| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of three coplanar forces |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question on resolving forces and finding resultants. It requires Pythagoras' theorem, basic trigonometry (tan, sin, cos), and component addition. While multi-part with several steps, each step follows routine procedures taught in mechanics: finding magnitude from components, finding angles from components, and resolving perpendicular forces. Slightly easier than average A-level due to straightforward application of standard techniques without requiring problem-solving insight. |
| Spec | 3.03a Force: vector nature and diagrams3.03e Resolve forces: two dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | M1 | For using \( |
| \(P = 10\) | A1 | From \(P^2 = (-2.8)^2 + 9.6^2\) |
| \(R = 26.9\) | A1ft | 3 |
| (ii) | M1 | For using \(\tan \alpha = 9.6/(+ 2.8)\) or equivalent; may be scored in (i) |
| \(\alpha = 73.7\) | A1 | From c.w.o.; may be scored in (i) |
| (a) 24N | A1ft | \(fl\) \(25\cos(90 - \alpha)°\) for \(\alpha > 0\) |
| (b) 7N | A1ft | 4 |
| (iii) | M1 | For using \(\cos \theta = X/R\), \(\sin \theta = Y/R\) or \(\tan \theta = Y/X\), finding X or Y or X and Y as necessary |
| \(\cos \theta = (24 - 2.8)26.9 \ldots\) or | A1ft | |
| \(\sin \theta = (7 + 9.6)/26.9 \ldots\) or | ||
| \(\tan \theta = (7 + 9.6)/(24 - 2.8)\) | ||
| Alternative for the above 2 marks: For using \(\theta = \tan^{-1}(Y/X) + \tan^{-1}(P/25)\) | M1 | |
| \(\theta = \tan^{-1}(7/24) + \tan^{-1}(10/25)\) | A1ft | |
| \(\theta = 38.1\) | A1 | 3 |
(i) | M1 | For using $|X| = \sqrt{x_1^2 + x_2^2}$ for P or R
$P = 10$ | A1 | From $P^2 = (-2.8)^2 + 9.6^2$
$R = 26.9$ | A1ft | 3 | From $R^2 = 10^2 + 25^2$ or $R^2 = 21.2^2 + 16.6^2$
(ii) | M1 | For using $\tan \alpha = 9.6/(+ 2.8)$ or equivalent; may be scored in (i)
$\alpha = 73.7$ | A1 | From c.w.o.; may be scored in (i)
(a) 24N | A1ft | $fl$ $25\cos(90 - \alpha)°$ for $\alpha > 0$
(b) 7N | A1ft | 4 | $fl$ $25\sin(90 - \alpha)°$ for $\alpha > 0$
(iii) | M1 | For using $\cos \theta = X/R$, $\sin \theta = Y/R$ or $\tan \theta = Y/X$, finding X or Y or X and Y as necessary
$\cos \theta = (24 - 2.8)26.9 \ldots$ or | A1ft |
$\sin \theta = (7 + 9.6)/26.9 \ldots$ or |
$\tan \theta = (7 + 9.6)/(24 - 2.8)$ |
| | | Alternative for the above 2 marks: For using $\theta = \tan^{-1}(Y/X) + \tan^{-1}(P/25)$ | M1
$\theta = \tan^{-1}(7/24) + \tan^{-1}(10/25)$ | | A1ft
$\theta = 38.1$ | A1 | 36\\
\includegraphics[max width=\textwidth, alt={}, center]{d0fa61eb-f320-427e-8883-de224d293933-4_474_831_269_657}
Forces of magnitudes $P \mathrm {~N}$ and 25 N act at right angles to each other. The resultant of the two forces has magnitude $R \mathrm {~N}$ and makes an angle of $\theta ^ { \circ }$ with the $x$-axis (see diagram). The force of magnitude $P \mathrm {~N}$ has components - 2.8 N and 9.6 N in the $x$-direction and the $y$-direction respectively, and makes an angle of $\alpha ^ { \circ }$ with the negative $x$-axis.\\
(i) Find the values of $P$ and $R$.\\
(ii) Find the value of $\alpha$, and hence find the components of the force of magnitude 25 N in
\begin{enumerate}[label=(\alph*)]
\item the $x$-direction,
\item the $y$-direction.\\
(iii) Find the value of $\theta$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2006 Q6 [10]}}