| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| 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 standard 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 routine M1 content with no problem-solving insight required, making it easier than average but not trivial due to the multi-step calculation. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\downarrow \quad 20 + 16\cos 60 = 28\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\rightarrow \quad 16\sin 60\) | B1 | Any form. May be seen in (i). Accept any appropriate equivalent resolution. |
| Magnitude \(\sqrt{28^2 + 192} = 31.2409\ldots\) so \(31.2\) N (3 s.f.) | M1 | Use of Pythagoras with 2 distinct components (but not 16 and \(\pm 20\)) |
| \(= 31.2\) N (3 s.f.) | F1 | Allow \(34.788\ldots\) only as FT |
| or Cosine rule: \(\text{mag}^2 = 16^2 + 20^2 - 2\times16\times20\times\cos 120\) | M1 | Must be used with 20 N, 16 N and \(60°\) or \(120°\) |
| Correct substitution | A1 | |
| \(31.2\) N (3 s.f.) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Magnitude of acceleration is \(15.620\ldots\) m s\(^{-2}\) so \(15.6\) m s\(^{-2}\) (3 s.f.) | B1 | Award only for their \(F \div 2\) |
| Angle with 20 N force is \(\arctan\!\left(\dfrac{16\sin 60}{28}\right)\) | M1 | Or equiv. May use force or acceleration. Allow use of sine or cosine rules. FT only \(s \leftrightarrow c\) and sign errors. Accept reciprocal of the fraction. |
| so \(26.3295\ldots\) so \(26.3°\) (3 s.f.) | A1 | cao |
# Question 4:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\downarrow \quad 20 + 16\cos 60 = 28$ | B1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\rightarrow \quad 16\sin 60$ | B1 | Any form. May be seen in (i). Accept any appropriate equivalent resolution. |
| Magnitude $\sqrt{28^2 + 192} = 31.2409\ldots$ so $31.2$ N (3 s.f.) | M1 | Use of Pythagoras with 2 distinct components (but not 16 and $\pm 20$) |
| $= 31.2$ N (3 s.f.) | F1 | Allow $34.788\ldots$ only as FT |
| **or** Cosine rule: $\text{mag}^2 = 16^2 + 20^2 - 2\times16\times20\times\cos 120$ | M1 | Must be used with 20 N, 16 N and $60°$ or $120°$ |
| Correct substitution | A1 | |
| $31.2$ N (3 s.f.) | A1 | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Magnitude of acceleration is $15.620\ldots$ m s$^{-2}$ so $15.6$ m s$^{-2}$ (3 s.f.) | B1 | Award only for **their** $F \div 2$ |
| Angle with 20 N force is $\arctan\!\left(\dfrac{16\sin 60}{28}\right)$ | M1 | Or equiv. May use force or acceleration. Allow use of sine or cosine rules. FT only $s \leftrightarrow c$ and sign errors. Accept reciprocal of the fraction. |
| so $26.3295\ldots$ so $26.3°$ (3 s.f.) | A1 | cao |
---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]{94f23528-931c-47b6-89aa-4b6edd25cc30-3_193_351_261_895}
\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 Q4 [7]}}