Momentum and Newton’s Laws of Motion – Uniform Gravity with Air‑Resistance
Learning objective
Describe and explain qualitatively the motion of objects moving in a uniform gravitational field when air‑resistance (drag) is present, and relate the description to the concepts of linear momentum and Newton’s laws as required by the Cambridge AS & A Level Physics syllabus (3.1).
Key concepts
- Linear momentum \( \mathbf{p}=m\mathbf{v} \)
- Newton’s First Law – inertia
- Newton’s Second Law – \( \mathbf{F}_{\text{net}} = \dfrac{d\mathbf{p}}{dt}=m\mathbf{a} \)
- Newton’s Third Law – equal and opposite reaction forces
- Weight as the effect of a uniform gravitational field \( \mathbf{W}=m\mathbf{g} \) (field strength \( g=9.81\;\text{m s}^{-2} \) near the Earth’s surface)
- Air‑resistance (drag) – a force opposite to the direction of motion, usually proportional to \(v\) (linear) or \(v^{2}\) (quadratic)
Why the gravitational field is taken as uniform
The field is assumed uniform because the change in \(g\) over the height of typical school‑level experiments (< 100 m) is less than 0.3 %. Over such a small region the field lines are essentially parallel and the field strength \(g\) can be treated as constant.
Linear momentum – quick quantitative reminder
For a 0.5 kg ball moving downwards at 5 m s\(^{-1}\):
\[
\mathbf{p}=m\mathbf{v}=0.5\;\text{kg}\times5\;\text{m s}^{-1}=2.5\;\text{kg m s}^{-1}\;\text{(downwards)}.
\]
Momentum is a vector (unit kg m s\(^{-1}\)). The change of momentum per unit time equals the net external force acting on the object.
Newton’s three laws – statements and relevance to falling bodies
- First law (inertia): A body remains at rest or moves with constant velocity unless acted on by a net external force. In free fall the net force is not zero, so the body accelerates.
- Second law: \( \mathbf{F}_{\text{net}} = m\mathbf{a} = \dfrac{d\mathbf{p}}{dt} \). This links the weight‑drag imbalance to the acceleration of the falling object.
- Third law: If the falling object exerts a drag force on the air, the air exerts an equal and opposite reaction on the object – this reaction is precisely the drag force that appears in the free‑fall force diagram.
Weight and the uniform gravitational field
Weight is the force exerted on a mass by a uniform gravitational field:
\[
\mathbf{W}=m\mathbf{g},
\]
where \( \mathbf{g} \) is the constant field strength (direction downwards). In syllabus language the field is “uniform” because \( \mathbf{g} \) does not vary over the region of interest.
Forces on a vertically falling object
- Weight \( \mathbf{W}=m\mathbf{g} \) (downwards)
- Air‑resistance (drag) \( \mathbf{F}_{\text{drag}} \) (upwards, opposite to the velocity)
Typical forms of drag
- Linear drag (low speeds, very small Reynolds number, Re < 1): \( F_{\text{drag}} = k\,v \)
For a sphere moving in a viscous fluid, \( k \approx 6\pi\eta r \) (Stokes’ law), where \( \eta \) is the dynamic viscosity and \( r \) the radius.
- Quadratic drag (higher speeds, Re ≫ 1): \( F{\text{drag}} = \tfrac{1}{2} C{d}\,\rho\,A\,v^{2} \)
Typical drag coefficients: \( C{d}\approx0.47 \) for a smooth sphere, \( C{d}\approx1.0\!-\!1.2 \) for a flat plate or a feather.
Qualitative description of the motion
At any instant the net force is
\[
\mathbf{F}{\text{net}} = m\mathbf{g} - \mathbf{F}{\text{drag}}(v).
\]
Three regimes can be identified:
| Regime | Dominant forces | Resulting acceleration | Velocity behaviour |
|---|
| Initial phase (very small \(v\)) | \( mg \gg F_{\text{drag}} \) | \( a \approx g \) (downwards) | Speed increases rapidly |
| Intermediate phase | \( mg \) comparable to \( F_{\text{drag}} \) | \( a \) decreases as \(v\) grows | Speed continues to rise but more slowly |
| Terminal‑velocity phase | \( mg = F{\text{drag}}(v{t}) \) | \( a = 0 \) (net force zero) | Constant speed \( v_{t} \) |
Terminal velocity
Setting \( mg = F{\text{drag}}(v{t}) \) gives:
- Linear drag: \( v_{t}= \dfrac{mg}{k} \)
- Quadratic drag: \( v{t}= \sqrt{\dfrac{2mg}{C{d}\,\rho\,A}} \)
Momentum perspective
From Newton’s second law
\[
\frac{d\mathbf{p}}{dt}= \mathbf{F}{\text{net}} = mg\,\hat{\mathbf{y}} - \mathbf{F}{\text{drag}}(v).
