| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Resultant of three coplanar forces |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question requiring resolution of forces into components and basic trigonometry. Part (i) involves routine component resolution and Pythagoras/tan for the resultant; part (ii) requires setting the y-component to zero and solving for P. The given tan ratio simplifies calculations. While multi-step, it follows a well-practiced procedure with no novel insight required, making it slightly easier than average. |
| Spec | 3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For resolving forces horizontally | |
| \(R_x = 40 \times (24/25) - 30 \times (7/25) = [30]\) | A1 | Allow \(R_x = 40\cos 16.3 - 30\sin 16.3\) |
| M1 | For resolving forces vertically | |
| \(R_y = 50 - 40 \times (7/25) - 30 \times (24/25) = [10]\) | A1 | Allow \(R_y = 50 - 40\sin 16.3 - 30\cos 16.3\) |
| \(R = \sqrt{R_x^2 + R_y^2}\) and \(\theta = \tan^{-1}\!\left(\dfrac{R_y}{R_x}\right)\) | M1 | For using Pythagoras to find resultant force \(R\) and trigonometry to find angle \(\theta\) with \(x\)-axis |
| \(R = 31.6\) N and \(\theta = 18.4°\) with positive \(x\)-axis | A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R_1 = 40 - 50 \times (7/25) = [26]\) | M1, A1 | Resolve forces along 40 N direction; allow \(R_1 = 40 - 50\sin 16.3\) |
| \(R_2 = 30 - 50 \times (24/25) = [-18]\) | M1, A1 | Resolve forces along 30 N direction; allow \(R_2 = 30 - 50\cos 16.3\) |
| \(R^2 = R_1^2 + R_2^2\) and \(\arctan(-R_2/R_1)\) | M1 | Use Pythagoras and trigonometry |
| \(R = 31.6\) N and direction is \(34.7 - \alpha = 18.4°\) with positive \(x\)-axis | A1 | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P = 40\) | B1 | 1 |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces horizontally |
| $R_x = 40 \times (24/25) - 30 \times (7/25) = [30]$ | A1 | Allow $R_x = 40\cos 16.3 - 30\sin 16.3$ |
| | M1 | For resolving forces vertically |
| $R_y = 50 - 40 \times (7/25) - 30 \times (24/25) = [10]$ | A1 | Allow $R_y = 50 - 40\sin 16.3 - 30\cos 16.3$ |
| $R = \sqrt{R_x^2 + R_y^2}$ **and** $\theta = \tan^{-1}\!\left(\dfrac{R_y}{R_x}\right)$ | M1 | For using Pythagoras to find resultant force $R$ **and** trigonometry to find angle $\theta$ with $x$-axis |
| $R = 31.6$ N **and** $\theta = 18.4°$ with positive $x$-axis | A1 | 6 | |
**Alternative method for 3(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R_1 = 40 - 50 \times (7/25) = [26]$ | M1, A1 | Resolve forces along 40 N direction; allow $R_1 = 40 - 50\sin 16.3$ |
| $R_2 = 30 - 50 \times (24/25) = [-18]$ | M1, A1 | Resolve forces along 30 N direction; allow $R_2 = 30 - 50\cos 16.3$ |
| $R^2 = R_1^2 + R_2^2$ **and** $\arctan(-R_2/R_1)$ | M1 | Use Pythagoras **and** trigonometry |
| $R = 31.6$ N **and** direction is $34.7 - \alpha = 18.4°$ with positive $x$-axis | A1 | 6 | Using $\arctan(18/26) = 34.7°$ is angle between $R$ and 40 N force |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = 40$ | B1 | 1 | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{9a99a969-db40-4d29-bb37-ea7ac15cdc2d-2_476_659_897_742}
Coplanar forces of magnitudes $50 \mathrm {~N} , 40 \mathrm {~N}$ and 30 N act at a point $O$ in the directions shown in the diagram, where $\tan \alpha = \frac { 7 } { 24 }$.\\
(i) Find the magnitude and direction of the resultant of the three forces.\\
(ii) The force of magnitude 50 N is replaced by a force of magnitude $P \mathrm {~N}$ acting in the same direction. The resultant of the three forces now acts in the positive $x$-direction. Find the value of $P$.
\hfill \mbox{\textit{CAIE M1 2016 Q3 [7]}}