the Cambridge 9702 syllabus, mass is defined functionally rather than just as "quantity of matter."

• Inertial Mass: This is a measure of how difficult it is to change the velocity of an object. If two different masses are subjected to the same resultant force, the larger mass will experience a smaller acceleration ($a \propto 1/m$).
• Scalar Nature: Mass is a scalar quantity, measured in kilograms ($\text{kg}$). It is an intrinsic property; it does not change regardless of whether the object is on Earth, in deep space, or on the Moon.
• Inertia in Practice: An object with high inertia (large mass) requires a significant resultant force to start moving, stop moving, or change direction.

3.2 Linear Momentum ($p$)

Momentum is a vector quantity that quantifies the "motion" of an object.

The Equation: $p = mv$ (Must be memorised)

• Units: The SI unit is $\text{kg m s}^{-1}$. It is equivalent to the Newton-second ($\text{N s}$).
• Vector Direction: Momentum always acts in the same direction as the velocity vector.
• Change in Momentum ($\Delta p$): In collisions or rebounds, calculating the change is critical. $$\Delta p = p_{final} - p_{initial} = m(v - u)$$ Crucial Note: Because it is a vector, if an object reverses direction, one direction must be defined as negative. For a ball of mass $m$ hitting a wall at speed $v$ and rebounding at speed $v$: $$\Delta p = m(-v) - m(v) = -2mv$$ The magnitude of the change is $2mv$.

3.3 Newton’s First Law (N1L)

Newton’s First Law defines the condition of equilibrium.

• Statement: If the resultant force ($\sum F$) is zero, the acceleration is zero.
• Implications:
  1. If $v = 0$, the object remains at rest (Static Equilibrium).
  2. If $v \neq 0$, the object continues at a constant speed in a straight line (Dynamic Equilibrium).
• Exam Context: Whenever a question mentions "terminal velocity," "constant velocity," or "uniform motion," you must immediately state that the resultant force is zero.

3.4 Newton’s Second Law (N2L)

This law provides the mathematical definition of force.

The Universal Form: The resultant force is the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$ (Must be memorised)

Deriving $F = ma$:

1. Start with $F = \frac{\Delta (mv)}{\Delta t}$.
2. If the mass $m$ is constant, it can be factored out: $F = m \frac{\Delta v}{\Delta t}$.
3. Since $\frac{\Delta v}{\Delta t} = a$, we arrive at: $F = ma$ (Found on Data Sheet)

Directionality: The syllabus specifically requires you to understand that acceleration and resultant force are always in the same direction.

• If an object is moving North but slowing down, the acceleration is South, which means the resultant force must be acting South.
• In $F = ma$, $F$ represents the vector sum of all forces acting on the object.

3.5 Newton’s Third Law (N3L)

N3L describes the interaction between two objects. To identify a "Newton’s Third Law Pair," the forces must satisfy the "S.O.M.E." criteria:

1. Same magnitude.
2. Opposite direction.
3. Magnitude acts on different objects (A on B, B on A).
4. Equal type (e.g., both are gravitational, both are contact forces).

Common Misconception: The "Book on Table" A book sits on a table. The weight ($W$) acts down, and the normal contact force ($R$) acts up. These are NOT an N3L pair because:

• They act on the same object (the book).
• They are different types (Weight is gravitational; $R$ is electromagnetic/contact).
• They only happen to be equal because the book is in equilibrium (N1L).

The Correct N3L Pairs:

• Pair 1: Earth pulls book down (Gravitational) $\leftrightarrow$ Book pulls Earth up (Gravitational).
• Pair 2: Book pushes table down (Contact) $\leftrightarrow$ Table pushes book up (Contact).

3.6 Weight and Gravitational Fields

Weight is the force exerted on a mass due to a gravitational field.

The Equation: $W = mg$ (Must be memorised)

• $g$: On Earth, $g = 9.81 \text{ m s}^{-2}$ (the acceleration of free fall) or $9.81 \text{ N kg}^{-1}$ (gravitational field strength).
• Mass vs. Weight:
  • Mass is measured in $\text{kg}$ using a balance; it is constant everywhere.
  • Weight is measured in $\text{N}$ using a spring balance (newtonmeter); it varies depending on the local value of $g$.

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Worked Example 1 — Momentum Change and Average Force

A jet of water flows horizontally at $15.0 \text{ m s}^{-1}$ from a hose and hits a vertical wall. The water does not rebound but runs down the wall. The cross-sectional area of the jet is $4.0 \times 10^{-4} \text{ m}^2$. The density of water is $1000 \text{ kg m}^{-3}$. Calculate the force exerted by the water on the wall.

Step 1: Calculate the mass of water hitting the wall per second ($\Delta m / \Delta t$). Volume per second $= \text{Area} \times \text{velocity} = (4.0 \times 10^{-4}) \times 15.0 = 6.0 \times 10^{-3} \text{ m}^3 \text{ s}^{-1}$ Mass per second $= \text{Volume per second} \times \text{density} = (6.0 \times 10^{-3}) \times 1000 = 6.0 \text{ kg s}^{-1}$

Step 2: Identify the change in velocity ($\Delta v$). Initial velocity ($u$) $= 15.0 \text{ m s}^{-1}$ Final velocity ($v$) $= 0 \text{ m s}^{-1}$ (since it doesn't rebound) $\Delta v = 0 - 15.0 = -15.0 \text{ m s}^{-1}$

Step 3: Apply Newton's Second Law ($F = \frac{\Delta p}{\Delta t}$). Since $p = mv$, and here the velocity change is constant for a continuous stream: $F = \frac{\Delta m}{\Delta t} \times \Delta v$ $F = 6.0 \times (-15.0) = -90 \text{ N}$

Answer: The magnitude of the force is $90 \text{ N}$.

