| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Bearing and compass direction problems |
| Difficulty | Moderate -0.3 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions followed by a basic application of F=ma. The bearing notation adds minimal complexity, and the calculations involve simple trigonometry and arithmetic. Slightly easier than average due to the routine nature of the problem-solving steps. |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2.5\sin\theta = 2.4\) | M1 | \(2.5\text{CorS}\theta = 2.4\); \(2.5\cos\theta = 2.4\) M1 hence \(\theta = 16.3\) A0 |
| \(\theta = 73.7\) | A1 | Accept 74 |
| \(2.5\cos\theta = F\) | M1 | \(F = 2.5\ \text{SorC}\theta\), opposite to that above; \(2.5\sin\theta = F\) M1 hence |
| \(F = 0.7\) | A1 | Exact, but allow \(0.702\) (3 sf) \(\theta = 73.7\); \(F = 0.7(00)\) A1 SC |
| *OR* | ||
| \(2.4^2 + F^2 = 2.5^2\) or \(F^2 = 2.5^2 - 2.4^2\) | M1 | \(F\) can then be used to find \(\theta\) |
| \(F = 0.7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2.4 = 0.2a\) | M1 | N2L, any horizontal force other than \(F\), \(0.7\), \(2.5\) (do not treat removing/using \(2.5\) as a MR); Including \(g\), automatically M0 |
| \(a = 12 \text{ ms}^{-2}\) | A1 | \(12.0\) from \(2.5\sin73.7/0.2\) |
| Bearing \((0)90°\) OR "To right", opposite old \(2.4\) N force" etc | B1 | Angle value other than exactly \(90°\) or \(0°\) B0; Horizontal is B0 (ambiguous) |
| [3] | Allow B1 for force direction, if acceleration not found |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2.5\sin\theta = 2.4$ | M1 | $2.5\text{CorS}\theta = 2.4$; $2.5\cos\theta = 2.4$ M1 hence $\theta = 16.3$ A0 |
| $\theta = 73.7$ | A1 | Accept 74 |
| $2.5\cos\theta = F$ | M1 | $F = 2.5\ \text{SorC}\theta$, opposite to that above; $2.5\sin\theta = F$ M1 hence |
| $F = 0.7$ | A1 | Exact, but allow $0.702$ (3 sf) $\theta = 73.7$; $F = 0.7(00)$ A1 SC |
| *OR* | | |
| $2.4^2 + F^2 = 2.5^2$ or $F^2 = 2.5^2 - 2.4^2$ | M1 | $F$ can then be used to find $\theta$ |
| $F = 0.7$ | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2.4 = 0.2a$ | M1 | N2L, any horizontal force other than $F$, $0.7$, $2.5$ (do not treat removing/using $2.5$ as a MR); Including $g$, automatically M0 |
| $a = 12 \text{ ms}^{-2}$ | A1 | $12.0$ from $2.5\sin73.7/0.2$ |
| Bearing $(0)90°$ OR "To right", opposite old $2.4$ N force" etc | B1 | Angle value other than exactly $90°$ or $0°$ B0; Horizontal is B0 (ambiguous) |
| | [3] | Allow B1 for force direction, if acceleration not found |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{66eb8290-3a80-40bf-be40-a936ed7d5a1b-2_309_520_941_744}
A particle rests on a smooth horizontal surface. Three horizontal forces of magnitudes $2.5 \mathrm {~N} , F \mathrm {~N}$ and 2.4 N act on the particle on bearings $\theta ^ { \circ } , 180 ^ { \circ }$ and $270 ^ { \circ }$ respectively (see diagram). The particle is in equilibrium.\\
(i) Find $\theta$ and $F$.
The 2.4 N force suddenly ceases to act on the particle, which has mass 0.2 kg .\\
(ii) Find the magnitude and direction of the acceleration of the particle.
\hfill \mbox{\textit{OCR M1 2014 Q2 [7]}}