Moderate -0.8 This is a straightforward statics problem requiring resolution of forces in two directions (or using triangle of forces) with standard angles. The setup is clear, only three forces act on the particle, and it's a direct application of equilibrium conditions with no conceptual subtlety—easier than average A-level mechanics questions.
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\includegraphics[max width=\textwidth, alt={}, center]{c3246fbe-6f77-48f7-98eb-19e9166008bc-03_721_622_296_724}
A particle of mass 0.2 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point on a vertical wall. The particle is held in equilibrium by a force of magnitude \(X \mathrm {~N}\), perpendicular to the string, with the string taut and making an angle of \(30 ^ { \circ }\) with the wall (see diagram).
Find the tension in the string and the value of \(X\).
First M1 for any two fractions. Second M1 for all three fractions or another pair. Allow \(\dfrac{X}{\sin120}\) and \(\dfrac{T}{\sin150}\) for M1 marks.
\(X = 1\), Tension \(= 1.73\) N or \(\sqrt{3}\) N
(A1)
For both.
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## Question 2:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X\cos30 - T\sin30 = 0$ | M1 | Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. |
| $X\sin30 + T\cos30 - 0.2g = 0$ | M1 | Must have correct number of relevant terms. Allow sin/cos mix but must be consistent with their other equation. Allow sign errors. |
| $X = 1$, Tension $= 1.73$ N $[1.7320..]$ or $\sqrt{3}$ N | A1 | For both. |
**Alternative: Resolving in directions of X and T**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X - 0.2g\cos60 = 0$ | (M1) | Must have correct number of relevant terms. Allow sin/cos mix. Allow sign errors. |
| $T - 0.2g\sin60 = 0$ | (M1) | Must have correct number of relevant terms. Allow sin/cos mix but must be consistent with their other equation. Allow sign errors. |
| $X = 1$, Tension $= 1.73$ N $[1.7320..]$ or $\sqrt{3}$ N | (A1) | For both. |
**Alternative: Lami's theorem**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\dfrac{0.2g}{\sin90} = \dfrac{X}{\sin150} = \dfrac{T}{\sin120}$ | (M1M1) | First M1 for any two fractions. Second M1 for all three fractions or another pair. Allow $\dfrac{X}{\sin120}$ and $\dfrac{T}{\sin150}$ for M1 marks. |
| $X = 1$, Tension $= 1.73$ N or $\sqrt{3}$ N | (A1) | For both. |
| | **3** | |
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\includegraphics[max width=\textwidth, alt={}, center]{c3246fbe-6f77-48f7-98eb-19e9166008bc-03_721_622_296_724}
A particle of mass 0.2 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point on a vertical wall. The particle is held in equilibrium by a force of magnitude $X \mathrm {~N}$, perpendicular to the string, with the string taut and making an angle of $30 ^ { \circ }$ with the wall (see diagram).
Find the tension in the string and the value of $X$.\\
\hfill \mbox{\textit{CAIE M1 2024 Q2 [3]}}