Forces, Density and Pressure – Cambridge A‑Level Physics (9702)

1. Newton’s Laws and Linear Momentum

1.1 Newton’s three laws (syllabus requirement)

First law (inertia): An object remains at rest or moves with constant velocity unless acted on by a net external force.
Second law: The net external force on a body equals the rate of change of its linear momentum \[ \mathbf{F}_{\text{net}}=\frac{d\mathbf{p}}{dt}. \] For a body of constant mass this reduces to the familiar form \(\mathbf{F}=m\mathbf{a}\).
Third law (action–reaction): For every action force there is an equal and opposite reaction force acting on a different body.

1.2 Linear momentum

Vector quantity, direction same as velocity.
Definition: \(\displaystyle\mathbf{p}=m\mathbf{v}\)
Magnitude: \(p=mv\)
Units: kg·m·s⁻¹

1.3 Example – Applying Newton’s second law

A 2 kg block slides on a frictionless horizontal table. A constant horizontal force of 6 N acts on it for 3 s. Find the final speed if the block starts from rest.

\[ \mathbf{F}=m\mathbf{a}\;\Rightarrow\;a=\frac{F}{m}= \frac{6}{2}=3\;\text{m s}^{-2}. \] \[ v = a t = 3\times3 = 9\;\text{m s}^{-1}. \]

Momentum after 3 s: \(p = mv = 2\times9 = 18\;\text{kg m s}^{-1}\).

2. Impulse–Momentum Theorem

2.1 Definition of impulse

Impulse \(\mathbf{J}\) is the time‑integral of the net force: \(\displaystyle\mathbf{J}= \int \mathbf{F}\,dt\).
For a constant force over a time interval \(\Delta t\): \(\mathbf{J}= \mathbf{F}\,\Delta t\).
Impulse equals the change in momentum: \(\displaystyle\mathbf{J}= \Delta\mathbf{p}= \mathbf{p}_{\text{final}}-\mathbf{p}_{\text{initial}}\).
Graphically, impulse is the area under a force–time graph.

2.2 Example – Stopping a moving cart

A 5 kg cart moving at 4 m s⁻¹ is brought to rest by a constant braking force applied for 0.8 s. Find the magnitude of the braking force.

\[ \Delta p = m(v_f - v_i)=5(0-4)=-20\;\text{kg m s}^{-1}. \] \[ J = |\Delta p| = 20\;\text{N·s}=F\Delta t \;\Rightarrow\;F=\frac{20}{0.8}=25\;\text{N}. \]

3. Collisions

3.1 Elastic collisions (kinetic energy conserved)

One‑dimensional elastic collision between masses \(m_1\) and \(m_2\) (initial speeds \(v_1, v_2\)):

\[ v_1'=\frac{m_1-m_2}{m_1+m_2}\,v_1+\frac{2m_2}{m_1+m_2}\,v_2,\qquad v_2'=\frac{2m_1}{m_1+m_2}\,v_1+\frac{m_2-m_1}{m_1+m_2}\,v_2. \]

3.2 Inelastic collisions (kinetic energy not conserved)

Perfectly inelastic (bodies stick together):

\[ v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}. \]

3.3 Two‑dimensional collisions

Resolve all vectors into orthogonal components (e.g. \(x\) and \(y\)).
Apply conservation of momentum separately to each component.
Only components with no external impulse are conserved.

