| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.5 This is a straightforward mechanics question requiring resolution of forces into components and application of Pythagoras' theorem. The question guides students through the process step-by-step (finding parallel and perpendicular components, then using the given resultant magnitude), making it slightly easier than average. The algebra is simple and the method is standard textbook material for M1. |
| Spec | 3.03a Force: vector nature and diagrams3.03e Resolve forces: two dimensions |
| Answer | Marks |
|---|---|
| \(10 - 8\cos\theta\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(8\sin\theta\) | B1 | [2 marks total] |
| Answer | Marks | Guidance |
|---|---|---|
| \((10 - 8\cos\theta)^2 + (8\sin\theta)^2 = 8^2\) | M1 | For using \(X^2 + Y^2 = R^2\) or for using the cosine rule in the relevant triangle |
| \(10^2 + 8^2 - 2\times10\times8\cos\theta = 8^2\) | A1ft | |
| \(\cos\theta = \frac{5}{8}\) | A1 | [3] AG |
| Answer | Marks | Guidance |
|---|---|---|
| \([\cos\varphi = (10 - 8\cos\theta)/8\) and \(\sin\varphi = 8\sin\theta/8]\) | M1 | For using \(\cos\varphi = X/R\) and \(\sin\varphi = Y/R\) |
| \(8\cos\varphi = (10 - 8\cos\theta)\) and \(\varphi = \theta\) | A1ft | |
| \(\cos\theta = \frac{5}{8}\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \([5, \sqrt{39}, 64]\) | M1 | For assuming \(\cos\theta = 5/8\) and hence finding exact values of \(\sin\theta\), X, Y and \(X^2 + Y^2\) |
| \(R = 8\) | A1 | |
| \(\rightarrow\) assumption correct | A1 |
| Answer | Marks |
|---|---|
| M1 | For assuming \(\cos\theta = 5/8\) and hence finding \(\theta = 51.3°\) and the values of X, Y and \(X^2 + Y^2\) |
| \(R = 8\) or \(8.0\) or \(8.00\) or \(7.997...\) | |
| \(\rightarrow\) assumption correct | A1 |
# Question 1:
## Part (i)(a)
$10 - 8\cos\theta$ | B1 |
## Part (i)(b)
$8\sin\theta$ | B1 | [2 marks total]
## Part (ii)
$(10 - 8\cos\theta)^2 + (8\sin\theta)^2 = 8^2$ | M1 | For using $X^2 + Y^2 = R^2$ or for using the cosine rule in the relevant triangle
$10^2 + 8^2 - 2\times10\times8\cos\theta = 8^2$ | A1ft |
$\cos\theta = \frac{5}{8}$ | A1 | [3] AG
**First alternative for (ii):**
$[\cos\varphi = (10 - 8\cos\theta)/8$ and $\sin\varphi = 8\sin\theta/8]$ | M1 | For using $\cos\varphi = X/R$ and $\sin\varphi = Y/R$
$8\cos\varphi = (10 - 8\cos\theta)$ and $\varphi = \theta$ | A1ft |
$\cos\theta = \frac{5}{8}$ | A1 | AG
**Second alternative for (ii):**
$[5, \sqrt{39}, 64]$ | M1 | For assuming $\cos\theta = 5/8$ and hence finding exact values of $\sin\theta$, X, Y and $X^2 + Y^2$
$R = 8$ | A1 |
$\rightarrow$ assumption correct | A1 |
**SR for (ii) (max 2/3):**
| M1 | For assuming $\cos\theta = 5/8$ and hence finding $\theta = 51.3°$ and the values of X, Y and $X^2 + Y^2$
$R = 8$ or $8.0$ or $8.00$ or $7.997...$ | |
$\rightarrow$ assumption correct | A1 |
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\includegraphics[max width=\textwidth, alt={}, center]{a4cb105b-55d2-4793-95d2-3d791990a1f6-2_341_929_269_609}
Forces of magnitudes 10 N and 8 N act in directions as shown in the diagram.\\
(i) Write down in terms of $\theta$ the component of the resultant of the two forces
\begin{enumerate}[label=(\alph*)]
\item parallel to the force of magnitude 10 N ,
\item perpendicular to the force of magnitude 10 N .\\
(ii) The resultant of the two forces has magnitude 8 N . Show that $\cos \theta = \frac { 5 } { 8 }$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2008 Q1 [5]}}