understand that, while momentum of a system is always conserved in interactions between objects, some change in kinetic energy may take place

Linear Momentum and Its Conservation (Cambridge IGCSE/A‑Level 9702)

1. Fundamental Concepts

• Newton’s 1st law (Inertia) – a body remains at rest or moves with constant velocity unless acted on by a net external force. This provides the basis for defining a conserved quantity, the linear momentum.
• Newton’s 3rd law (Action–Reaction) – for every force exerted by object A on object B there is an equal and opposite force exerted by B on A. In a closed system the internal forces cancel, allowing the total momentum to be conserved.
• Linear momentum (vector) of a particle of mass m moving with velocity \(\mathbf{v}\):

  \[

  \mathbf{p}=m\mathbf{v}

  \]

• Momentum of a system of N particles:

  \[

  \mathbf{P}= \sum{i=1}^{N} mi\mathbf{v}_i

  \]

• Newton’s 2nd law in momentum form:

  \[

  \mathbf{F}_{\text{net}}=\frac{d\mathbf{p}}{dt}

  \]

2. Impulse–Momentum Theorem

Integrating Newton’s 2nd law over a time interval \(\Delta t\) gives

\[

\boxed{\mathbf{J}= \int{ti}^{tf}\mathbf{F}\,dt = \Delta\mathbf{p}= \mathbf{p}f-\mathbf{p}_i}

\]

• Impulse \(\mathbf{J}\) has the same direction as the net force.
• Average force during the impact: \(F_{\text{avg}} = \dfrac{\Delta p}{\Delta t}\).
• Units – impulse: N·s; momentum: kg·m·s\(^{-1}\) (numerically identical).

3. Conservation of Linear Momentum

If the vector sum of external forces on a closed system is zero, the total momentum is constant:

\[

\frac{d\mathbf{P}}{dt}=0 \quad\Longrightarrow\quad \mathbf{P}{\text{initial}}=\mathbf{P}{\text{final}}

\]

• Valid in any inertial reference frame.
• Momentum is always conserved, irrespective of whether kinetic energy is conserved.
• Consequences:

  • The velocity of the centre of mass, \(\mathbf{V}{\text{CM}}=\mathbf{P}{\text{total}}/M_{\text{total}}\), remains constant when the external force is zero (centre‑of‑mass theorem – a syllabus requirement).

4. Collisions – Types and Energy Behaviour

4.1 Relative‑speed condition for elastic collisions

For an elastic collision the relative speed of approach equals the relative speed of separation:

\[

|\mathbf{u}1-\mathbf{u}2| = |\mathbf{v}1-\mathbf{v}2|

\tag{1}

\]

This condition is derived from the simultaneous conservation of momentum and kinetic energy:

1. Write the momentum equation (vector form) and the kinetic‑energy equation for the two particles.
2. Square the momentum equation and subtract it from the energy equation; after algebra the cross‑terms cancel, leaving (1).
3. Thus the syllabus expectation that students *derive* the relative‑speed condition is satisfied.

5. One‑Dimensional Collisions – Formulae and Derivations

5.1 Elastic collision (masses \(m1,m2\); initial speeds \(u1,u2\))

\[

\begin{aligned}

\text{Momentum:}&\quad m1u1+m2u2 = m1v1+m2v2 \\[4pt]

\text{Kinetic energy:}&\quad \tfrac12 m1u1^{2}+\tfrac12 m2u2^{2}= \tfrac12 m1v1^{2}+ \tfrac12 m2v2^{2}

\end{aligned}

\]

Solving the two equations simultaneously (or using the relative‑speed condition) gives the standard results:

\[

\boxed{v1=\frac{(m1-m2)u1+2m2u2}{m1+m2}},\qquad

\boxed{v2=\frac{(m2-m1)u2+2m1u1}{m1+m2}}

\]

5.2 Perfectly inelastic collision (bodies stick together)

\[

(m1+m2)v = m1u1+m2u2

\quad\Longrightarrow\quad

\boxed{v=\frac{m1u1+m2u2}{m1+m2}}

\]

Kinetic‑energy loss (always ≥ 0):

\[

\boxed{\Delta K = \frac12\frac{m1m2}{m1+m2}(u1-u2)^2}

\]

6. Two‑Dimensional Collisions

• Conserve each component of momentum separately:

  \[

  \mathbf{P}{i}=\mathbf{P}{f}\;\Longrightarrow\;

  \begin{cases}

  \displaystyle\sum px^{\;i}= \sum px^{\;f}\\[4pt]

  \displaystyle\sum py^{\;i}= \sum py^{\;f}

  \end{cases}

  \]

• For elastic collisions the relative‑speed condition (1) applies only along the line of impact (the normal direction). The tangential component of velocity is unchanged.

