| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.3 This is a straightforward M1 mechanics question testing vector addition and equilibrium. Part (i) uses the cosine rule (a standard technique for finding angles between forces), and part (ii) applies basic equilibrium conditions with three forces. Both parts are routine applications of core mechanics principles with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10d Vector operations: addition and scalar multiplication3.03m Equilibrium: sum of resolved forces = 03.03p Resultant forces: using vectors |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| M1 | Resolve // Resultant |
| A1 | |
| M1 | |
| A1 | [4] |
| M1 | Cosine rule for triangle of forces; Cosine rule must give obtuse angle |
| A1 | Do not accept 82.8° from incorrect working |
| OR | |
| M1 | \(6^2 = (4\sin\theta)^2 + (4+4\sin\theta)^2\) |
| A1 | \(36 = 16 + 32\sin\theta + 16\) hence \(\theta = 7.2°\) |
| M1 | \(\theta = 90-7.2\) |
| A1 | \(\theta = 82.8°\) |
| Answer | Marks |
|---|---|
| Marks | Guidance |
| M1 | Must have signs correct |
| A1 | |
| B1√ | Ft(90- cv(angle in (i))/2) |
| [3] |
## Part i
**Answer/Working:**
$4\cos\theta + 4\cos\theta = 6$
$\cos\theta = 6/8$
Angle ($= 2\theta = 2\cos^{-1}0.75$) = 82.8°
OR
$6^2 = 4^2 + 4^2 - 2x4x4\cos\alpha$
$\alpha = 97.2°$
Angle = 180 − 97.2
Angle = 82.8°
OR
$6^2 = (4\sin\theta)^2 + (4+4\cos\theta)^2$
$36 = 16 + 32\cos\theta + 16$
$\cos\theta =4/32$
$\theta = 82.8°$
| **Marks** | **Guidance** |
|-----------|-------------|
| M1 | Resolve // Resultant |
| A1 | |
| M1 | |
| A1 | [4] |
| M1 | Cosine rule for triangle of forces; Cosine rule must give obtuse angle |
| A1 | Do not accept 82.8° from incorrect working |
| OR | |
| M1 | $6^2 = (4\sin\theta)^2 + (4+4\sin\theta)^2$ |
| A1 | $36 = 16 + 32\sin\theta + 16$ hence $\theta = 7.2°$ |
| M1 | $\theta = 90-7.2$ |
| A1 | $\theta = 82.8°$ |
## Part ii
**Answer/Working:**
$mg = 6 + 3$ OR $mg = 4\cos(\text{Ans}(i))/2 + 4\cos((\text{Ans}(i))/2 + 3$
$m = 0.918$
Angle = 48.6°
| **Marks** | **Guidance** |
|-----------|-------------|
| M1 | Must have signs correct |
| A1 | |
| B1√ | Ft(90- cv(angle in (i))/2) |
| [3] | |
---Two forces each of magnitude $4\text{ N}$ have a resultant of magnitude $6\text{ N}$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the angle between the two $4\text{ N}$ forces. [4]
\end{enumerate}
The two given forces of magnitude $4\text{ N}$ act on a particle of mass $m\text{ kg}$ which remains at rest on a smooth horizontal surface. The surface exerts a force of magnitude $3\text{ N}$ on the particle.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find $m$, and give the acute angle between the surface and one of the $4\text{ N}$ forces. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2016 Q3 [7]}}