| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring resolution of forces into components and finding the resultant direction using basic trigonometry. It involves routine application of standard techniques (resolving forces, Pythagoras, arctan) with no problem-solving insight needed, making it easier than average but not trivial since it requires careful angle work and multiple computational steps. |
| Spec | 3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \([X = 7 + 10\cos50° - 15\cos80°,\) | M1 | For obtaining an expression for X or Y |
| \(Y = 10\sin50° + 15\sin80°]\) | ||
| (a) x-component is 10.8 N | A1 | |
| (b) y-component is 22.4 N | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \([\theta = \tan^{-1}(22.4/10.8)]\) | M1 | For using \(\theta = \tan^{-1}(Y/X)\) |
| Direction 64.2° anticlockwise from x-axis | A1 | [2] Accept 64.3° |
**(i)**
$[X = 7 + 10\cos50° - 15\cos80°,$ | M1 | For obtaining an expression for X or Y
$Y = 10\sin50° + 15\sin80°]$ | |
(a) x-component is 10.8 N | A1 |
(b) y-component is 22.4 N | A1 | [3]
**(ii)**
$[\theta = \tan^{-1}(22.4/10.8)]$ | M1 | For using $\theta = \tan^{-1}(Y/X)$
Direction 64.2° anticlockwise from x-axis | A1 | [2] Accept 64.3°\includegraphics{figure_3}
Forces of magnitudes 7 N, 10 N and 15 N act on a particle in the directions shown in the diagram.
\begin{enumerate}[label=(\roman*)]
\item Find the component of the resultant of the three forces
\begin{enumerate}[label=(\alph*)]
\item in the $x$-direction,
\item in the $y$-direction.
\end{enumerate}
[3]
\item Hence find the direction of the resultant. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2009 Q3 [5]}}