understand that objects moving against a resistive force may reach a terminal (constant) velocity
Momentum and Newton’s Laws of Motion
Learning Objective
Understand that an object moving against a resistive force may reach a terminal (constant) velocity, and be able to apply Newton’s laws and the conservation of linear momentum to a range of situations (including collisions).
Key Definitions (Cambridge wording)
| Term | Definition |
|---|---|
| Linear momentum | The product of an object’s mass and its velocity; a vector quantity with units kg·m s⁻¹. \( \mathbf{p}=m\mathbf{v}\) |
| Impulse | The integral of a force over the time interval during which it acts; has the same units as momentum. \( \mathbf{J}= \displaystyle\int{t1}^{t_2}\mathbf{F}\,dt = \Delta\mathbf{p}\) |
| Terminal velocity | The constant speed attained when the net external force on a falling object becomes zero (weight = drag). Acceleration is zero but the object continues to move. |
| Reynolds number ( \(Re\) ) | A dimensionless quantity that predicts the flow regime: \( Re = \dfrac{\rho v L}{\mu}\) where \(\rho\) is fluid density, \(v\) the speed of the object, \(L\) a characteristic length (e.g. diameter) and \(\mu\) the dynamic viscosity. Rough guideline: \(Re \lesssim 2\times10^{3}\) → laminar (linear drag); \(Re \gtrsim 2\times10^{3}\) → turbulent (quadratic drag). |
1. Newton’s Laws of Motion
- First law (Inertia): A body remains at rest or moves with uniform straight‑line velocity unless acted on by a net external force.
- Second law:
- General (variable mass): \(\displaystyle \mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt}\).
- Constant‑mass form (used in almost all Cambridge AS & A Level problems):
\(\displaystyle \mathbf{F}=m\mathbf{a}\)
Assumptions: the mass of the object does not change during the interval and the reference frame is inertial (i.e. not accelerating).
- Third law: For every action there is an equal and opposite reaction.
2. Linear Momentum
- Definition: \(\mathbf{p}=m\mathbf{v}\) (vector, SI unit kg·m s⁻¹).
- Impulse–momentum theorem: \(\displaystyle \mathbf{J}= \int{t1}^{t_2}\mathbf{F}\,dt = \Delta\mathbf{p}\).
- Conservation of linear momentum: In the absence of external forces the total momentum of a closed system remains constant. This holds for both 1‑D and 2‑D interactions.
3. Collisions
3.1 One‑dimensional collisions
- Inelastic: Momentum conserved, kinetic energy not conserved. Example – a lump of clay sticks to a block.
- Elastic: Both momentum and kinetic energy conserved. Example – two steel balls colliding on a smooth track.
3.2 Two‑dimensional collisions
Momentum is conserved separately in the horizontal (\(x\)) and vertical (\(y\)) directions. Vector diagrams are essential for solving billiard‑ball or projectile‑impact problems.
Example – 1‑D elastic collision of identical spheres
Two identical spheres (mass \(m\)) approach each other with speeds \(u\) and \(-u\). Solving the simultaneous equations for momentum and kinetic‑energy conservation gives \(v1 = -u\) and \(v2 = u\); they simply exchange velocities.
4. Forces on a Falling Object
Consider a body of mass \(m\) moving vertically through a fluid (air, water, oil…). The relevant forces are:
| Force | Direction | Expression | Typical Units |
|---|---|---|---|
| Weight | Downwards | \(W = mg\) | N (kg·m s⁻²) |
| Resistive (drag) force | Opposes motion (upwards for a falling body) |
|
|
Flow regimes
- Viscous (laminar) flow – low Reynolds number, drag ∝ \(v\).
- Turbulent flow – high Reynolds number, drag ∝ \(v^{2}\).
5. Terminal (Constant) Velocity
When the net external force on the object becomes zero, its acceleration is zero and the speed stops changing. This constant speed is the terminal velocity \(v_t\).
5.1 Derivation – linear (viscous) drag
\[
mg - kv = 0 \;\Longrightarrow\; v_t = \frac{mg}{k}
\]
5.2 Derivation – quadratic (pressure) drag
\[
mg - \tfrac12 C\rho A v^{2}=0 \;\Longrightarrow\;
v_t = \sqrt{\frac{2mg}{C\rho A}}
\]
5.3 Qualitative picture
- Initially \(mg > F_d\) → downward acceleration.
- As \(v\) increases, \(Fd\) grows until \(Fd = mg\).
