understand that mass is the property of an object that resists change in motion

Momentum and Newton’s Laws of Motion – A‑Level Physics 9702

Learning Objective

Understand that mass is the quantitative measure of an object’s inertia – the property that resists any change in its state of motion.

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Recall: Physical Quantities, Vectors & Units

Quantity	Scalar / Vector	SI Unit	Typical Uncertainty
Displacement, velocity, acceleration, force, momentum	Vector	m, m s⁻¹, m s⁻², N, kg m s⁻¹	± 0.5 % (instrument) + ± last‑digit (reading)
Distance, speed, mass, time, kinetic energy	Scalar	m, m s⁻¹, kg, s, J	± 0.5 % (instrument) + ± last‑digit (reading)

• Always write quantities with their SI units and check dimensional consistency.
• Precision = repeatability of a measurement; accuracy = closeness to the true value.

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Kinematics Refresher (required before applying dynamics)

• Displacement Δx – vector; distance s – scalar.
• Velocity \(\mathbf{v}\) – gradient of a distance‑time graph; speed – magnitude of \(\mathbf{v}\).
• Acceleration \(\mathbf{a}\) – gradient of a velocity‑time graph; also the second derivative of displacement.
• Area under a velocity‑time graph gives displacement; area under an acceleration‑time graph gives change in velocity.

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Dynamics – Forces & Newton’s Laws

Newton’s First Law – Law of Inertia

An object remains at rest or moves with constant velocity unless acted upon by a net external force. The resistance to any change in motion is called inertia, and the magnitude of inertia is quantified by the object's mass (m).

Newton’s Second Law – From the Momentum Principle

The fundamental statement is the momentum principle:

\[ \mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt} \qquad\text{where}\qquad \mathbf{p}=m\mathbf{v} \]

For a body of constant mass:

\[ \mathbf{F}_{\text{net}} = \frac{d}{dt}(m\mathbf{v}) = m\frac{d\mathbf{v}}{dt}=m\mathbf{a} \]

Thus the familiar form \(\mathbf{F}_{\text{net}} = m\mathbf{a}\) is a special case of the more general momentum equation.

• Net force is the vector sum of all individual forces acting on the body.
• Vector addition can be performed graphically (tip‑to‑tail) or analytically using components.

Newton’s Third Law – Action & Reaction

For every interaction there is an equal and opposite pair of forces:

\[ \mathbf{F}_{AB} = -\mathbf{F}_{BA} \]

These forces act on different bodies, so they never cancel when analysing the motion of a single object.

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Mass, Inertia & Drag (Non‑uniform Motion)

• Mass is the proportionality constant that relates force to acceleration (inertia).
• When a body moves through a fluid, it experiences a resistive (drag) force that depends on speed.

Simple drag models (Cambridge syllabus requirement)

Two common approximations:

\[ \mathbf{F}_{\text{drag}} = -k\mathbf{v}\qquad\text{(linear drag, low speeds)} \] \[ \mathbf{F}_{\text{drag}} = -k v^{2}\,\hat{\mathbf{v}}\qquad\text{(quadratic drag, high speeds)} \]

For a falling object of mass \(m\) under gravity \(mg\) with linear drag, the terminal velocity is obtained when the net force is zero:

\[ mg - kv_{t}=0\;\;\Longrightarrow\;\;v_{t}= \frac{mg}{k} \]

Example: A sky‑diver (mass 80 kg) with a linear drag constant \(k=160\;\text{N s m}^{-1}\) reaches a terminal speed \(v_{t}= \frac{80\times9.8}{160}=4.9\;\text{m s}^{-1}\).

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Linear Momentum

Definition (vector)

\[ \mathbf{p}=m\mathbf{v} \]

Because mass appears directly, a larger mass gives a larger momentum for the same speed, and therefore a greater resistance to a change in motion.

Momentum principle (already introduced)

\[ \mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt} \]

Impulse–Momentum Theorem

\[ \mathbf{J} = \int_{t_{1}}^{t_{2}}\mathbf{F}\,dt = \Delta\mathbf{p} \]

Impulse \(\mathbf{J}\) is the area under a force‑time graph and equals the change in momentum.

