| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| 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 coplanar forces problem requiring resolution into perpendicular components and basic trigonometry. Part (i) involves routine vector addition, while part (ii) requires setting one component to zero and recalculating—both are textbook exercises with no novel insight needed, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03p Resultant forces: using vectors |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For resolving forces horizontally or vertically o.e. | |
| \(25\cos 30 - 15\cos 40\ (= 10.1599\ldots)\) | A1 | |
| \(25\sin 30 + 15\sin 40 - 30\ (= -7.8581\ldots)\) | A1 | |
| M1 | For using a method for either magnitude or direction | |
| Magnitude \(= \sqrt{(10.15\ldots^2 + 7.858\ldots^2)} = 12.8\text{ N}\) | A1 | Magnitude \(= 12.844\ldots\) |
| Angle \(37.7°\) below the horizontal in the direction \(BA\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F\cos 40 = 25\cos 30\) | M1 | For equating forces in the direction \(BC\) to zero |
| \(F = 28.3\) | A1 | \(F = 28.2628\ldots\) |
| New resultant force \(= 28.26\ldots\sin 40 + 25\sin 30 - 30 = 0.667\text{ N upwards}\) | B1 |
## Question 5(i):
| | M1 | For resolving forces horizontally or vertically o.e. |
|---|---|---|
| $25\cos 30 - 15\cos 40\ (= 10.1599\ldots)$ | A1 | |
| $25\sin 30 + 15\sin 40 - 30\ (= -7.8581\ldots)$ | A1 | |
| | M1 | For using a method for either magnitude or direction |
| Magnitude $= \sqrt{(10.15\ldots^2 + 7.858\ldots^2)} = 12.8\text{ N}$ | A1 | Magnitude $= 12.844\ldots$ |
| Angle $37.7°$ below the horizontal in the direction $BA$ | A1 | |
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## Question 5(ii):
| $F\cos 40 = 25\cos 30$ | M1 | For equating forces in the direction $BC$ to zero |
|---|---|---|
| $F = 28.3$ | A1 | $F = 28.2628\ldots$ |
| New resultant force $= 28.26\ldots\sin 40 + 25\sin 30 - 30 = 0.667\text{ N upwards}$ | B1 | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{98a5537b-d503-4a42-bbfe-0bd221084ee0-06_449_654_260_742}
Coplanar forces, of magnitudes $15 \mathrm {~N} , 25 \mathrm {~N}$ and 30 N , act at a point $B$ on the line $A B C$ in the directions shown in the diagram.\\
(i) Find the magnitude and direction of the resultant force.\\
(ii) The force of magnitude 15 N is now replaced by a force of magnitude $F \mathrm {~N}$ acting in the same direction. The new resultant force has zero component in the direction $B C$. Find the value of $F$, and find also the magnitude and direction of the new resultant force.\\
\hfill \mbox{\textit{CAIE M1 2018 Q5 [9]}}