| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of two forces (triangle/parallelogram law) |
| Difficulty | Moderate -0.3 This is a straightforward application of the parallelogram/triangle law for forces using the cosine rule, followed by basic trigonometry for components. The question requires standard formulas (R² = P² + Q² + 2PQ cos θ) with given values, making it slightly easier than average but still requiring proper method and calculation accuracy. |
| Spec | 3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct diagram (may be implied), or recognising that resultant acts along bisector, or \(12\cos\beta = 10 + 10\cos\theta\) and \(12\sin\beta = 10\sin\theta\), or \(X = 10 - 10\cos\alpha\) and \(Y = 10\sin\alpha\) | B1 | |
| Complete method for \(\alpha\): \([\alpha = 2\sin^{-1}\frac{6}{10}]\) or \([12^2 = 10^2 + 10^2 - 2\times10^2\cos\alpha]\), or resolving forces along the bisector \([2\times10\cos\frac{\theta}{2} = 12]\), or squaring and adding using \(c^2\beta + s^2\beta = 1\) and \(c^2\theta + s^2\theta = 1\), \([144 = 100 + 200\cos\theta + 100]\) | M1 | Wrong diagram case: Triangle sides 10, 10, 12 with angle \(\theta\) opposite the 12: (i) M0, \(12^2 = 10^2 + 10^2 - 2\times10^2\cos\theta\) M1 A0 (max 1 out of 3) |
| \(\theta = 106.3°\) or \(1.85\) rads | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For using component \(= 12\cos\frac{\theta}{2}\) [\(12\times0.6\)] or \(10 - 10\cos\alpha\) | M1 | |
| Component is \(7.2\) N (ft only when B1 in part (i) is scored) | A1 ft | Accept \(7.19\), \(7.20\), or \(7.21\) for final A1 |
| SR for candidates whose diagram has triangle with sides 10, 10, 12 and angle \(\theta\) opposite the 12 (max 1 out of 2): Component is \(\pm7.2\) N | B1 | |
| Alternative (scale diagram): As for first mark in scheme above; value of \(\theta\) in range \(105°\) to \(107°\) obtained; \(\theta = 106.3°\); for drawing relevant perpendicular and measuring appropriate length; Component is \(7.2\) N | B1, B1, B1, M1, A1 |
# Question 3:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct diagram (may be implied), or recognising that resultant acts along bisector, or $12\cos\beta = 10 + 10\cos\theta$ and $12\sin\beta = 10\sin\theta$, or $X = 10 - 10\cos\alpha$ and $Y = 10\sin\alpha$ | B1 | |
| Complete method for $\alpha$: $[\alpha = 2\sin^{-1}\frac{6}{10}]$ or $[12^2 = 10^2 + 10^2 - 2\times10^2\cos\alpha]$, or resolving forces along the bisector $[2\times10\cos\frac{\theta}{2} = 12]$, or squaring and adding using $c^2\beta + s^2\beta = 1$ and $c^2\theta + s^2\theta = 1$, $[144 = 100 + 200\cos\theta + 100]$ | M1 | Wrong diagram case: Triangle sides 10, 10, 12 with angle $\theta$ opposite the 12: (i) M0, $12^2 = 10^2 + 10^2 - 2\times10^2\cos\theta$ M1 A0 (max 1 out of 3) |
| $\theta = 106.3°$ or $1.85$ rads | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For using component $= 12\cos\frac{\theta}{2}$ [$12\times0.6$] or $10 - 10\cos\alpha$ | M1 | |
| Component is $7.2$ N (ft only when B1 in part (i) is scored) | A1 ft | Accept $7.19$, $7.20$, or $7.21$ for final A1 |
| SR for candidates whose diagram has triangle with sides 10, 10, 12 and angle $\theta$ opposite the 12 (max 1 out of 2): Component is $\pm7.2$ N | B1 | |
| Alternative (scale diagram): As for first mark in scheme above; value of $\theta$ in range $105°$ to $107°$ obtained; $\theta = 106.3°$; for drawing relevant perpendicular and measuring appropriate length; Component is $7.2$ N | B1, B1, B1, M1, A1 | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{430f1f9a-7a3a-47a0-b742-daf74e68adfd-2_368_584_1302_794}
Two forces, each of magnitude 10 N , act at a point $O$ in the directions of $O A$ and $O B$, as shown in the diagram. The angle between the forces is $\theta$. The resultant of these two forces has magnitude 12 N .\\
(i) Find $\theta$.\\
(ii) Find the component of the resultant force in the direction of $O A$.
\hfill \mbox{\textit{CAIE M1 2002 Q3 [5]}}