| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| 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 application of resolving forces into components. Part (a) requires setting the y-component to zero and solving a simple trigonometric equation. Part (b) involves resolving both forces, finding the resultant components, then calculating magnitude and direction using Pythagoras and inverse tan. Standard M1 mechanics with no conceptual challenges beyond basic resolution of forces. |
| Spec | 3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(F = 20\sin 60\) | M1 | Attempt to resolve in \(y\)-direction; 2 terms; must be \(20\cos 60\) or \(20\sin 60\) and must be linked to \(F\) (can be implied by the correct answer seen only). |
| \(= 17.3\text{ N}\) | A1 | AWRT 17.3 \((17.320508\ldots)\) or \(10\sqrt{3}\). |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Horizontal component \(= X = R\cos\theta = \pm(20\cos 60)\) \([\pm 10]\) | *M1 | Correct number of terms; allow sin/cos mix; allow sign errors |
| Vertical component \(= Y = R\sin\theta = \pm(20\sin 60 - 10)\) \([= \pm7.3205...]\) | A1 | For both correct |
| Magnitude \(= \sqrt{(20\sin 60 - 10)^2 + (20\cos 60)^2}\) \([=12.393136...]\) | DM1 | OE – correct number of terms |
| Angle \(= \tan^{-1}\left(\dfrac{20\sin 60 - 10}{20\cos 60}\right)\) \([=36.206023...]\) | DM1 | OE (e.g. reciprocal) - correct number of terms |
| Magnitude \(= 12.4\text{ N}\) and Direction \(= 36.2°\) above (positive) \(x\)-axis | A1 | OE for direction e.g. \(36.2°\) anticlockwise from (positive) \(x\)-direction, \(36.2°\) above the horizontal. Possibly seen on a diagram. (Radians: \(0.63191...\) to 3sf or better) |
| 5 |
## Question 2:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $F = 20\sin 60$ | **M1** | Attempt to resolve in $y$-direction; 2 terms; must be $20\cos 60$ or $20\sin 60$ and must be linked to $F$ (can be implied by the correct answer seen only). |
| $= 17.3\text{ N}$ | **A1** | AWRT 17.3 $(17.320508\ldots)$ or $10\sqrt{3}$. |
| | **[2]** | |
## Question 2(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Horizontal component $= X = R\cos\theta = \pm(20\cos 60)$ $[\pm 10]$ | *M1 | Correct number of terms; allow sin/cos mix; allow sign errors |
| Vertical component $= Y = R\sin\theta = \pm(20\sin 60 - 10)$ $[= \pm7.3205...]$ | A1 | For both correct |
| Magnitude $= \sqrt{(20\sin 60 - 10)^2 + (20\cos 60)^2}$ $[=12.393136...]$ | DM1 | OE – correct number of terms |
| Angle $= \tan^{-1}\left(\dfrac{20\sin 60 - 10}{20\cos 60}\right)$ $[=36.206023...]$ | DM1 | OE (e.g. reciprocal) - correct number of terms |
| Magnitude $= 12.4\text{ N}$ and Direction $= 36.2°$ above (positive) $x$-axis | A1 | OE for direction e.g. $36.2°$ anticlockwise from (positive) $x$-direction, $36.2°$ above the horizontal. Possibly seen on a diagram. (Radians: $0.63191...$ to 3sf or better) |
| | **5** | |
---2\\
\begin{tikzpicture}[>=stealth, scale=1.2]
% Dashed x and y axes
\draw[dashed] (-1.5,0) -- (0,0);
\draw[->, dashed] (0,0) -- (3,0) node[right] {$x$};
\draw[->, dashed] (0,0) -- (0,3) node[above] {$y$};
% Point P
\node[below left] at (0,0) {$P$};
% 20 N force arrow (60 degrees from x-axis, into first quadrant)
\draw[->, thick] (0,0) -- ({2.5*cos(60)},{2.5*sin(60)}) node[right] {$20\,\text{N}$};
% Angle arc between x-axis and 20 N force
\draw (0.7,0) arc[start angle=0, end angle=60, radius=0.7];
\node at (0.95,0.4) {$60°$};
% F N force arrow (straight down)
\draw[->, thick] (0,0) -- (0,-2.5) node[below] {$F\,\text{N}$};
\end{tikzpicture}
Two forces of magnitudes 20 N and $F \mathrm {~N}$ act at a point $P$ in the directions shown in the diagram.
\begin{enumerate}[label=(\alph*)]
\item Given that the resultant force has no component in the $y$-direction, calculate the value of $F$.
\item Given instead that $F = 10$, find the magnitude and direction of the resultant force.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q2 [7]}}