recall F = ma and solve problems using it, understanding that acceleration and resultant force are always in the same direction

Momentum, Newton’s Laws of Motion and Non‑Uniform Dynamics (Cambridge 9702 – AS/A‑Level)

Learning Objectives (Cambridge Topic 3)

• State and apply Newton’s 1st, 2nd and 3rd laws in vector form.
• Recall that the resultant (net) force and the acceleration of a particle are always collinear and point in the same direction.
• Draw accurate free‑body diagrams (FBDs) and use Σ F = ma to solve quantitative problems.
• Define linear momentum, use the impulse–momentum theorem and apply conservation of linear momentum for collisions and explosions.
• Analyse non‑uniform motion when friction, linear air‑drag or other variable forces act.
• Carry out a simple experiment to determine the acceleration due to gravity, g.
• Follow the Cambridge problem‑solving checklist for dynamics.

Syllabus Mapping – How the Notes Cover Each Outcome

1. Newton’s Laws of Motion and Momentum

1.1 First Law – Inertia

A particle remains at rest or moves with constant velocity (zero acceleration) unless acted upon by a non‑zero resultant force.

1.2 Second Law – \( \mathbf{F}_{\text{net}} = m\mathbf{a} \)

• Vector form: the net force \(\mathbf{F}_{\text{net}}\) and the acceleration \(\mathbf{a}\) are parallel and point in the same direction.
• Scalar mass: \(m>0\) and is constant for a classical particle.
• If \(\Sigma\mathbf{F}=0\) then \(\mathbf{a}=0\) (uniform motion).

1.3 Third Law – Action & Reaction

For every action force \(\mathbf{F}{AB}\) on body B there is an equal and opposite reaction force \(\mathbf{F}{BA}\) on body A:

\[

\mathbf{F}{AB}= -\mathbf{F}{BA}

\]

Because the two forces act on different bodies they never cancel in the \(\mathbf{F}=m\mathbf{a}\) equation for a single object.

1.4 Linear Momentum, Impulse and Conservation

• Momentum: \(\displaystyle \mathbf{p}=m\mathbf{v}\) (vector, same direction as \(\mathbf{v}\)).
• Impulse–Momentum Theorem: \(\displaystyle \mathbf{J}= \int{t1}^{t2}\mathbf{F}{\text{net}}\,dt \approx \mathbf{F}_{\text{avg}}\Delta t = \Delta\mathbf{p}\).
• Conservation of Linear Momentum: In the absence of external forces, \(\displaystyle \sum\mathbf{p}{\text{initial}} = \sum\mathbf{p}{\text{final}}\).

2. Free‑Body Diagrams (FBDs)

1. Represent the object as a point (or simple shape).
2. Draw every external force as an arrow starting at the centre of mass.
3. Label each force (e.g. \(W\) for weight, \(N\) for normal, \(f\) for friction, \(F\) for applied force).
4. Choose a convenient coordinate system (e.g. +x right, +y up).
5. Write the component equations: \(\Sigma Fx = m ax\), \(\Sigma Fy = m ay\).

3. Non‑Uniform Motion

3.1 Friction

• Static friction: \(fs \le \mus N\). It prevents motion up to its maximum value.
• Kinetic friction: \(fk = \muk N\). It opposes actual motion with a constant magnitude.
• Both act opposite to the direction of relative motion (or the tendency to move).

3.2 Linear Air‑Drag (low‑speed model)

\[

\mathbf{F}_{\text{drag}} = -b\,\mathbf{v}

\]

where \(b>0\) (kg s⁻¹) and the negative sign guarantees the drag always opposes the velocity.

3.3 Terminal Velocity (vertical fall with linear drag)

\[

mg - b vt = 0 \;\;\Longrightarrow\;\; vt = \frac{mg}{b}

\]

This follows directly from \(\Sigma F_y =0\) when the net force vanishes.

3.4 Variable‑Force Problems (Differential Approach)

When the net force depends on velocity or position, write the appropriate differential equation and integrate:

\[

m\frac{dv}{dt}=F{\text{net}}(v)\qquad\text{or}\qquad m v\frac{dv}{dx}=F{\text{net}}(x)

\]

Apply the limits that correspond to the physical situation (e.g. \(v=0\) at \(t=0\)).

4. Simple Experimental Determination of the Acceleration of Free Fall (\(g\))

Although the syllabus only requires a qualitative description, the following short procedure provides a quantitative estimate suitable for classroom work.

1. Set up a vertical metre‑stick (or a calibrated ruler) fixed to a wall.
2. Place a small dense ball at a known height (e.g. 1.0 m) and release it without any push.
3. Use a stopwatch to time the fall over a measured distance (preferably the whole 1 m). Repeat three times and take the average time \(t\).
4. Assuming the ball starts from rest, use the constant‑acceleration formula \(s = \tfrac12 g t^{2}\) to solve for \(g\):

\[

g = \frac{2s}{t^{2}}

\]

5. Compare the experimental value with the accepted 9.81 m s⁻² and discuss sources of error (reaction time, air‑drag, alignment of the ruler, etc.).

