| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| 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 two-particle equilibrium problem requiring resolution of forces in two directions for each particle. While it involves multiple steps (finding tension in AB, then θ and tension in BQ), the method is straightforward: resolve vertically and horizontally for each particle, then solve simultaneous equations. The 45° angle simplifies calculations. This is slightly easier than average as it's a textbook application of equilibrium with no novel insight required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T_{AB}\sin 45=0.2g\cos 45\) OR \(\tan 45=\frac{T_{AB}}{0.2g}\) OR \(\tan 45=\frac{0.2g}{T_{AB}}\) OR \(T_{AB}=T_{AP}\cos 45\) and \(T_{AP}\sin 45=0.2g\) | *B1 | BOD for using \(\sin 45\) instead of \(\cos 45\), particularly if using wrong components in (b). |
| \(T_{AB}=2\ \text{N}\) | DB1 | Condone \(T_{AB}=0.2g\) (with or without working) for full marks. WWW. DO NOT ISW. |
| Alternative using Lami's Theorem: | ||
| \(\frac{T_{AB}}{\sin 135}=\frac{0.2g}{\sin 135}\) | B1 | |
| \(T_{AB}=2\ \text{N}\) | B1 | Condone \(T_{AB}=0.2g\). WWW. DO NOT ISW. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T_{BQ}\cos\theta - \text{their }T_{AB}=0\) and \(T_{BQ}\sin\theta-0.1g=0\) | *M1 | For resolving either horizontally OR vertically; 2 terms; allow sin/cos mix; allow their \(T_{AB}\). M0 for any use of \(T_{AB}=T_{BQ}=T_{AP}\). |
| Both equations correct | A1FT | FT their \(T_{AB}\) ONLY; \(T_{AB}\neq T_{BQ}\). Sight of \((\text{their }T_{AB})\tan\theta=0.1g\) is M1 only without seeing an equation for \(T_{BQ}\). |
| \(\theta=\tan^{-1}\!\left(\frac{0.1g}{\text{their }T_{AB}}\right)\) or \(T_{BQ}=\sqrt{(0.1g)^2+(\text{their }T_{AB})^2}\) | DM1 | Solve for \(\theta\) or solve for \(T_{BQ}\) from equations with the correct number of relevant terms. Using their \(T_{AB}\). |
| \(\theta=26.6\) AND \(T_{BQ}=\sqrt{5}\) or \(2.24\) | A1 | \(\theta=26.56505\ldots\), \(T_{BQ}=2.236067\ldots\). AWRT 26.6 and 2.24. |
| First Alternative for 4(b): | ||
| \(T_{BQ}=(\text{their }T_{AB})\cos\theta+0.1g\sin\theta\) and \((\text{their }T_{AB})\sin\theta=0.1g\cos\theta\) | M1 | For resolving either parallel or perpendicular to \(BQ\); 2 terms; allow sign errors on 3-term equation only; allow sin/cos mix; allow their \(T_{AB}\). Sight of \((\text{their }T_{AB})\tan\theta=0.1g\) is M1 only. M0 for any use of \(T_{AB}=T_{BQ}=T_{AP}\). |
| Both correct | A1FT | FT their \(T_{AB}\) ONLY; \(T_{AB}\neq T_{BQ}\). |
| \(\theta=\tan^{-1}\!\left(\frac{0.1g}{\text{their }T_{AB}}\right)\) | DM1 | Solve for \(\theta\) (or \(T_{BQ}\)) from equations with correct number of relevant terms. Using their \(T_{AB}\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T_{BQ}^2 = (0.1g)^2 + (T_{AB})^2\) OR \(0.1g = T_{BQ}\sin(\theta)\) OR \((T_{AB}) = T_{BQ}\cos(\theta)\) | M1 | Using their \(T_{AB}\) ONLY; M0 for any use of \(T_{AB} = T_{BQ} = T_{AP}\) |
| For any 2 equations. FT their \(T_{AB}\) ONLY | A1FT | \(T_{AB} \neq T_{BQ}\). Sight of \((T_{AB})\tan\theta = 0.1g\) is M1 only without seeing an equation for \(T_{BQ}\) |
| Solve for \(T_{BQ}\) or \(\theta\): \(T_{BQ} = \sqrt{(0.1g)^2 + (T_{AB})^2}\), \(\theta = \tan^{-1}\left(\frac{0.1g}{T_{AB}}\right)\) | DM1 | Solve for \(\theta\) or solve for \(T_{BQ}\) from equations with correct number of relevant terms. Using their \(T_{AB}\) |
