Cambridge Syllabus Notes

Momentum and Newton’s Laws of Motion

Learning Objective

Understand that an object moving against a resistive force may reach a terminal (constant) velocity, and be able to apply Newton’s laws and the conservation of linear momentum to a range of situations (including collisions).

Key Definitions (Cambridge wording)

Term	Definition
Linear momentum	The product of an object’s mass and its velocity; a vector quantity with units kg·m s⁻¹. \( \mathbf{p}=m\mathbf{v}\)
Impulse	The integral of a force over the time interval during which it acts; has the same units as momentum. \( \mathbf{J}= \displaystyle\int_{t_1}^{t_2}\mathbf{F}\,dt = \Delta\mathbf{p}\)
Terminal velocity	The constant speed attained when the net external force on a falling object becomes zero (weight = drag). Acceleration is zero but the object continues to move.
Reynolds number ( \(Re\) )	A dimensionless quantity that predicts the flow regime: \( Re = \dfrac{\rho v L}{\mu}\) where \(\rho\) is fluid density, \(v\) the speed of the object, \(L\) a characteristic length (e.g. diameter) and \(\mu\) the dynamic viscosity. Rough guideline: \(Re \lesssim 2\times10^{3}\) → laminar (linear drag); \(Re \gtrsim 2\times10^{3}\) → turbulent (quadratic drag).

1. Newton’s Laws of Motion

First law (Inertia): A body remains at rest or moves with uniform straight‑line velocity unless acted on by a net external force.
Second law:
- General (variable mass): \(\displaystyle \mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt}\).
- Constant‑mass form (used in almost all Cambridge AS & A Level problems):
  \(\displaystyle \mathbf{F}=m\mathbf{a}\)
  Assumptions: the mass of the object does not change during the interval and the reference frame is inertial (i.e. not accelerating).
Third law: For every action there is an equal and opposite reaction.

2. Linear Momentum

Definition: \(\mathbf{p}=m\mathbf{v}\) (vector, SI unit kg·m s⁻¹).
Impulse–momentum theorem: \(\displaystyle \mathbf{J}= \int_{t_1}^{t_2}\mathbf{F}\,dt = \Delta\mathbf{p}\).
Conservation of linear momentum: In the absence of external forces the total momentum of a closed system remains constant. This holds for both 1‑D and 2‑D interactions.

3. Collisions

3.1 One‑dimensional collisions

Inelastic: Momentum conserved, kinetic energy not conserved. Example – a lump of clay sticks to a block.
Elastic: Both momentum and kinetic energy conserved. Example – two steel balls colliding on a smooth track.

3.2 Two‑dimensional collisions

Momentum is conserved separately in the horizontal (\(x\)) and vertical (\(y\)) directions. Vector diagrams are essential for solving billiard‑ball or projectile‑impact problems.

Example – 1‑D elastic collision of identical spheres

Two identical spheres (mass \(m\)) approach each other with speeds \(u\) and \(-u\). Solving the simultaneous equations for momentum and kinetic‑energy conservation gives \(v_1 = -u\) and \(v_2 = u\); they simply exchange velocities.

4. Forces on a Falling Object

Consider a body of mass \(m\) moving vertically through a fluid (air, water, oil…). The relevant forces are:

Force	Direction	Expression	Typical Units
Weight	Downwards	\(W = mg\)	N (kg·m s⁻²)
Resistive (drag) force	Opposes motion (upwards for a falling body)	Linear (viscous) drag: \(F_d = kv\) Quadratic (pressure) drag: \(F_d = \tfrac12 C\rho A v^{2}\)	\(k\): N·s m⁻¹ \(C\): dimensionless, \(\rho\): kg m⁻³, \(A\): m²

Flow regimes

Viscous (laminar) flow – low Reynolds number, drag ∝ \(v\).
Turbulent flow – high Reynolds number, drag ∝ \(v^{2}\).

5. Terminal (Constant) Velocity

When the net external force on the object becomes zero, its acceleration is zero and the speed stops changing. This constant speed is the terminal velocity \(v_t\).

5.1 Derivation – linear (viscous) drag

\[ mg - kv = 0 \;\Longrightarrow\; v_t = \frac{mg}{k} \]

5.2 Derivation – quadratic (pressure) drag

\[ mg - \tfrac12 C\rho A v^{2}=0 \;\Longrightarrow\; v_t = \sqrt{\frac{2mg}{C\rho A}} \]

5.3 Qualitative picture

Initially \(mg > F_d\) → downward acceleration.
As \(v\) increases, \(F_d\) grows until \(F_d = mg\).
Beyond this point the forces balance; the object continues to fall at the constant speed \(v_t\).

