| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of three coplanar forces |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring Pythagoras' theorem, basic trigonometry (tan, sin, cos), and vector component addition. Part (i) is routine calculation from components; part (ii) involves resolving a perpendicular force and finding the resultant, which is standard M1 material with no conceptual challenges or novel problem-solving required. |
| Spec | 3.03a Force: vector nature and diagrams3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([F^2 = 27.5^2 + (-24)^2]\) \(F = 36.5\) \([\tan\alpha° = -(−24/27.5)]\) \(\alpha = 41.1\) | M1, A1, M1, A1 | For using \(F^2 = X^2 + Y^2\) (may be scored in (ii)) or For using \(\tan\alpha° = -Y/X\) |
| (ii) \(R = 94.9\) \([\alpha° + \theta° = \tan^{-1}(87.6/36.5);\) or \((\alpha° + \theta°) = \cos^{-1}(36.5/94.9)\) or \(\theta° = \tan^{-1}(87.6\sin48.9° - 24)/(27.5 + 87.6\cos48.9°)]\) \(\theta = 26.3\) | B1, M1, A1 ft | For using \(\tan(\alpha° + \theta°) = 87.6/F\) or \(\cos(\alpha° + \theta°) = F/R\) or \(\tan\theta° = Y/X\) |
(i) $[F^2 = 27.5^2 + (-24)^2]$ $F = 36.5$ $[\tan\alpha° = -(−24/27.5)]$ $\alpha = 41.1$ | M1, A1, M1, A1 | For using $F^2 = X^2 + Y^2$ (may be scored in (ii)) or For using $\tan\alpha° = -Y/X$ | [4]
(ii) $R = 94.9$ $[\alpha° + \theta° = \tan^{-1}(87.6/36.5);$ or $(\alpha° + \theta°) = \cos^{-1}(36.5/94.9)$ or $\theta° = \tan^{-1}(87.6\sin48.9° - 24)/(27.5 + 87.6\cos48.9°)]$ $\theta = 26.3$ | B1, M1, A1 ft | For using $\tan(\alpha° + \theta°) = 87.6/F$ or $\cos(\alpha° + \theta°) = F/R$ or $\tan\theta° = Y/X$ | [3] | ft 67.4 – incorrect $\alpha$
---5 A force of magnitude $F \mathrm {~N}$ acts in a horizontal plane and has components 27.5 N and - 24 N in the $x$-direction and the $y$-direction respectively. The force acts at an angle of $\alpha ^ { \circ }$ below the $x$-axis.\\
(i) Find the values of $F$ and $\alpha$.
A second force, of magnitude 87.6 N , acts in the same plane at $90 ^ { \circ }$ anticlockwise from the force of magnitude $F \mathrm {~N}$. The resultant of the two forces has magnitude $R \mathrm {~N}$ and makes an angle of $\theta ^ { \circ }$ with the positive $x$-axis.\\
(ii) Find the values of $R$ and $\theta$.
\hfill \mbox{\textit{CAIE M1 2010 Q5 [7]}}