| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Resultant of coplanar forces |
| Difficulty | Moderate -0.3 This is a standard M1 forces question requiring resolution of forces into perpendicular components and use of Pythagoras/trigonometry. Parts (i) and (ii) are routine applications of component methods, while part (iii) requires the simple insight that maximum resultant occurs when forces are parallel (sum = 43N) and minimum when opposed optimally (20-15-8 = 0N, though achieving exactly zero requires specific arrangement). The question is slightly easier than average due to clear structure and standard techniques, though the triangle of forces setup requires careful angle work. |
| Spec | 3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((X=) 15 - 20\cos 60,\; 15 - 20\sin 30\) OR \((Y=) 8 - 20\cos 30,\; 8 - 20\sin 60\) | M1 | Accept \((X=) 15 + 20\cos 120\), \((Y=) 8 + 20\cos 150\), and R A \(= 100°\) |
| \((X=) 5 \text{ N}\) \quad (34.048.. if in rad mode) | A1 | Must be \(+\)ve |
| \((Y=) -9.32 \text{ N}\) \quad (4.9149.. if in rad mode) | A1 [3] | Must be \(-\)ve. Allow \(8 - 10\sqrt{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R^2 = (+/-9.32)^2 + 5^2\) | M1 | Uses Pythagoras on ans(i), neither component 8 or 15 |
| \(R = 10.6 \text{ N}\) | A1 ft | \(\sqrt{X(\mathbf{i})^2 + Y(\mathbf{i})^2}\) |
| \(\tan\theta = (+/-9.32)/5\) | M1 | Finds any relevant angle with 8 N or 15 N, neither component 8 or 15 |
| Angle \(= 152°\) | A1 [4] | CAO, must be 3sf or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (Greatest \(=\)) 43 N | B1 | |
| (Least \(=\)) 0 N | B1 [2] |
# Question 4:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(X=) 15 - 20\cos 60,\; 15 - 20\sin 30$ OR $(Y=) 8 - 20\cos 30,\; 8 - 20\sin 60$ | M1 | Accept $(X=) 15 + 20\cos 120$, $(Y=) 8 + 20\cos 150$, and R A $= 100°$ |
| $(X=) 5 \text{ N}$ \quad (34.048.. if in rad mode) | A1 | Must be $+$ve |
| $(Y=) -9.32 \text{ N}$ \quad (4.9149.. if in rad mode) | A1 **[3]** | Must be $-$ve. Allow $8 - 10\sqrt{3}$ |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R^2 = (+/-9.32)^2 + 5^2$ | M1 | Uses Pythagoras on ans(i), neither component 8 or 15 |
| $R = 10.6 \text{ N}$ | A1 ft | $\sqrt{X(\mathbf{i})^2 + Y(\mathbf{i})^2}$ |
| $\tan\theta = (+/-9.32)/5$ | M1 | Finds any relevant angle with 8 N or 15 N, neither component 8 or 15 |
| Angle $= 152°$ | A1 **[4]** | CAO, must be 3sf or better |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| (Greatest $=$) 43 N | B1 | |
| (Least $=$) 0 N | B1 **[2]** | |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{2b3457b6-1fe9-4e67-91d4-a8bc4a5b1709-2_325_481_1699_792}
Three horizontal forces of magnitudes $8 \mathrm {~N} , 15 \mathrm {~N}$ and 20 N act at a point. The 8 N and 15 N forces are at right angles. The 20 N force makes an angle of $150 ^ { \circ }$ with the 8 N force and an angle of $120 ^ { \circ }$ with the 15 N force (see diagram).\\
(i) Calculate the components of the resultant force in the directions of the 8 N and 15 N forces.\\
(ii) Calculate the magnitude of the resultant force, and the angle it makes with the direction of the 8 N force.
The directions in which the three horizontal forces act can be altered.\\
(iii) State the greatest and least possible magnitudes of the resultant force.
\hfill \mbox{\textit{OCR M1 2012 Q4 [9]}}