| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle with string at angle to wall |
| Difficulty | Moderate -0.8 This is a straightforward statics problem requiring resolution of forces in two perpendicular directions with three forces (tension, weight, horizontal force). Part (a) involves solving two simultaneous equations, and part (b) is essentially the same process with different given values. This is a standard textbook exercise testing basic equilibrium concepts with no novel problem-solving required, making it easier than average. |
| Spec | 3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to resolve in one direction and form equation | M1 | Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. If only one equation shown and it involves 32, it must be 32, not \(P\). |
| \(T\sin\theta = 32\) and \(T\cos\theta = 80\), or \(0 = 80\sin\theta - 32\cos\theta\) and \(T = 80\cos\theta + 32\sin\theta\) | A1 | For both horizontal and vertical, or both parallel and perpendicular. |
| Attempt to solve for \(T\) or \(\theta\) | M1 | Must get to \(T\) or \(\theta\); e.g. \(T = \sqrt{32^2 + 80^2}\) or \(\theta = \tan^{-1}\left(\frac{32}{80}\right)\). Condone e.g. \(\theta = \tan^{-1}\left(\frac{80}{32}\right)\). Must come from equations with correct number of relevant terms. |
| \(T = 86.2\) [N 86.1626...] or \(16\sqrt{29}\) or \(\sqrt{7424}\) and \(\theta = 21.8°\) [21.801...] | A1 | For both. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T^2 = 80^2 + 32^2[-2\times80\times32\cos90]\) or \(T\sin\theta = 32\) or \(T\cos\theta = 80\) | M1 | For any of the five; allow sign errors. |
| \(80\tan\theta = 32\) or \(T = 80\cos\theta + 32\sin\theta\) | A1 | For any two equations. |
| Attempt to solve for \(T\) or \(\theta\) | M1 | Must get to \(T\) or \(\theta\); e.g. \(T = \sqrt{32^2 + 80^2}\) or \(\theta = \tan^{-1}\left(\frac{32}{80}\right)\). |
| \(T = 86.2\) [N 86.1626...] or \(16\sqrt{29}\) or \(\sqrt{7424}\) and \(\theta = 21.8°\) [21.801...] | A1 | For both. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{T}{\sin90} = \frac{32}{\sin(\theta)} = \frac{80}{\sin(90-\theta)}\) | M1 | For any two. |
| — | A1 | For all three. |
| Attempt to solve for \(T\) or \(\theta\) | M1 | e.g. \(\theta = \tan^{-1}\left(\frac{32}{80}\right)\). |
| \(T = 86.2\) [N 86.1626...] or \(16\sqrt{29}\) or \(\sqrt{7424}\) and \(\theta = 21.8°\) [21.801...] | A1 | For both. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to resolve in one direction and form equation | M1 | Must have correct number of relevant terms. Allow sin/cos mix. Allow sign errors. Must use 120, not \(T\). |
| \(120\sin\theta = P\) and \(120\cos\theta = 80\), or \(0 = 80\sin\theta - P\cos\theta\) and \(120 = 80\cos\theta + P\sin\theta\) | A1 | For both horizontal and vertical, or both parallel and perpendicular. |
| Attempt to solve for \(P\) or \(\theta\) | M1 | Must get to \(P\) or \(\theta\); e.g. \(P = \sqrt{120^2 - 80^2}\) or \(\theta = \cos^{-1}\left(\frac{80}{120}\right)\). Must come from equations with correct number of relevant terms. |
| \(P = 89.4\) [89.4427] or \(40\sqrt{5}\) or \(\sqrt{8000}\), \(\theta = 48.2°\) [48.1896...] | A1 | For both; allow \(P = 89.5\) (from \(120\sin48.2\)). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{120}{\sin90} = \frac{P}{\sin(\theta)} = \frac{80}{\sin(90-\theta)}\) | M1 | For any two. |
| — | A1 | For all three. |
| Attempt to solve for \(P\) or \(\theta\) | M1 | e.g. \(\theta = 90 - \sin^{-1}\left(\frac{80}{120}\right)\) or \(\theta = \cos^{-1}\left(\frac{80}{120}\right)\). |
| \(P = 89.4\) [89.4427] or \(40\sqrt{5}\) or \(\sqrt{8000}\), \(\theta = 48.2°\) [48.1896...] | A1 | For both; allow \(P = 89.5\) (from \(120\sin48.2\)). |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to resolve in one direction and form equation | M1 | Must have correct number of relevant terms (forces must have components as required). Allow sin/cos mix. Allow sign errors. If only one equation shown and it involves 32, it must be 32, not $P$. |
