Solution :-

Part (a): Finding the tension in AB

Using the equilibrium of forces for particle A:

Method 1: Resolving forces

\(T_{AB} \sin 45^\circ = 0.2g \cos 45^\circ\)

OR \( \tan 45^\circ = \frac{T_{AB}}{0.2g} \)

OR \( \tan 45^\circ = \frac{0.2g}{T_{AB}} \)

OR \( T_{AB} = T_{AP} \cos 45^\circ \) and \( T_{AP} \sin 45^\circ = 0.2g \)

\(T_{AB} = 2 \text{ N}\)

Method 2: Using Lami’s Theorem

\(\frac{T_{AB}}{\sin 135^\circ} = \frac{0.2g}{\sin 135^\circ}\)

\(T_{AB} = 2 \text{ N}\)

Note: Condone \(T_{AB} = 0.2g\) (with or without working) for full marks. DO NOT ISW (Ignore Subsequent Working).

Part (b): Finding θ and the tension in BQ

Method 1: Resolving forces horizontally and vertically

\(T_{BQ} \cos \theta – T_{AB} = 0\)

\(T_{BQ} \sin \theta – 0.1g = 0\)

\(\theta = \tan^{-1} \left( \frac{0.1g}{\text{their } T_{AB}} \right)\) or \(T_{BQ} = \sqrt{(0.1g)^2 + (\text{their } T_{AB})^2}\)

\(\theta = 26.6^\circ\) AND \(T_{BQ} = \sqrt{5} \text{ N} \text{ or } 2.24 \text{ N}\)

Method 2: Resolving parallel and perpendicular to BQ

\(T_{BQ} = (\text{their } T_{AB}) \cos \theta + 0.1g \sin \theta\)

\(\text{their } T_{AB}) \sin \theta = 0.1g \cos \theta \)

\(\theta = \tan^{-1} \left( \frac{0.1g}{\text{their } T_{AB}} \right)\)

\(\theta = 26.6^\circ\) AND \(T_{BQ} = \sqrt{5} \text{ N} \text{ or } 2.24 \text{ N}\)

Method 3: Using Triangle of Forces

\(0.1g = T_{BQ} \sin \theta\)

OR \((their T_{AB}) = T_{BQ} \cos \theta\)

\(\theta = \tan^{-1} \left( \frac{0.1g}{\text{their } T_{AB}} \right)\)

\(T_{BQ} = \sqrt{(0.1g)^2 + (\text{their } T_{AB})^2}\)

\(\theta = 26.6^\circ\) AND \(T_{BQ} = \sqrt{5} \text{ N} \text{ or } 2.24 \text{ N}\)

Method 4: Using Lami’s Theorem

\(\frac{T_{BQ}}{\sin 90^\circ} = \frac{0.1g}{\sin (180^\circ – \theta)} = \frac{\text{their } T_{AB}}{\sin (90^\circ + \theta)}\)

OR \(\frac{T_{BQ}}{\sin 90^\circ} = \frac{0.1g}{\sin \theta} = \frac{\text{their } T_{AB}}{\cos \theta}\)

\(\theta = \tan^{-1} \left( \frac{0.1g}{\text{their } T_{AB}} \right)\)

\(\theta = 26.6^\circ\) AND \(T_{BQ} = \sqrt{5} \text{ N} \text{ or } 2.24 \text{ N}\)

Method 5: Resolving on the Whole System

\(T_{BQ} \sin \theta + (\text{their } T_{AP}) \sin 45^\circ – 0.2g – 0.1g = 0\)

\(T_{BQ} \cos \theta – (\text{their } T_{AP}) \cos 45^\circ = 0\)

\(\theta = \tan^{-1} \left( \frac{0.1g + 0.2g – (\text{their } T_{AP}) \sin 45^\circ}{(\text{their } T_{AP}) \cos 45^\circ} \right)\)

\(T_{BQ} = \sqrt{(0.1g + 0.2g – (\text{their } T_{AP}) \sin 45^\circ)^2 + ((\text{their } T_{AP}) \cos 45^\circ)^2}\)

\(\theta = 26.6^\circ\) AND \(T_{BQ} = \sqrt{5} \text{ N} \text{ or } 2.24 \text{ N}\)

Note:

• For resolving either horizontally OR vertically: 2 terms; allow sin/cos mix; allow their \(T_{AB}\); MO for any use of \(T_{AB} = T_{BQ} = T_{AP}\).
• For both. FT their \(T_{AB}\) ONLY: \(T_{AB} \ne T_{BQ}\). Sight of \((\text{their } T_{AB}) \tan \theta = 0.1g\) is M1 only without seeing an equation for \(T_{BQ}\).
• Solve for θ (or solve for \(T_{BQ}\)) from equations with the correct number of relevant terms. Using their \(T_{AB}\).
• \(\theta = 26.56505…\), \(T_{BQ} = 2.236067…\). AWRT 26.6 and 2.24.