\$\mathbf{F}_{\text{net}} = m\mathbf{a}.\$
(The more general form \(\mathbf{F}_{\text{net}} = d\mathbf{p}/dt\) is useful for deeper study but is not required by the syllabus.)
Momentum is a vector defined by
\$\mathbf{p}=m\mathbf{v}\$
Its direction is the same as the velocity of the object.
\$\mathbf{p}=m\bigl(vx\hat{\mathbf{i}}+vy\hat{\mathbf{j}}\bigr).\$
\$\$\sum p{x,\text{initial}}=\sum p{x,\text{final}},\qquad
\sum p{y,\text{initial}}=\sum p{y,\text{final}}.\$\$
The impulse delivered by a force \(\mathbf{F}(t)\) over a time interval \(\Delta t\) is
\$\mathbf{J}= \int{t1}^{t_2}\mathbf{F}\,dt = \Delta\mathbf{p}.\$
Impulse changes the momentum of a body without needing the detailed force history.
\$\mathbf{p}{\text{initial}}=\mathbf{p}{\text{final}}.\$
\$\$\tfrac12 m1v1^{2}+\tfrac12 m2v2^{2}=
\tfrac12 m1v1'^{2}+\tfrac12 m2v2'^{2}.\$\$
Note: the syllabus does not require the coefficient of restitution; the kinetic‑energy equation is needed only for perfectly elastic collisions.
\$v'=\frac{m1v1+m2v2}{m1+m2}.\$
Cart A: \(mA=2.0\;\text{kg},\;vA=+3.0\;\text{m s}^{-1}\) (right).
Cart B: \(mB=3.0\;\text{kg},\;vB=-2.0\;\text{m s}^{-1}\) (left).
Find the velocities after a perfectly elastic collision.
\$mAvA+mBvB = mAvA'+mBvB'.\$
\$\$v_A' = -0.2\;\text{m s}^{-1},\qquad
v_B' = 2.8\;\text{m s}^{-1}.\$\$
The lighter cart rebounds, the heavier cart continues forward faster – a classic elastic‑collision outcome.
For completeness, the usual expressions are
\$\$F{\text{static}}^{\max}= \mus N,\qquad
F{\text{kinetic}} = \muk N,\$\$
where \(\mus\) and \(\muk\) are the static and kinetic coefficients of friction. These are useful for extension work but are not examined.

Conceptual form: \(F_{\text{drag}}\propto v\).
Conceptual form: \(F_{\text{drag}}\propto v^{2}\).
Beyond the syllabus – useful for extension work
For students who wish to explore further, the drag force can be written in a simple piece‑wise form:
\$\$%
F_{\text{drag}}(v)=%
\begin{cases}
k{1}\,v, & v\lesssim v{c}\\[4pt]
k{2}\,v^{2}, & v\gtrsim v{c}
\end{cases}%
\$\$
where \(k{1},k{2}>0\) depend on the object's size, shape and the fluid’s density, and \(v_{c}\) is the characteristic speed at which the flow changes regime.
Assume a sky‑diver of mass \(m=80\;\text{kg}\) reaches a terminal speed when the weight \(mg\) is balanced by a quadratic drag force \(F{\text{drag}}=k v^{2}\). If the measured terminal speed is \(vt=55\;\text{m s}^{-1}\), the drag constant is
\$k = \frac{mg}{v_t^{2}} = \frac{80\times9.8}{55^{2}}\approx 0.026\;\text{kg m}^{-1}.\$
This illustrates how the qualitative idea of “drag ∝ \(v^{2}\) at high speed” leads directly to a terminal‑velocity calculation.

| Aspect | Friction (solid–solid) | Viscous/Drag (solid–fluid) |
|---|---|---|
| Typical speed dependence | Nearly independent of speed (static/kinetic friction) | Increases with speed – roughly linear at low \(v\), quadratic at high \(v\) |
| Direction of force | Opposite to relative motion (or impending motion) | Opposite to the object's velocity |
| Energy dissipation | Converted to heat at the contact surface | Converted to heat in the fluid and, at high speed, to turbulent kinetic energy |
| Dependence on normal reaction | Approximately proportional to the normal reaction \(N\) | Independent of \(N\); depends on fluid density, object shape and speed |
| Typical syllabus‑level formula | \(F_f \approx \mu N\) (qualitative) | \(F_{\text{drag}}\) ∝ \(v\) (low speed) or ∝ \(v^{2}\) (high speed) |
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