| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions using given trigonometric ratios, followed by a straightforward application of F=ma. While it involves multiple forces and angles, the method is routine and well-practiced at this level, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For resolving forces in the \(x\) and \(y\) directions (or for sketching a marked triangle of forces) | |
| \(F\cos\theta = 2.5 \times 24 \div 25 + 2.6 \times 5 \div 13\) | A1 | \((= 3.4)\) |
| \(F\sin\theta = 2.6 \times 12 \div 13 - 2.5 \times 7 \div 25\) | A1 | \((= 1.7)\) |
| M1 | For using \(F^2 = (F\cos\theta)^2 + (F\sin\theta)^2\) to find \(F\) or \(\tan\theta = F\sin\theta \div F\cos\theta\) to find \(\theta\) | |
| For \(F = 3.80\) N or \(\tan\theta = 0.5\) | A1 | |
| For \(\tan\theta = 0.5\) or \(F = 3.80\) N | B1 | [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([3.80 = 0.5a]\) | M1 | For using Newton's \(2^{nd}\) law with the magnitude of the resultant force equal to the value of \(F\) found |
| Acceleration is \(7.60\ ms^{-2}\) | A1ft | ft value of \(F\) found in (i) |
| Direction is \(26.6°\) clockwise from \(+ve\) \(x\)-axis | B1ft | [3] ft value of \(\tan\theta\) found in (i) |
## Question 6:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For resolving forces in the $x$ and $y$ directions (or for sketching a marked triangle of forces) |
| $F\cos\theta = 2.5 \times 24 \div 25 + 2.6 \times 5 \div 13$ | A1 | $(= 3.4)$ |
| $F\sin\theta = 2.6 \times 12 \div 13 - 2.5 \times 7 \div 25$ | A1 | $(= 1.7)$ |
| | M1 | For using $F^2 = (F\cos\theta)^2 + (F\sin\theta)^2$ to find $F$ or $\tan\theta = F\sin\theta \div F\cos\theta$ to find $\theta$ |
| For $F = 3.80$ N or $\tan\theta = 0.5$ | A1 | |
| For $\tan\theta = 0.5$ or $F = 3.80$ N | B1 | [6] |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[3.80 = 0.5a]$ | M1 | For using Newton's $2^{nd}$ law with the magnitude of the resultant force equal to the value of $F$ found |
| Acceleration is $7.60\ ms^{-2}$ | A1ft | ft value of $F$ found in (i) |
| Direction is $26.6°$ clockwise from $+ve$ $x$-axis | B1ft | [3] ft value of $\tan\theta$ found in (i) |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{2c628138-0729-4160-a95c-d6ab0f199cc5-3_639_939_1260_603}
A particle $P$ of mass 0.5 kg lies on a smooth horizontal plane. Horizontal forces of magnitudes $F \mathrm {~N}$, 2.5 N and 2.6 N act on $P$. The directions of the forces are as shown in the diagram, where $\tan \alpha = \frac { 12 } { 5 }$ and $\tan \beta = \frac { 7 } { 24 }$.\\
(i) Given that $P$ is in equilibrium, find the values of $F$ and $\tan \theta$.\\
(ii) The force of magnitude $F \mathrm {~N}$ is removed. Find the magnitude and direction of the acceleration with which $P$ starts to move.
\hfill \mbox{\textit{CAIE M1 2013 Q6 [9]}}