5. Problem‑Solving Strategy (Cambridge Dynamics – Topic 3)

1. Read the question carefully – underline given data and what is required.
2. Sketch a clear diagram – show the object, motion direction, distances and time intervals.
3. Draw a free‑body diagram – label every external force, choose a sign convention.
4. Write the vector equations:

   • \(\Sigma Fx = m ax\), \(\Sigma Fy = m ay\).
   • Include friction, linear drag, or any other variable force.
   • For short‑duration forces use \(\mathbf{J}= \Delta\mathbf{p}\).

5. Identify the type of motion:

   • Constant acceleration → use kinematic equations (\(v^2 = u^2 + 2as\), \(s = ut + \tfrac12 at^2\), …).
   • Variable acceleration → set up the differential equation and integrate.

6. Solve algebraically for the unknown(s). Keep track of vector directions throughout.
7. Check the answer:

   • Resultant‑force direction ↔ acceleration direction.
   • Units, significant figures, and physical plausibility.

8. State the final result with appropriate units and a brief interpretation.

6. Key Equations Summary

7. Common Misconceptions (and How to Avoid Them)

• “Force and acceleration are always upward/downward.” – They point in the direction of the *net* force, which may be horizontal, diagonal or opposite to an individual applied force.
• “Action–reaction forces cancel.” – They act on different bodies; each must be considered in the FBD of the body of interest.
• “Mass can change during motion.” – In classical mechanics the mass of a particle is constant; only velocity changes (unless dealing with rockets or relativistic problems, which are outside the 9702 syllabus).
• “Friction always opposes motion.” – Static friction opposes the *tendency* to move (up to \(\mus N\)); kinetic friction opposes *actual* motion with magnitude \(\muk N\).
• “Drag is always proportional to \(v^2\).” – For the AS/A‑Level syllabus the simple linear model \(\mathbf{F}_{\text{drag}} = -b\mathbf{v}\) is sufficient for low speeds; the quadratic model belongs to higher‑level studies.

8. Worked Examples

Example 1 – Constant Horizontal Force (No Friction)

Question: A 2.0 kg cart is pulled horizontally by a constant force of 10 N. It starts from rest. Find its speed after it has moved 5.0 m.

1. FBD: one horizontal force \(F=10\;\text{N}\) to the right; weight and normal cancel vertically.
2. Net force: \(\Sigma F_x = 10\;\text{N}\).
3. Acceleration: \(a = F/m = 10/2.0 = 5.0\;\text{m s}^{-2}\) (rightwards).
4. Use \(v^{2}=u^{2}+2as\) with \(u=0\), \(s=5.0\;\text{m}\):

   \[

   v = \sqrt{2(5.0)(5.0)} = \sqrt{50}=7.07\;\text{m s}^{-1}

   \]

5. Check: \(\Sigma F\) points right, so \(a\) and \(v\) are rightwards – consistent.

Example 2 – Sliding Block with Kinetic Friction

Question: A 3.0 kg block slides down a smooth 30° incline. The coefficient of kinetic friction is \(\mu_k = 0.15\). Determine the block’s acceleration.

1. Choose axes parallel (\(x\)) and perpendicular (\(y\)) to the incline.
2. Forces:

   • Weight component down the plane: \(W_{\parallel}=mg\sin30^{\circ}=3.0\times9.81\times0.5=14.7\;\text{N}\).
   • Normal reaction: \(N = mg\cos30^{\circ}=3.0\times9.81\times0.866=25.5\;\text{N}\).
   • Kinetic friction up the plane: \(fk = \muk N =0.15\times25.5=3.83\;\text{N}\).

3. Net force down the plane: \(\Sigma Fx = W{\parallel} - f_k = 14.7 - 3.83 = 10.9\;\text{N}\).
4. Acceleration: \(a = \Sigma F_x / m = 10.9 / 3.0 = 3.63\;\text{m s}^{-2}\) down the incline.
5. Direction check – net force and acceleration both point down the plane.

Example 3 – Variable Force: Linear Drag on a Falling Sphere

Question: A small sphere of mass 0.20 kg falls vertically under gravity with a linear drag force \(\mathbf{F}_{\text{drag}} = -b\mathbf{v}\) where \(b = 0.10\;\text{kg s}^{-1}\). Find the speed after 4 s, assuming it starts from rest.

1. Net force: \(mg - b v = m\,\dfrac{dv}{dt}\).
2. Rearrange: \(\dfrac{dv}{dt} + \dfrac{b}{m}v = g\).
3. This is a first‑order linear ODE. Solution for \(v(0)=0\):

   \[

   v(t) = \frac{mg}{b}\left(1 - e^{-\frac{b}{m}t}\right)

   \]

4. Insert numbers: \(\displaystyle \frac{mg}{b}= \frac{0.20\times9.81}{0.10}=19.62\;\text{m s}^{-1}\);

   \(\frac{b}{m}= \frac{0.10}{0.20}=0.5\;\text{s}^{-1}\).

5. Thus \(v(4) = 19.62\bigl(1-e^{-0.5\times4}\bigr)=19.62\bigl(1-e^{-2}\bigr)=19.62(1-0.135)=16.9\;\text{m s}^{-1}\).
6. Direction: downward, matching the net force direction.