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Conservation of Linear Momentum

Isolated system criteria (Cambridge requirement)

• External net force on the system is zero (or the forces cancel such that \(\sum\mathbf{F}_{\text{ext}}=0\)).
• Typical laboratory approximation: frictionless air‑track, smooth ice, or short interaction time so that gravity and normal forces produce no net horizontal impulse.

Mathematical statement

\[ \sum_{\text{system}}\mathbf{p}_{\text{initial}} = \sum_{\text{system}}\mathbf{p}_{\text{final}} \]

1‑D Collisions

• Elastic collision – both momentum and kinetic energy are conserved.
• Inelastic collision – momentum conserved, kinetic energy not conserved.
• Perfectly inelastic collision – bodies stick together after impact.
• Optional – coefficient of restitution (e): \[ e = \frac{v_{2}'-v_{1}'}{v_{1}-v_{2}}\quad(0\le e\le1) \] where \(e=1\) for a perfectly elastic impact and \(e=0\) for a perfectly inelastic impact.

2‑D Collisions (vector treatment)

Momentum conservation must be applied separately to each component:

\[ \sum p_{x,\text{initial}} = \sum p_{x,\text{final}},\qquad \sum p_{y,\text{initial}} = \sum p_{y,\text{final}} \]

This allows analysis of glancing blows, billiard‑ball problems, and explosions.

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Worked Example – Elastic 1‑D Collision

Two gliders on a frictionless air‑track:

• Glider A: \(m_{1}=0.5\;\text{kg},\;v_{1}=4.0\;\text{m s}^{-1}\) (right).
• Glider B: \(m_{2}=1.5\;\text{kg},\;v_{2}=0\;\text{m s}^{-1}\) (at rest).

1. Conservation of momentum:
   \[ m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}' \]
2. Conservation of kinetic energy (elastic):
   \[ \tfrac12 m_{1}v_{1}^{2}+\tfrac12 m_{2}v_{2}^{2} =\tfrac12 m_{1}{v_{1}'}^{2}+\tfrac12 m_{2}{v_{2}'}^{2} \]
3. Solving the two equations gives \[ v_{1}'=-2.0\;\text{m s}^{-1},\qquad v_{2}'=+2.0\;\text{m s}^{-1} \] Glider A rebounds, Glider B moves forward with the same speed.

Worked Example – Perfectly Inelastic 1‑D Collision

Same masses, but the gliders stick together.

• Initial momentum: \(p_{i}=m_{1}v_{1}=0.5\times4.0=2.0\;\text{kg m s}^{-1}\).
• Total mass after impact: \(m_{\text{tot}}=m_{1}+m_{2}=2.0\;\text{kg}\).

Using momentum conservation:

\[ p_{i}=m_{\text{tot}}v_{f}\;\Longrightarrow\;v_{f}= \frac{2.0}{2.0}=1.0\;\text{m s}^{-1} \]

Kinetic energy loss:

\[ K_{i}= \tfrac12 m_{1}v_{1}^{2}=4.0\;\text{J},\qquad K_{f}= \tfrac12 m_{\text{tot}}v_{f}^{2}=1.0\;\text{J} \]

The missing 3 J is transformed into internal energy (sound, deformation).

Worked Example – 2‑D Elastic Collision (Billiard Balls)

Ball 1 (\(m\)) moves at \(5.0\;\text{m s}^{-1}\) along the +x‑axis, strikes identical stationary ball 2. After impact, ball 1 moves at \(3.0\;\text{m s}^{-1}\) at \(30^{\circ}\) above the x‑axis. Find the speed of ball 2.

1. Apply momentum conservation in x‑direction:
   \[ mv_{1}=m v_{1}'\cos30^{\circ}+m v_{2}'\cos\theta \]
2. Apply momentum conservation in y‑direction (initial y‑momentum = 0):
   \[ 0=m v_{1}'\sin30^{\circ}+m v_{2}'\sin\theta \]
3. Since the masses are equal, solving the two equations gives \(v_{2}'=4.0\;\text{m s}^{-1}\) directed at \(60^{\circ}\) below the x‑axis.

