| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 7 |
| 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 basic resolution of forces and application of F=ma. Part (i) involves simple component addition, part (ii) uses Pythagoras or the cosine rule, and part (iii) applies Newton's second law. All techniques are standard M1 procedures with no problem-solving insight required, making it easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Component in direction of 20 N force: \(20 + 16\cos 60° = 28\) N | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Perpendicular component: \(16\sin 60° = 13.86\) N | M1 | |
| Resultant \(= \sqrt{28^2 + (16\sin 60°)^2}\) | M1 | |
| \(= \sqrt{784 + 192} = \sqrt{976} = 31.2\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = \frac{31.24}{2} = 15.6\) m s\(^{-2}\) | M1 A1 | |
| \(\theta = \arctan\!\left(\frac{16\sin 60°}{28}\right) = 26.3°\) with the 20 N force | M1 A1 |
# Question 4:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Component in direction of 20 N force: $20 + 16\cos 60° = 28$ N | B1 | |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Perpendicular component: $16\sin 60° = 13.86$ N | M1 | |
| Resultant $= \sqrt{28^2 + (16\sin 60°)^2}$ | M1 | |
| $= \sqrt{784 + 192} = \sqrt{976} = 31.2$ N | A1 | |
**(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = \frac{31.24}{2} = 15.6$ m s$^{-2}$ | M1 A1 | |
| $\theta = \arctan\!\left(\frac{16\sin 60°}{28}\right) = 26.3°$ with the 20 N force | M1 A1 | |4 Fig. 4 shows forces of magnitudes 20 N and 16 N inclined at $60 ^ { \circ }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{52d6c914-b204-4587-a82e-fbab6693fcf8-3_191_346_328_858}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Calculate the component of the resultant of these two forces in the direction of the 20 N force.\\
(ii) Calculate the magnitude of the resultant of these two forces.
These are the only forces acting on a particle of mass 2 kg .\\
(iii) Find the magnitude of the acceleration of the particle and the angle the acceleration makes with the 20 N force.
\hfill \mbox{\textit{OCR MEI M1 2007 Q4 [7]}}