3.4 Example – Elastic collision on a frictionless track

Two blocks: \(m_1=2\;\text{kg}\) with \(v_1=4\;\text{m s}^{-1}\) and \(m_2=3\;\text{kg}\) initially at rest.

\[ v_1'=\frac{2-3}{2+3}\times4=-0.8\;\text{m s}^{-1},\qquad v_2'=\frac{2\times2}{5}\times4=3.2\;\text{m s}^{-1}. \]

3.5 Example – Perfectly inelastic collision

A 1 kg cart at 2 m s⁻¹ collides with a 2 kg cart at rest and sticks together.

\[ v'=\frac{1\times2+2\times0}{1+2}= \frac{2}{3}\;\text{m s}^{-1}. \]

4. Turning Effects of Forces (Moments)

4.1 Moment (torque)

Moment about point O: \(\displaystyle\boldsymbol{\tau}= \mathbf{r}\times\mathbf{F}\).
Magnitude: \(\tau = Fr\sin\theta\) (or \(F\) times the perpendicular distance \(r_\perp\)).
Units: N·m (equivalent to joule).
Direction given by the right‑hand rule.

4.2 Couples

Two equal, opposite forces whose lines of action do not coincide.
Resultant force = 0, but a net moment exists: \(\tau_{\text{couple}} = F\,d\) where \(d\) is the perpendicular separation.
Produces pure rotation without translation.

4.3 Centre of gravity (CG)

Point at which the weight of a body may be considered to act.
For a uniform body, CG coincides with the geometric centre.
In equilibrium: \(\displaystyle\sum\tau = 0\) about any point.

4.4 Example – Seesaw balance

Child A: \(m=30\;\text{kg}\) sits 1.5 m from the pivot. Child B: \(m=20\;\text{kg}\) sits on the opposite side at distance \(x\).

\[ 30g(1.5)=20g\,x\;\Rightarrow\;x=\frac{30\times1.5}{20}=2.25\;\text{m}. \]

5. Fluid Mechanics – Density and Pressure

5.1 Density

Mass per unit volume:

\[ \rho=\frac{m}{V}\qquad\text{Units: kg m}^{-3}. \]

5.2 Pressure

Force per unit area (scalar):

\[ p=\frac{F}{A}\qquad\text{Units: Pa = N m}^{-2}. \]

Gauge pressure: measured relative to atmospheric pressure.
Absolute pressure: \(p_{\text{abs}} = p_{\text{gauge}} + p_{\text{atm}}\).

5.3 Hydrostatic pressure

Pressure at depth \(h\) in a fluid of density \(\rho\):

\[ p = p_0 + \rho g h, \] where \(p_0\) is the pressure at the free surface (often atmospheric). The relation is linear with depth.

5.4 Dynamic pressure and Bernoulli’s principle

Dynamic pressure (kinetic energy per unit volume of a moving fluid):

\[ p_{\text{dyn}}=\frac{1}{2}\rho v^{2}. \]

For an ideal, incompressible, non‑viscous flow:

\[ p_{\text{static}}+\frac{1}{2}\rho v^{2}+ \rho g h = \text{constant}. \]

5.5 Momentum flux in a fluid

When a fluid of density \(\rho\) and speed \(v\) strikes a surface and is brought to rest, the force exerted is

\[ F = \dot{m} v = \rho A v^{2}, \] where \(\dot{m}= \rho A v\) is the mass‑flow rate and \(A\) the cross‑sectional area.

5.6 Example – Water jet on a flat plate

Water jet: \(\rho = 1000\;\text{kg m}^{-3}\), area \(A = 0.01\;\text{m}^{2}\), speed \(v = 20\;\text{m s}^{-1}\).

\[ F = \rho A v^{2}= 1000\times0.01\times20^{2}= 4000\;\text{N}. \]

6. Non‑Uniform Motion & Air Resistance (required by the syllabus)

6.1 Frictional and drag forces

Friction (solid contact): Approximate model \(F_{\text{fr}} = \mu N\) (static or kinetic coefficient \(\mu\)).
Viscous (linear) drag: \(F_{\text{drag}} = k v\) – useful for low speeds or small objects in a viscous medium.
Quadratic (turbulent) drag: \(F_{\text{drag}} = \tfrac{1}{2}C_D \rho A v^{2}\) – dominant at higher speeds.