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Mini‑Lab: Verifying Conservation of Momentum (Air‑Track)

1. Aim: Test momentum conservation for elastic and perfectly inelastic collisions.
2. Equipment: Air‑track, two gliders (different masses), photogates or motion sensors, set of calibrated masses, ruler, data sheet.
3. Method (outline):
   • Measure and record each glider’s mass (\(m_{1}, m_{2}\)) with uncertainties.
   • Release glider 1 from a fixed height to give a repeatable initial speed \(v_{1}\) (measure with photogates).
   • Record the initial speed of glider 2 (\(v_{2}\) – usually zero).
   • For an elastic impact (spring‑loaded bumper) record post‑collision speeds \(v_{1}'\) and \(v_{2}'\).
   • Replace the bumper with a Velcro‑covered one so the gliders stick together; record the common final speed \(v_{f}\).
   • Calculate total initial and final momentum, propagate uncertainties, and compare.
4. Data analysis:
   • Momentum of each body: \(p=m v\).
   • Uncertainty propagation: \(\displaystyle\frac{\Delta p}{p}= \sqrt{\left(\frac{\Delta m}{m}\right)^{2}+\left(\frac{\Delta v}{v}\right)^{2}}\).
   • Percentage difference: \(\displaystyle\frac{|p_{\text{i}}-p_{\text{f}}|}{p_{\text{i}}}\times100\%.\)
   • Discuss sources of error (air‑track friction, timing resolution, imperfect elasticity) and suggest improvements.

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Summary Table

Law / Principle	Mathematical Form	Physical Meaning	Role of Mass
Newton’s First Law	\(\mathbf{F}_{\text{net}}=0\;\Rightarrow\;\mathbf{v}= \text{constant}\)	Objects maintain their state of motion unless a net external force acts.	Mass quantifies the inertia that resists any change in motion.
Newton’s Second Law (derived)	\(\displaystyle\mathbf{F}_{\text{net}}=\frac{d\mathbf{p}}{dt}=m\mathbf{a}\)	Force produces an acceleration proportional to the mass.	Mass is the proportionality constant linking force and acceleration.
Newton’s Third Law	\(\mathbf{F}_{AB}=-\mathbf{F}_{BA}\)	Forces occur in equal‑and‑opposite pairs on interacting bodies.	Each body’s mass determines the acceleration produced by the pair of forces.
Momentum Definition	\(\mathbf{p}=m\mathbf{v}\)	Momentum combines mass and velocity into a conserved vector quantity.	Mass directly scales momentum, increasing resistance to change.
Impulse–Momentum Theorem	\(\displaystyle\mathbf{J}= \int\mathbf{F}\,dt = \Delta\mathbf{p}\)	Impulse (area under a force‑time graph) equals the change in momentum.	Mass appears in \(\Delta\mathbf{p}=m\Delta\mathbf{v}\).
Conservation of Linear Momentum	\(\displaystyle\sum\mathbf{p}_{\text{initial}} = \sum\mathbf{p}_{\text{final}}\) (isolated system)	Total momentum of a closed system remains constant.	Distribution of mass among the objects dictates how momentum is shared after interaction.
Linear Drag (non‑uniform motion)	\(\mathbf{F}_{\text{drag}}=-k\mathbf{v}\) or \(\mathbf{F}_{\text{drag}}=-k v^{2}\hat{\mathbf{v}}\)	Resistive force proportional to speed (low‑speed) or speed squared (high‑speed).	Mass determines the terminal speed \(v_{t}=mg/k\) (linear model).

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Key Take‑aways

• Mass is the quantitative measure of inertia – the larger the mass, the greater the resistance to any change in speed or direction.
• Newton’s second law (\(\mathbf{F}=m\mathbf{a}\)) follows directly from the momentum principle \(\mathbf{F}=d\mathbf{p}/dt\).
• Momentum (\(\mathbf{p}=m\mathbf{v}\)) and the impulse–momentum theorem provide a powerful way to analyse forces that act over short times.
• Conservation of linear momentum applies to both 1‑D and 2‑D collisions, provided the system is isolated.
• Drag forces introduce non‑uniform motion; simple linear and quadratic models give the concept of terminal velocity.
• Accurate experimental work—including uncertainties, error analysis and clear identification of an isolated system—is essential for the A‑Level exam (Paper 5, AO3).