| $T\sin\theta = 32$ and $T\cos\theta = 80$, or $0 = 80\sin\theta - 32\cos\theta$ and $T = 80\cos\theta + 32\sin\theta$ | A1 | For both horizontal and vertical, or both parallel and perpendicular. |
| Attempt to solve for $T$ or $\theta$ | M1 | Must get to $T$ or $\theta$; e.g. $T = \sqrt{32^2 + 80^2}$ or $\theta = \tan^{-1}\left(\frac{32}{80}\right)$. Condone e.g. $\theta = \tan^{-1}\left(\frac{80}{32}\right)$. Must come from equations with correct number of relevant terms. |
| $T = 86.2$ [N 86.1626...] or $16\sqrt{29}$ or $\sqrt{7424}$ and $\theta = 21.8°$ [21.801...] | A1 | For both. |
**Alternative method using triangle of forces:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T^2 = 80^2 + 32^2[-2\times80\times32\cos90]$ or $T\sin\theta = 32$ or $T\cos\theta = 80$ | M1 | For any of the five; allow sign errors. |
| $80\tan\theta = 32$ or $T = 80\cos\theta + 32\sin\theta$ | A1 | For any two equations. |
| Attempt to solve for $T$ or $\theta$ | M1 | Must get to $T$ or $\theta$; e.g. $T = \sqrt{32^2 + 80^2}$ or $\theta = \tan^{-1}\left(\frac{32}{80}\right)$. |
| $T = 86.2$ [N 86.1626...] or $16\sqrt{29}$ or $\sqrt{7424}$ and $\theta = 21.8°$ [21.801...] | A1 | For both. |
**Alternative Triangle of forces method using sine rule:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{T}{\sin90} = \frac{32}{\sin(\theta)} = \frac{80}{\sin(90-\theta)}$ | M1 | For any two. |
| — | A1 | For all three. |
| Attempt to solve for $T$ or $\theta$ | M1 | e.g. $\theta = \tan^{-1}\left(\frac{32}{80}\right)$. |
| $T = 86.2$ [N 86.1626...] or $16\sqrt{29}$ or $\sqrt{7424}$ and $\theta = 21.8°$ [21.801...] | A1 | For both. |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to resolve in one direction and form equation | M1 | Must have correct number of relevant terms. Allow sin/cos mix. Allow sign errors. Must use 120, not $T$. |
| $120\sin\theta = P$ and $120\cos\theta = 80$, or $0 = 80\sin\theta - P\cos\theta$ and $120 = 80\cos\theta + P\sin\theta$ | A1 | For both horizontal and vertical, or both parallel and perpendicular. |
| Attempt to solve for $P$ or $\theta$ | M1 | Must get to $P$ or $\theta$; e.g. $P = \sqrt{120^2 - 80^2}$ or $\theta = \cos^{-1}\left(\frac{80}{120}\right)$. Must come from equations with correct number of relevant terms. |
| $P = 89.4$ [89.4427] or $40\sqrt{5}$ or $\sqrt{8000}$, $\theta = 48.2°$ [48.1896...] | A1 | For both; allow $P = 89.5$ (from $120\sin48.2$). |
**Alternative Triangle of forces method using sine rule:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{120}{\sin90} = \frac{P}{\sin(\theta)} = \frac{80}{\sin(90-\theta)}$ | M1 | For any two. |
| — | A1 | For all three. |
| Attempt to solve for $P$ or $\theta$ | M1 | e.g. $\theta = 90 - \sin^{-1}\left(\frac{80}{120}\right)$ or $\theta = \cos^{-1}\left(\frac{80}{120}\right)$. |
| $P = 89.4$ [89.4427] or $40\sqrt{5}$ or $\sqrt{8000}$, $\theta = 48.2°$ [48.1896...] | A1 | For both; allow $P = 89.5$ (from $120\sin48.2$). |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{99f20949-471d-4da3-a680-ec24abf6baa5-06_438_463_264_840}
A light string $A B$ is fixed at $A$ and has a particle of weight 80 N attached at $B$. A horizontal force of magnitude $P \mathrm {~N}$ is applied at $B$ such that the string makes an angle $\theta ^ { \circ }$ to the vertical (see diagram).
\begin{enumerate}[label=(\alph*)]
\item It is given that $P = 32$ and the system is in equilibrium.
Find the tension in the string and the value of $\theta$.
\item It is given instead that the tension in the string is 120 N and that the particle attached at $B$ still has weight 80 N .
Find the value of $P$ and the value of $\theta$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q5 [8]}}