\]
When the object approaches \( v{t} \), \( \mathbf{F}{\text{net}}\to0 \) and therefore \( d\mathbf{p}/dt\to0 \); the momentum becomes constant.
Factors that affect the terminal speed
- Mass \(m\) – larger \(m\) raises \(v_{t}\) because weight grows faster than drag.
- Cross‑sectional area \(A\) – larger \(A\) increases drag, lowering \(v_{t}\).
- Shape (drag coefficient \(C{d}\)) – streamlined bodies have smaller \(C{d}\) and higher \(v_{t}\).
- Air density \(\rho\) – higher \(\rho\) (e.g. at lower altitude) increases drag, reducing \(v_{t}\).
Illustrative example – feather vs. steel ball
- Both are released from the same height in still air.
- The feather’s tiny mass means \( mg \) is quickly balanced by a relatively small drag force; it reaches a low terminal speed almost immediately.
- The steel ball, being much heavier, experiences a large net downward force for a longer interval, so its acceleration stays close to \(g\) before drag becomes comparable to its weight.
- Consequently the ball attains a much higher \(v_{t}\) and hits the ground far sooner than the feather.
Experiment to determine the acceleration of free fall (\(g\))
A simple classroom experiment uses a motion sensor (or a high‑frame‑rate video) to record the vertical position of a falling object.
| Step | Procedure |
|---|
| 1 | Set up a motion‑sensor or a high‑frame‑rate camera aimed vertically. |
| 2 | Measure the initial height \(h\) from the sensor to the release point. |
| 3 | Release a small dense sphere (e.g. a steel ball) from rest and record its position \(y(t)\) as it falls. |
| 4 | Plot \(y\) versus \(t\) and fit the data to \( y = \tfrac{1}{2} a t^{2} \) (initial velocity zero). |
| 5 | The fitted coefficient gives the experimental acceleration \(a\); compare with the accepted value \(g=9.81\;\text{m s}^{-2}\). |
Data‑table sketch (to be completed by students)
| Time \(t\) (s) | Position \(y\) (m) |
|---|
| 0.00 | — |
| 0.10 | — |
| 0.20 | — |
Common misconceptions
- “Air resistance always reduces acceleration to zero.” – It reduces the acceleration; only at terminal velocity does the net acceleration become zero.
- “Heavier objects fall faster because they are heavier.” – In a vacuum the answer is yes; in air the greater weight means a larger drag force is required to balance it, giving a higher terminal speed, not a higher instantaneous acceleration.
- “Momentum is conserved in free fall.” – Momentum changes because external forces (gravity and drag) act on the object; only the combined system (object + air) conserves momentum.
Suggested diagrams (to be drawn by students)
- Free‑fall force diagram: arrows for weight \(mg\) (down) and drag \(F_{\text{drag}}\) (up); label the reaction pair from Newton’s third law.
- Velocity‑versus‑time graph showing the rapid rise, gradual curvature, and asymptotic approach to \(v_{t}\).
- Sketch of the experimental set‑up with a motion sensor, release point, and measured distances.
Summary
- Linear momentum \( \mathbf{p}=m\mathbf{v} \) changes according to the net external force (Newton’s second law).
- Weight \( \mathbf{W}=m\mathbf{g} \) is the constant force exerted by a uniform gravitational field.
- Air‑resistance opposes motion; its magnitude grows with speed (linear for low‑Re, quadratic for high‑Re).
- The competition between weight and drag creates three distinct motion regimes, ending in a constant‑speed terminal velocity where \( \mathbf{F}_{\text{net}}=0 \) and \( d\mathbf{p}/dt=0 \).
- Mass, cross‑sectional area, shape (drag coefficient) and air density determine the numerical value of the terminal speed.
- An experiment using a motion sensor or video analysis can determine the free‑fall acceleration \(g\) and reinforces the link between theory and real data.