| \(\theta = 26.6\) AND \(T_{BQ} = \sqrt{5}\) or \(2.24\) | A1 | \(\theta = 26.56505...\), \(T_{BQ} = 2.236067...\). AWRT 26.6 and 2.24 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{T_{BQ}}{\sin 90} = \frac{0.1g}{\sin(180-\theta)} = \frac{T_{AB}}{\sin(90+\theta)}\) OR \(\frac{T_{BQ}}{\sin 90} = \frac{0.1g}{\sin\theta} = \frac{T_{AB}}{\cos\theta}\) | *M1 | For any 2 fractions correct; M0 for any use of \(T_{AB} = T_{BQ} = T_{AP}\) |
| All 3 fractions correct. FT their \(T_{AB}\) ONLY | A1FT | Sight of \((T_{AB})\tan\theta = 0.1g\) is M1 only without seeing an equation for \(T_{BQ}\) |
| \(\theta = \tan^{-1}\left(\frac{0.1g}{T_{AB}}\right)\) | DM1 | Solve for \(\theta\) |
| \(\theta = 26.6\) AND \(T_{BQ} = \sqrt{5}\) or \(2.24\) | A1 | AWRT 26.6 and 2.24 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T_{BQ}\sin\theta + (T_{AP})\sin 45 = 0.2g + 0.1g\) and \(T_{BQ}\cos\theta = (T_{AP})\cos 45\) | M1 | For resolving either horizontally or vertically on the whole system using their \(T_{AP}\). M0 for \(T_{AB} = T_{BQ} = T_{AP}\). If \(T_{AP}\) not found in part (a), must have \(T_{AP}\sin 45 = 0.2g\) or \((T_{AB}) = T_{AP}\cos 45\). Correct number of terms; allow sign errors on 4-term equation ONLY; allow sin/cos mix |
| Both equations correct. May see \(T_{BQ}\sin\theta = 1\) and \(T_{BQ}\cos\theta = 2\) | A1FT | Sight of \((T_{AB})\tan\theta = 0.1g\) is M1 only without seeing equation for \(T_{BQ}\) |
| \(\theta = \tan^{-1}\left(\frac{0.1g + 0.2g - (T_{AP})\sin 45}{(T_{AP})\cos 45}\right)\) or \(T_{BQ} = \sqrt{(0.1g + 0.2g - (T_{AP})\sin 45)^2 + ((T_{AP})\cos 45)^2}\) | DM1 | Solve for \(\theta\) or solve for \(T_{BQ}\) from equations with correct number of relevant terms. Using their \(T_{AP}\) |
| \(\theta = 26.6\) AND \(T_{BQ} = \sqrt{5}\) or \(2.24\) | A1 | AWRT 26.6 and 2.24 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T_{AB}\sin 45=0.2g\cos 45$ OR $\tan 45=\frac{T_{AB}}{0.2g}$ OR $\tan 45=\frac{0.2g}{T_{AB}}$ OR $T_{AB}=T_{AP}\cos 45$ and $T_{AP}\sin 45=0.2g$ | *B1 | BOD for using $\sin 45$ instead of $\cos 45$, particularly if using wrong components in (b). |
| $T_{AB}=2\ \text{N}$ | DB1 | Condone $T_{AB}=0.2g$ (with or without working) for full marks. WWW. DO NOT ISW. |
| **Alternative using Lami's Theorem:** | | |
| $\frac{T_{AB}}{\sin 135}=\frac{0.2g}{\sin 135}$ | B1 | |
| $T_{AB}=2\ \text{N}$ | B1 | Condone $T_{AB}=0.2g$. WWW. DO NOT ISW. |
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T_{BQ}\cos\theta - \text{their }T_{AB}=0$ and $T_{BQ}\sin\theta-0.1g=0$ | *M1 | For resolving either horizontally OR vertically; 2 terms; allow sin/cos mix; allow their $T_{AB}$. M0 for any use of $T_{AB}=T_{BQ}=T_{AP}$. |
| Both equations correct | A1FT | FT their $T_{AB}$ ONLY; $T_{AB}\neq T_{BQ}$. Sight of $(\text{their }T_{AB})\tan\theta=0.1g$ is M1 only without seeing an equation for $T_{BQ}$. |
| $\theta=\tan^{-1}\!\left(\frac{0.1g}{\text{their }T_{AB}}\right)$ or $T_{BQ}=\sqrt{(0.1g)^2+(\text{their }T_{AB})^2}$ | DM1 | Solve for $\theta$ or solve for $T_{BQ}$ from equations with the correct number of relevant terms. Using their $T_{AB}$. |
| $\theta=26.6$ AND $T_{BQ}=\sqrt{5}$ or $2.24$ | A1 | $\theta=26.56505\ldots$, $T_{BQ}=2.236067\ldots$. AWRT 26.6 and 2.24. |
| **First Alternative for 4(b):** | | |
| $T_{BQ}=(\text{their }T_{AB})\cos\theta+0.1g\sin\theta$ and $(\text{their }T_{AB})\sin\theta=0.1g\cos\theta$ | M1 | For resolving either parallel or perpendicular to $BQ$; 2 terms; allow sign errors on 3-term equation only; allow sin/cos mix; allow their $T_{AB}$. Sight of $(\text{their }T_{AB})\tan\theta=0.1g$ is M1 only. M0 for any use of $T_{AB}=T_{BQ}=T_{AP}$. |
| Both correct | A1FT | FT their $T_{AB}$ ONLY; $T_{AB}\neq T_{BQ}$. |