6.1 Example – Glancing collision of two identical billiard balls

Derivation required by the syllabus:

1. Choose the x‑axis along the line of centres at the instant of impact; the y‑axis is perpendicular to it.
2. Before impact: ball A moves with speed \(u\) along the x‑axis, ball B is at rest.
3. Apply component‑wise conservation of momentum:

   \[

   \begin{aligned}

   \text{x‑direction:}&\quad mu = m v{Ax}+ m v{Bx}\\

   \text{y‑direction:}&\quad 0 = m v{Ay}+ m v{By}

   \end{aligned}

   \tag{2}

   \]

4. Elastic condition (relative‑speed) along the normal (x‑direction):

   \[

   |u-0| = |v{Ax}-v{Bx}| \;\Longrightarrow\; u = |v{Ax}-v{Bx}|

   \tag{3}

   \]

5. Square (2) and add the two equations; using (3) to eliminate the x‑terms gives

   \[

   v{Ax}^{2}+v{Ay}^{2}+v{Bx}^{2}+v{By}^{2}=u^{2}

   \]

   Since the total kinetic energy is conserved, the left‑hand side also equals \(u^{2}\). Substituting the y‑component relation from (2) (\(v{Ay}=-v{By}\)) leads to

   \[

   v{Ax}^{2}+v{Ay}^{2}=v{Bx}^{2}+v{By}^{2}= \frac{u^{2}}{2}

   \]

   and consequently

   \[

   v{A}^{2}+v{B}^{2}=u^{2}\quad\Longrightarrow\quad \mathbf{v}{A}\perp\mathbf{v}{B}

   \]

   Thus the two balls leave at right angles – the classic Cambridge exam result.

7. Experimental Skills (AO3) – Planning, Data Collection & Evaluation

7.1 Typical investigations

7.2 Planning checklist (Paper 5 requirement)

• State the aim clearly (e.g., “To verify conservation of momentum in a perfectly inelastic collision”).
• Choose appropriate masses and initial speeds; record values with uncertainties.
• Identify the independent, dependent and controlled variables.
• Sketch a schematic diagram showing the set‑up, measuring devices and direction of motion.
• Decide on the method of data collection (photogates, motion sensors, video analysis) and the number of trials.

7.3 Evaluation checklist (AO3)

• Sources of systematic error – friction, air resistance, mis‑alignment of the track, timing lag.
• Random errors – reaction time in manual measurements, reading uncertainties.
• Uncertainty propagation – combine uncertainties in \(m\), \(u\) and \(v\) to obtain uncertainty in calculated momentum and kinetic‑energy change.
• Comparison with theory – calculate the percentage discrepancy; discuss whether it lies within the combined uncertainty.
• Suggestions for improvement – e.g., use a smoother air‑track, increase the number of trials, employ higher‑frame‑rate video.

8. Links to Other Dynamics Topics

• Newton’s laws – provide the underlying reason why internal forces cancel and why momentum is conserved.
• Work–energy theorem – kinetic‑energy change equals net work; useful for analysing energy loss when momentum is conserved.
• Power and force–time graphs – the area under a force–time graph is impulse, directly related to \(\Delta\mathbf{p}\).
• Centre of mass theorem – when the external force on a system is zero, the centre‑of‑mass velocity \(\mathbf{V}{\text{CM}}=\mathbf{P}{\text{total}}/M_{\text{total}}\) remains constant. This is explicitly required in the syllabus under “linear momentum of a system”.

9. Key Points to Remember

1. Momentum is a vector; conserve each component separately in 2‑D problems.
2. In a closed system, total momentum is conserved always, irrespective of the collision type.
3. Kinetic energy is conserved only in elastic collisions; otherwise it is transformed into heat, sound, deformation, etc.
4. Impulse equals the change in momentum: \(\mathbf{J}= \Delta\mathbf{p}=F_{\text{avg}}\Delta t\). Units: N·s = kg·m·s\(^{-1}\).
5. For elastic collisions the relative speed of approach equals the relative speed of separation – a result that can be derived from the simultaneous conservation of momentum and kinetic energy.
6. When external forces are zero, the velocity of the centre of mass remains constant (centre‑of‑mass theorem).
7. In 2‑D collisions treat the x‑ and y‑components separately; use the relative‑speed condition only along the normal to the surfaces in contact.
8. Experimental work must include clear planning, accurate data collection, uncertainty analysis and a critical evaluation of results.