6.2 Terminal velocity

When the downward weight equals the upward drag, the net force is zero and the object moves at constant speed \(v_t\).

\[ mg = \tfrac{1}{2}C_D \rho A v_t^{2}\quad\Rightarrow\quad v_t = \sqrt{\frac{2mg}{C_D \rho A}}. \]

6.3 Qualitative description of motion in a uniform gravitational field with air resistance

Initial acceleration \(a = g - F_{\text{drag}}/m\) (less than \(g\)).
Velocity increases, drag grows, acceleration decreases.
When \(F_{\text{drag}} = mg\), acceleration becomes zero → terminal velocity.

6.4 Example – Skydiver (quadratic drag)

Mass \(m = 80\;\text{kg}\), \(C_D = 1.0\), cross‑sectional area \(A = 0.7\;\text{m}^{2}\), air density \(\rho = 1.2\;\text{kg m}^{-3}\).

\[ v_t = \sqrt{\frac{2mg}{C_D \rho A}} = \sqrt{\frac{2\times80\times9.81}{1.0\times1.2\times0.7}} \approx 53\;\text{m s}^{-1}. \]

7. Summary of Key Quantities

Quantity	Symbol	Units	Formula
Linear momentum	\(\mathbf{p}\)	kg·m·s⁻¹	\(\mathbf{p}=m\mathbf{v}\)
Impulse	\(\mathbf{J}\)	N·s	\(\displaystyle\mathbf{J}= \int\mathbf{F}\,dt\)
Force (Newton’s 2nd law)	\(\mathbf{F}\)	N	\(\displaystyle\mathbf{F}= \frac{d\mathbf{p}}{dt}=m\mathbf{a}\)
Moment (torque)	\(\boldsymbol{\tau}\)	N·m	\(\boldsymbol{\tau}= \mathbf{r}\times\mathbf{F}\)
Density	\(\rho\)	kg·m⁻³	\(\rho=m/V\)
Pressure	\(p\)	Pa (N·m⁻²)	\(p=F/A\)
Hydrostatic pressure	\(p\)	Pa	\(p=p_0+\rho g h\)
Dynamic pressure	\(p_{\text{dyn}}\)	Pa	\(p_{\text{dyn}}=\tfrac{1}{2}\rho v^{2}\)
Bernoulli constant	–	Pa	\(p_{\text{static}}+\tfrac{1}{2}\rho v^{2}+\rho g h = \text{constant}\)
Force from fluid momentum flux	\(F\)	N	\(F=\rho A v^{2}\)
Linear drag	\(F_{\text{drag}}\)	N	\(F_{\text{drag}} = k v\)
Quadratic drag	\(F_{\text{drag}}\)	N	\(F_{\text{drag}} = \tfrac{1}{2}C_D \rho A v^{2}\)
Terminal velocity (quadratic drag)	\(v_t\)	m·s⁻¹	\(v_t = \sqrt{\dfrac{2mg}{C_D \rho A}}\)

8. Quick Revision Checklist

State and apply Newton’s 1st, 2nd and 3rd laws.
Write the momentum definition \(\mathbf{p}=m\mathbf{v}\) and the impulse–momentum theorem.
Distinguish elastic, inelastic and perfectly inelastic collisions; use the 1‑D formulas.
Apply the component method for two‑dimensional collisions.
Calculate moments: \(\tau = Fr_{\perp}\); recognise couples and the role of the centre of gravity.
Identify when forces are non‑uniform (friction, air drag) and choose an appropriate drag model.
Derive terminal velocity by setting weight equal to drag.
Remember pressure is scalar: \(p=F/A\); differentiate gauge and absolute pressure.
Use \(p=p_0+\rho g h\) for hydrostatic problems.
Apply dynamic pressure \(\tfrac{1}{2}\rho v^{2}\) and Bernoulli’s equation for fluid‑flow questions.
Compute forces from fluid momentum flux: \(F=\rho A v^{2}\).

Forces, density and pressure