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Worked Example 2 — Vertical Motion in a Lift

A student of mass $65 \text{ kg}$ stands on a scale in a lift that is accelerating downwards at $2.2 \text{ m s}^{-2}$. Calculate the reading on the scale.

Step 1: Identify the forces. Weight ($W$) acting down: $W = mg = 65 \times 9.81 = 637.65 \text{ N}$ Normal contact force ($R$) acting up: This is the scale reading.

Step 2: Set up the N2L equation ($F_{res} = ma$). The acceleration is downwards, so the downward force must be larger. $F_{res} = W - R$ $ma = W - R$

Step 3: Substitute and solve for $R$. $65 \times 2.2 = 637.65 - R$ $143 = 637.65 - R$ $R = 637.65 - 143 = 494.65 \text{ N}$

Answer: The scale reading is $490 \text{ N}$ (to 2 significant figures).

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Worked Example 3 — Rebound and Impulse

A rubber ball of mass $0.20 \text{ kg}$ is dropped from a height and hits the floor at $6.0 \text{ m s}^{-1}$. It rebounds vertically at $4.5 \text{ m s}^{-1}$. The ball is in contact with the floor for $0.040 \text{ s}$. Calculate the average resultant force acting on the ball during the collision.

Step 1: Define directions. Let Up be positive and Down be negative. $u = -6.0 \text{ m s}^{-1}$ $v = +4.5 \text{ m s}^{-1}$

Step 2: Calculate change in momentum ($\Delta p$). $\Delta p = m(v - u)$ $\Delta p = 0.20 \times (4.5 - (-6.0))$ $\Delta p = 0.20 \times (10.5) = 2.1 \text{ kg m s}^{-1}$

Step 3: Calculate average force ($F = \Delta p / \Delta t$). $F = \frac{2.1}{0.040} = 52.5 \text{ N}$

Answer: The average resultant force is $53 \text{ N}$ acting upwards.

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Key Equations

Equation	Symbols	SI Units	Status
$p = mv$	$p$: momentum, $m$: mass, $v$: velocity	$\text{kg m s}^{-1}$, $\text{kg}$, $\text{m s}^{-1}$	Memorise
$F = \frac{\Delta p}{\Delta t}$	$F$: resultant force, $\Delta p$: change in momentum, $\Delta t$: time	$\text{N}$, $\text{kg m s}^{-1}$, $\text{s}$	Memorise
$F = ma$	$F$: resultant force, $m$: mass, $a$: acceleration	$\text{N}$, $\text{kg}$, $\text{m s}^{-2}$	Data Sheet
$W = mg$	$W$: weight, $m$: mass, $g$: acceleration of free fall	$\text{N}$, $\text{kg}$, $\text{m s}^{-2}$	Memorise

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Common Mistakes to Avoid

• ❌ Wrong: Forgetting that momentum is a vector.
  • ✓ Right: Always assign a positive and negative direction. If an object reverses, the change in velocity is the sum of the speeds ($v - (-u) = v + u$).
• ❌ Wrong: Stating $F=ma$ is Newton's Second Law.
  • ✓ Right: $F = \frac{\Delta p}{\Delta t}$ is the law. $F=ma$ is a special case derived from it, valid only when mass is constant.
• ❌ Wrong: Confusing mass and weight in $F=ma$.
  • ✓ Right: Mass ($m$) is used in the formula. If you are given weight, you must divide by $9.81$ to find the mass before calculating acceleration.
• ❌ Wrong: Thinking N3L pairs act on the same object and cancel out.
  • ✓ Right: N3L forces act on different objects. They can never cancel each other out. Only forces acting on the same object (resultant force) determine its acceleration.
• ❌ Wrong: Ignoring weight in vertical motion problems.
  • ✓ Right: For a rising rocket, the resultant force is $(\text{Thrust} - \text{Weight})$. For a falling object with air resistance, it is $(\text{Weight} - \text{Drag})$.

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Exam Tips

1. The "Constant Mass" Condition: If asked when $F=ma$ is applicable, the required answer is "when mass is constant." This is a frequent 1-mark question.
2. Force-Time Graphs: The area under a Force-time graph represents the change in momentum (also known as Impulse).
3. Significant Figures: Always use $g = 9.81 \text{ m s}^{-2}$. Since this is 3 s.f., your final answers should generally be given to 2 or 3 s.f. Never provide more than the data allows.
4. Defining the System: In N3L questions, use the template: "The force of [Object A] on [Object B] is equal and opposite to the force of [Object B] on [Object A]."
5. Vector Diagrams: If forces are not collinear (e.g., a block on a slope), you must resolve forces into components (usually parallel and perpendicular to the plane) before applying $F=ma$.
6. Units: Remember that $1 \text{ N} = 1 \text{ kg m s}^{-2}$. Checking base units is a great way to verify if your derived equation is correct.