| $\theta=\tan^{-1}\!\left(\frac{0.1g}{\text{their }T_{AB}}\right)$ | DM1 | Solve for $\theta$ (or $T_{BQ}$) from equations with correct number of relevant terms. Using their $T_{AB}$. |
## Question 4(b):
**First/Second Alternative (Triangle of Forces):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $T_{BQ}^2 = (0.1g)^2 + (T_{AB})^2$ OR $0.1g = T_{BQ}\sin(\theta)$ OR $(T_{AB}) = T_{BQ}\cos(\theta)$ | M1 | Using their $T_{AB}$ ONLY; M0 for any use of $T_{AB} = T_{BQ} = T_{AP}$ |
| For any 2 equations. FT their $T_{AB}$ ONLY | A1FT | $T_{AB} \neq T_{BQ}$. Sight of $(T_{AB})\tan\theta = 0.1g$ is M1 only without seeing an equation for $T_{BQ}$ |
| Solve for $T_{BQ}$ or $\theta$: $T_{BQ} = \sqrt{(0.1g)^2 + (T_{AB})^2}$, $\theta = \tan^{-1}\left(\frac{0.1g}{T_{AB}}\right)$ | DM1 | Solve for $\theta$ or solve for $T_{BQ}$ from equations with correct number of relevant terms. Using their $T_{AB}$ |
| $\theta = 26.6$ AND $T_{BQ} = \sqrt{5}$ or $2.24$ | A1 | $\theta = 26.56505...$, $T_{BQ} = 2.236067...$. AWRT 26.6 and 2.24 |
**Third Alternative (Lami's Theorem):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{T_{BQ}}{\sin 90} = \frac{0.1g}{\sin(180-\theta)} = \frac{T_{AB}}{\sin(90+\theta)}$ OR $\frac{T_{BQ}}{\sin 90} = \frac{0.1g}{\sin\theta} = \frac{T_{AB}}{\cos\theta}$ | *M1 | For any 2 fractions correct; M0 for any use of $T_{AB} = T_{BQ} = T_{AP}$ |
| All 3 fractions correct. FT their $T_{AB}$ ONLY | A1FT | Sight of $(T_{AB})\tan\theta = 0.1g$ is M1 only without seeing an equation for $T_{BQ}$ |
| $\theta = \tan^{-1}\left(\frac{0.1g}{T_{AB}}\right)$ | DM1 | Solve for $\theta$ |
| $\theta = 26.6$ AND $T_{BQ} = \sqrt{5}$ or $2.24$ | A1 | AWRT 26.6 and 2.24 |
**Fourth Alternative (Resolving on whole system):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $T_{BQ}\sin\theta + (T_{AP})\sin 45 = 0.2g + 0.1g$ and $T_{BQ}\cos\theta = (T_{AP})\cos 45$ | M1 | For resolving either horizontally or vertically on the whole system using their $T_{AP}$. M0 for $T_{AB} = T_{BQ} = T_{AP}$. If $T_{AP}$ not found in part (a), must have $T_{AP}\sin 45 = 0.2g$ or $(T_{AB}) = T_{AP}\cos 45$. Correct number of terms; allow sign errors on 4-term equation ONLY; allow sin/cos mix |
| Both equations correct. May see $T_{BQ}\sin\theta = 1$ and $T_{BQ}\cos\theta = 2$ | A1FT | Sight of $(T_{AB})\tan\theta = 0.1g$ is M1 only without seeing equation for $T_{BQ}$ |
| $\theta = \tan^{-1}\left(\frac{0.1g + 0.2g - (T_{AP})\sin 45}{(T_{AP})\cos 45}\right)$ or $T_{BQ} = \sqrt{(0.1g + 0.2g - (T_{AP})\sin 45)^2 + ((T_{AP})\cos 45)^2}$ | DM1 | Solve for $\theta$ or solve for $T_{BQ}$ from equations with correct number of relevant terms. Using their $T_{AP}$ |
| $\theta = 26.6$ AND $T_{BQ} = \sqrt{5}$ or $2.24$ | A1 | AWRT 26.6 and 2.24 |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-06_389_1134_258_468}
The diagram shows two particles, $A$ and $B$, of masses 0.2 kg and 0.1 kg respectively. The particles are suspended below a horizontal ceiling by two strings, $A P$ and $B Q$, attached to fixed points $P$ and $Q$ on the ceiling. The particles are connected by a horizontal string, $A B$. Angle $A P Q = 45 ^ { \circ }$ and $B Q P = \theta ^ { \circ }$. Each string is light and inextensible. The particles are in equilibrium.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the tension in the string $A B$.\\
\includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-06_2715_44_110_2006}\\
\includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-07_2721_34_101_20}
\item Find the value of $\theta$ and the tension in the string $B Q$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q4 [6]}}