| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Standard +0.3 This is a standard statics problem requiring resolution of forces in two directions and consideration of limiting cases. Part (a) is routine force resolution with given X; part (b) requires recognizing that minimum X occurs when one string goes slack (T=0), which is a common textbook scenario. The 60° angle simplifies calculations. Slightly above average due to the conceptual step in part (b), but well within standard M1 material. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Horizontal: \(100 - T_U\sin 60 - T_L\sin 60 = 0\); Vertical: \(T_U\cos 60 - T_L\cos 60 - 5g = 0\); Perp to \(T_U\): \(T_L\cos 30 + 5g\cos 30 = 100\cos 60\) | M1 | Resolve horizontally or vertically or perpendicular to the upper string to reach an equation. Correct number of terms. Allow \(X\) for 100 in horizontal equation. |
| Either horizontal and vertical equations correct or perpendicular correct. Must see \(X = 100\) used for A1. | A1 | |
| Solve for either \(T_L\) or \(T_U\) using equation(s) with no missing term. | M1 | May see \(T_U = 107.74\) |
| \(T_L = 7.74\) N | A1 | Allow 7.73 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Horizontal: \(X - T_{up}\sin 60 = 0\); Vertical: \(T_{up}\cos 60 - 5g = 0\); Perp to \(T_{up}\): \(5g\cos 30 = X\cos 60\) | M1 | Resolve either horizontally or vertically or perpendicular to the upper string. Must be using the tension \(T_{low} = 0\). Equivalent to Lami as: \(\dfrac{5g}{\sin 150} = \dfrac{X}{\sin 120}\left(= \dfrac{T_{up}}{\sin 90}\right)\) |
| Either horizontal and vertical equations correct or perpendicular correct. | A1 | |
| Eliminate \(T_{up}\) and/or solve for \(X\) | M1 | \(T_{up} = 100\) |
| Least value of \(X = 86.6\) | A1 | Allow \(X = 50\sqrt{3}\) |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Horizontal: $100 - T_U\sin 60 - T_L\sin 60 = 0$; Vertical: $T_U\cos 60 - T_L\cos 60 - 5g = 0$; Perp to $T_U$: $T_L\cos 30 + 5g\cos 30 = 100\cos 60$ | M1 | Resolve horizontally or vertically or perpendicular to the upper string to reach an equation. Correct number of terms. Allow $X$ for 100 in horizontal equation. |
| Either horizontal and vertical equations correct or perpendicular correct. Must see $X = 100$ used for A1. | A1 | |
| Solve for either $T_L$ or $T_U$ using equation(s) with no missing term. | M1 | May see $T_U = 107.74$ |
| $T_L = 7.74$ N | A1 | Allow 7.73 |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Horizontal: $X - T_{up}\sin 60 = 0$; Vertical: $T_{up}\cos 60 - 5g = 0$; Perp to $T_{up}$: $5g\cos 30 = X\cos 60$ | M1 | Resolve either horizontally or vertically or perpendicular to the upper string. Must be using the tension $T_{low} = 0$. Equivalent to Lami as: $\dfrac{5g}{\sin 150} = \dfrac{X}{\sin 120}\left(= \dfrac{T_{up}}{\sin 90}\right)$ |
| Either horizontal and vertical equations correct or perpendicular correct. | A1 | |
| Eliminate $T_{up}$ and/or solve for $X$ | M1 | $T_{up} = 100$ |
| Least value of $X = 86.6$ | A1 | Allow $X = 50\sqrt{3}$ |6\\
\includegraphics[max width=\textwidth, alt={}, center]{083d3e44-1e42-461f-aa8d-a1a22047a47e-08_412_588_260_776}
A block of mass 5 kg is held in equilibrium near a vertical wall by two light strings and a horizontal force of magnitude $X \mathrm {~N}$, as shown in the diagram. The two strings are both inclined at $60 ^ { \circ }$ to the vertical.
\begin{enumerate}[label=(\alph*)]
\item Given that $X = 100$, find the tension in the lower string.
\item Find the least value of $X$ for which the block remains in equilibrium in the position shown. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q6 [8]}}