| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Component of resultant in given direction |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem requiring resolution of forces and basic trigonometry. Students need to resolve one force along OA (giving 8 + 8cos θ = 9), solve for θ, then find the resultant magnitude using standard vector addition. It's slightly easier than average as it's a direct application of standard techniques with no conceptual surprises, though it does require careful execution of multiple steps. |
| Spec | 1.10d Vector operations: addition and scalar multiplication3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([8 + 8\cos\theta = 9]\) | M1 | For an equation in \(\theta\) using component 9N |
| \(\theta = 82.8\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| For showing \(\theta\) or \((180° - \theta)\) or \(\theta/2\), in a triangle representing the two forces and the resultant, or for using \(Y = 8\sin\theta\) in \(R^2 = X^2 + Y^2\) | B1 | This mark may be implied by a correct equation for \(R(\theta)\) in the subsequent working |
| \([R^2 = 8^2 + 8^2 - 2\times8\times8\cos(180-\theta)]\), \(R^2 = 8^2 + 8^2 + 2\times8\times8\cos\theta\), \(\cos(\theta/2) = (R/2) \div 8\), \(R\cos(\theta/2) = 9\), \(R\sin(\theta/2) = 8\sin\theta\), \(R^2 = 9^2 + (8\sin\theta)^2\), \(R^2 = (8 + 8\cos\theta)^2 + (8\sin\theta)^2\) | M1 | For an equation in \(R\) or \(R^2\) |
| Magnitude is 12 N | A1 |
## Question 2:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[8 + 8\cos\theta = 9]$ | M1 | For an equation in $\theta$ using component 9N |
| $\theta = 82.8$ | A1 | |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| For showing $\theta$ or $(180° - \theta)$ or $\theta/2$, in a triangle representing the two forces and the resultant, or for using $Y = 8\sin\theta$ in $R^2 = X^2 + Y^2$ | B1 | This mark may be implied by a correct equation for $R(\theta)$ in the subsequent working |
| $[R^2 = 8^2 + 8^2 - 2\times8\times8\cos(180-\theta)]$, $R^2 = 8^2 + 8^2 + 2\times8\times8\cos\theta$, $\cos(\theta/2) = (R/2) \div 8$, $R\cos(\theta/2) = 9$, $R\sin(\theta/2) = 8\sin\theta$, $R^2 = 9^2 + (8\sin\theta)^2$, $R^2 = (8 + 8\cos\theta)^2 + (8\sin\theta)^2$ | M1 | For an equation in $R$ or $R^2$ |
| Magnitude is 12 N | A1 | |
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\includegraphics[max width=\textwidth, alt={}, center]{f7a22c07-44e3-4891-be60-cbab772f45df-2_549_589_934_778}
Two forces, each of magnitude 8 N , act at a point in the directions $O A$ and $O B$. The angle between the forces is $\theta ^ { \circ }$ (see diagram). The resultant of the two forces has component 9 N in the direction $O A$. Find\\
(i) the value of $\theta$,\\
(ii) the magnitude of the resultant of the two forces.
\hfill \mbox{\textit{CAIE M1 2007 Q2 [5]}}