6. Worked Example – Terminal Velocity with Linear Drag

Problem: A steel sphere of mass \(0.15\;\text{kg}\) falls through air. The linear drag coefficient is \(k = 0.025\;\text{N·s m}^{-1}\). Find its terminal velocity.

Solution:

\[ v_t = \frac{mg}{k} = \frac{0.15 \times 9.81}{0.025} \approx 5.89 \times 10^{1}\;\text{m s}^{-1} \;(\approx 59\;\text{m s}^{-1}) \]

7. Worked Example – 1‑D Inelastic Collision

Problem: A \(0.8\;\text{kg}\) cart moving at \(2.0\;\text{m s}^{-1}\) collides and sticks to a \(0.5\;\text{kg}\) cart initially at rest. Find the speed of the combined system after the collision.

Solution (conservation of momentum):

\[ m_1u_1 + m_2u_2 = (m_1+m_2)v \quad\Longrightarrow\quad (0.8)(2.0) + (0.5)(0) = (1.3)v \] \[ v = \frac{1.6}{1.3} \approx 1.23\;\text{m s}^{-1} \]

8. Practical Skills – Determining Terminal Velocity

Experiment design
1. Set up a motion sensor (or high‑speed camera) to record the vertical position \(y(t)\) of a small dense sphere dropped from a known height.
2. Differentiate the data numerically to obtain the velocity \(v(t)\).
3. Plot \(v\) versus \(t\); the curve will asymptote to a constant value – the terminal speed.

Sources of error
• Air currents in the laboratory (systematic).
• Timing resolution of the sensor (random).
• Uncertainty in the mass and radius of the sphere (affects \(k\) or \(C\rho A\)).

Uncertainty analysis
For linear drag, propagate uncertainties using \[ v_t = \frac{mg}{k}\;\; \Rightarrow\;\; \frac{\Delta v_t}{v_t}= \sqrt{\left(\frac{\Delta m}{m}\right)^2+\left(\frac{\Delta g}{g}\right)^2+\left(\frac{\Delta k}{k}\right)^2} \] (similar propagation applies to the quadratic form).

9. Summary Table of Equations

Concept	Equation	When to Use
Linear momentum	\(\mathbf{p}=m\mathbf{v}\)	Any moving object
Impulse–momentum theorem	\(\mathbf{J}= \displaystyle\int\mathbf{F}\,dt = \Delta\mathbf{p}\)	Force acting over a short time interval
Newton’s 2nd law (constant mass)	\(\mathbf{F}=m\mathbf{a}\)	Most A‑Level problems (mass constant, inertial frame)
Linear (viscous) drag	\(F_d = kv\)	Low speeds, laminar flow (\(Re \lesssim 2\times10^{3}\))
Quadratic (pressure) drag	\(F_d = \tfrac12 C\rho A v^{2}\)	Higher speeds, turbulent flow (\(Re \gtrsim 2\times10^{3}\))
Terminal velocity – linear drag	\(v_t = \dfrac{mg}{k}\)	When \(F_d = kv\)
Terminal velocity – quadratic drag	\(v_t = \sqrt{\dfrac{2mg}{C\rho A}}\)	When \(F_d = \tfrac12 C\rho A v^{2}\)
Conservation of momentum (1‑D)	\(m_1u_1+m_2u_2 = m_1v_1+m_2v_2\)	Collisions with no external horizontal force
Conservation of kinetic energy (elastic)	\(\tfrac12 m_1u_1^{2}+\tfrac12 m_2u_2^{2}= \tfrac12 m_1v_1^{2}+\tfrac12 m_2v_2^{2}\)	Elastic collisions only

10. Key Points to Remember

Newton’s second law in the constant‑mass form is \(\mathbf{F}=m\mathbf{a}\) (mass constant, inertial frame).
Linear momentum \(\mathbf{p}=m\mathbf{v}\) is conserved when no external forces act.
In collisions momentum is always conserved; kinetic energy is conserved only for elastic collisions.
Terminal velocity occurs when the resistive drag force exactly balances the weight.
Linear drag (\(F_d = kv\)) applies to low‑speed, laminar flow; quadratic drag (\(F_d = \tfrac12 C\rho A v^{2}\)) applies to higher‑speed, turbulent flow.
Even at terminal speed the object is still moving – acceleration is zero, not velocity.

11. Suggested Diagram

Falling object with weight and drag forces — Weight \(mg\) acting downwards, drag \(F_d\) acting upwards, and the constant velocity vector \(\mathbf{v}=v_t\) when the forces are balanced.

understand that objects moving against a resistive force may reach a terminal (constant) velocity