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(vf - vi)=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 \(m1\) and \(m2\) (initial speeds \(v1, v2\)):

\[

v1'=\frac{m1-m2}{m1+m2}\,v1+\frac{2m2}{m1+m2}\,v2,\qquad

v2'=\frac{2m1}{m1+m2}\,v1+\frac{m2-m1}{m1+m2}\,v2.

\]

3.2 Inelastic collisions (kinetic energy not conserved)

Perfectly inelastic (bodies stick together):

\[

v'=\frac{m1v1+m2v2}{m1+m2}.

\]

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: \(m1=2\;\text{kg}\) with \(v1=4\;\text{m s}^{-1}\) 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: \(F{\text{drag}} = \tfrac{1}{2}CD \rho A v^{2}\) – 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}CD \rho A vt^{2}\quad\Rightarrow\quad vt = \sqrt{\frac{2mg}{CD \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 \(\rho = 1.2\;\text{kg m}^{-3}\).

\[

vt = \sqrt{\frac{2mg}{CD \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

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: \(F=\rho A v^{2}\).