- Beyond this point the forces balance; the object continues to fall at the constant speed \(v_t\).
6. Worked Example – Terminal Velocity with Linear Drag
Problem: A steel sphere of mass \(0.15\;\text{kg}\) falls through air. The linear drag coefficient is \(k = 0.025\;\text{N·s m}^{-1}\). Find its terminal velocity.
Solution:
\[
v_t = \frac{mg}{k}
= \frac{0.15 \times 9.81}{0.025}
\approx 5.89 \times 10^{1}\;\text{m s}^{-1}
\;(\approx 59\;\text{m s}^{-1})
\]
7. Worked Example – 1‑D Inelastic Collision
Problem: A \(0.8\;\text{kg}\) cart moving at \(2.0\;\text{m s}^{-1}\) collides and sticks to a \(0.5\;\text{kg}\) cart initially at rest. Find the speed of the combined system after the collision.
Solution (conservation of momentum):
\[
m1u1 + m2u2 = (m1+m2)v
\quad\Longrightarrow\quad
(0.8)(2.0) + (0.5)(0) = (1.3)v
\]
\[
v = \frac{1.6}{1.3} \approx 1.23\;\text{m s}^{-1}
\]
8. Practical Skills – Determining Terminal Velocity
Experiment design
1. Set up a motion sensor (or high‑speed camera) to record the vertical position \(y(t)\) of a small dense sphere dropped from a known height.
2. Differentiate the data numerically to obtain the velocity \(v(t)\).
3. Plot \(v\) versus \(t\); the curve will asymptote to a constant value – the terminal speed.
Sources of error
• Air currents in the laboratory (systematic).
• Timing resolution of the sensor (random).
• Uncertainty in the mass and radius of the sphere (affects \(k\) or \(C\rho A\)).
Uncertainty analysis
For linear drag, propagate uncertainties using
\[
v_t = \frac{mg}{k}\;\; \Rightarrow\;\;
\frac{\Delta vt}{vt}= \sqrt{\left(\frac{\Delta m}{m}\right)^2+\left(\frac{\Delta g}{g}\right)^2+\left(\frac{\Delta k}{k}\right)^2}
\]
(similar propagation applies to the quadratic form).
9. Summary Table of Equations
| Concept | Equation | When to Use |
|---|---|---|
| Linear momentum | \(\mathbf{p}=m\mathbf{v}\) | Any moving object |
| Impulse–momentum theorem | \(\mathbf{J}= \displaystyle\int\mathbf{F}\,dt = \Delta\mathbf{p}\) | Force acting over a short time interval |
| Newton’s 2nd law (constant mass) | \(\mathbf{F}=m\mathbf{a}\) | Most A‑Level problems (mass constant, inertial frame) |
| Linear (viscous) drag | \(F_d = kv\) | Low speeds, laminar flow (\(Re \lesssim 2\times10^{3}\)) |
| Quadratic (pressure) drag | \(F_d = \tfrac12 C\rho A v^{2}\) | Higher speeds, turbulent flow (\(Re \gtrsim 2\times10^{3}\)) |
| Terminal velocity – linear drag | \(v_t = \dfrac{mg}{k}\) | When \(F_d = kv\) |
| Terminal velocity – quadratic drag | \(v_t = \sqrt{\dfrac{2mg}{C\rho A}}\) | When \(F_d = \tfrac12 C\rho A v^{2}\) |
| Conservation of momentum (1‑D) | \(m1u1+m2u2 = m1v1+m2v2\) | Collisions with no external horizontal force |
| Conservation of kinetic energy (elastic) | \(\tfrac12 m1u1^{2}+\tfrac12 m2u2^{2}= \tfrac12 m1v1^{2}+\tfrac12 m2v2^{2}\) | Elastic collisions only |
10. Key Points to Remember
- Newton’s second law in the constant‑mass form is \(\mathbf{F}=m\mathbf{a}\) (mass constant, inertial frame).
- Linear momentum \(\mathbf{p}=m\mathbf{v}\) is conserved when no external forces act.
- In collisions momentum is always conserved; kinetic energy is conserved only for elastic collisions.
- Terminal velocity occurs when the resistive drag force exactly balances the weight.
- Linear drag (\(Fd = kv\)) applies to low‑speed, laminar flow; quadratic drag (\(Fd = \tfrac12 C\rho A v^{2}\)) applies to higher‑speed, turbulent flow.
- Even at terminal speed the object is still moving – acceleration is zero, not velocity.
11. Suggested Diagram
