Linear momentum of a particle of mass m moving with velocity v is
\[
\mathbf{p}=m\mathbf{v}
\]
\[
\mathbf{F}_{\text{net}}=\frac{d\mathbf{p}}{dt}.
\]
For constant mass this reduces to \(\mathbf{F}=m\mathbf{a}\).
The same form also applies to variable‑mass systems (e.g. rockets) because the change of mass is automatically included in \(\mathbf{p}=m\mathbf{v}\).
\[
\sum\mathbf{p}{\text{initial}}=\sum\mathbf{p}{\text{final}}.
\]
Two carts on a frictionless track:
After they stick together, the common velocity \(v\) is found from momentum conservation:
\[
(m{A}+m{B})\,v = m{A}v{A}+m{B}v{B}
\;\;\Longrightarrow\;\;
v = \frac{2.0\times3.0}{2.0+1.0}=2.0\;\text{m s}^{-1}\;\text{(to the right)}.
\]
The kinetic energy before collision is \( \tfrac12 m{A}v{A}^{2}=9\;\text{J}\); after collision it is \(\tfrac12 (m{A}+m{B})v^{2}=6\;\text{J}\). The loss of 3 J appears as internal energy (deformation, heat, sound).
Resistive forces always act opposite to the instantaneous velocity of the body, thereby reducing its momentum.
\[
\mathbf{F}{\text{k}}=-F{k}\,\hat{\mathbf{v}},\qquad F_{k}\approx\text{constant}.
\]
When an object moves through a liquid or a gas the fluid exerts a resistive force that depends on the speed.
\[
\mathbf{F}{\text{drag}}=-k{1}v\,\hat{\mathbf{v}}\qquad (F\propto v)
\]
where \(k_{1}\) contains the fluid’s viscosity, the object’s characteristic size and shape.
\[
\mathbf{F}{\text{drag}}=-k{2}v^{2}\,\hat{\mathbf{v}}\qquad (F\propto v^{2})
\]
where \(k{2}= \tfrac12 C{d}\rho A\) (drag coefficient \(C_{d}\), fluid density \(\rho\), cross‑sectional area \(A\)).
For most A‑Level problems a simple two‑regime model suffices:
\[
\mathbf{F}_{\text{air}}=
\begin{cases}
-k{1}v\,\hat{\mathbf{v}}, & v
-k{2}v^{2}\,\hat{\mathbf{v}}, & v\ge v{c}\ (\text{turbulent})
\end{cases}
\]
When the downward weight of a falling body is exactly balanced by the upward drag, the net force is zero and the speed becomes constant – this is the terminal velocity \(v_{t}\).
\[
mg = k{1}v{t}\;\;\Longrightarrow\;\;v{t}= \frac{mg}{k{1}}.
\]
\[
mg = k{2}v{t}^{2}\;\;\Longrightarrow\;\;v{t}= \sqrt{\frac{mg}{k{2}}}.
\]
Starting from rest, the equation of motion is
\[
m\frac{dv}{dt}=mg-k_{1}v.
\]
Solving gives
\[
v(t)=v_{t}\bigl(1-e^{-t/\tau}\bigr),\qquad
\tau=\frac{m}{k_{1}}.
\]
The speed rises exponentially and asymptotically approaches \(v_{t}\). This behaviour is a common AO2 expectation.
Suggested experiment (linear drag)
\[
s = v{t}\,t - \tau v{t}\bigl(1-e^{-t/\tau}\bigr)
\]
or, for the steady‑state region, simply use \(v\approx v{t}=mg/k{1}\) to obtain
\[
k{1}= \frac{mg}{v{t}}.
\]
From Newton’s second law in momentum form:
\[
\frac{d\mathbf{p}}{dt}= \mathbf{F}{\text{net}} = -\mathbf{F}{\text{res}},
\]
where \(\mathbf{F}{\text{res}}\) is any of the forces listed above. Because \(\mathbf{F}{\text{res}}\) always points opposite to \(\mathbf{v}\), the magnitude of \(\mathbf{p}\) decreases. The rate of decrease is:
| Force type | Speed dependence | Typical example | Direction | Syllabus outcome |
|---|---|---|---|---|
| Static friction | None (zero speed) | Block at rest on a rough table | Opposes impending motion | 3.1 – describe friction qualitatively |
| Kinetic friction | ≈ constant (independent of speed) | Sliding crate on a floor | Opposes direction of motion | 3.1 – describe friction qualitatively |
| Viscous (laminar) drag | \(\propto v\) | Sphere sinking slowly in oil | Opposes velocity | 3.2 – explain effect of drag on motion |
| Inertial (turbulent) drag | \(\propto v^{2}\) | Car at highway speed, sky‑diver | Opposes velocity | 3.2 – explain effect of drag on motion |
| Air resistance (piece‑wise model) | Low \(v\): \(\propto v\); High \(v\): \(\propto v^{2}\) | Anything moving through air | Opposes velocity | 3.2 – describe terminal velocity |
\[
mg = k{2}v{t}^{2}\;\;\Longrightarrow\;\;v{t}= \sqrt{\frac{mg}{k{2}}}\approx 55\;\text{m s}^{-1}
\]
(typical value). Acceleration decreases continuously and becomes zero at \(v_{t}\).
The work done by a resistive force over a displacement \(x\) is
\[
W{\text{res}} = \int \mathbf{F}{\text{res}}\cdot d\mathbf{x}.
\]
For kinetic friction \(W = -F_{k}x\) (linear loss).
For linear drag \(W = -k{1}\int v\,dx = -k{1}\int v^{2}\,dt\) (exponential loss).
For quadratic drag \(W = -k{2}\int v^{2}\,dx = -k{2}\int v^{3}\,dt\) (stronger loss at high speed).
These expressions link the momentum analysis to the energy‑conservation part of the syllabus (Section 4 “Work, energy and power”).
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