| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| 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 equilibrium problem requiring resolution of forces in two perpendicular directions, followed by algebraic manipulation to reach a given result and verification. The trigonometry is straightforward (tan α = 4/3 gives sin/cos via 3-4-5 triangle), and part (b) is purely verification by substitution. Slightly easier than average due to the guided nature and standard techniques. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resolving either direction | M1 | Correct number of terms, allow sign errors, allow sin/cos mix. Do not allow with just \(\sin\alpha\) and \(\cos\alpha\) |
| \((33+15)\times\frac{3}{5} = P\cos\theta + 30\cos\theta\) OR \((33+15)\cos\left(\tan^{-1}\frac{4}{3}\right) = P\cos\theta + 30\cos\theta\) OR \(19.8+9 = P\cos\theta + 30\cos\theta\) | A1 | OE, but see note for final A1. Allow: \(28.8=(P+30)\cos\theta\); \((33+15)\cos53(.1)=P\cos\theta+30\cos\theta\); \(19.81+9.01=P\cos\theta+30\cos\theta\); \(19.86+9.03=P\cos\theta+30\cos\theta\) |
| \(15\times\frac{4}{5}+30\sin\theta = 33\times\frac{4}{5}+P\sin\theta\) OR \(15\sin\left(\tan^{-1}\frac{4}{3}\right)+30\sin\theta = 33\sin\left(\tan^{-1}\frac{4}{3}\right)+P\sin\theta\) OR \(12+30\sin\theta = 26.4+P\sin\theta\) | A1 | OE, but see note for final A1. Allow: \(14.4=(30-P)\sin\theta\); \(15\sin53(.1)+30\sin\theta=33\sin53(.1)+P\sin\theta\); \(12.00+30\sin\theta=26.39+P\sin\theta\); \(11.98+30\sin\theta=26.35+P\sin\theta\) |
| \([\text{Use } \cos^2\theta+\sin^2\theta=1 \text{ with}]\ \cos\theta=\frac{28.8}{P+30}\) and \(\sin\theta=\frac{14.4}{30-P}\) to get \(\left(\frac{14.4}{30-P}\right)^2+\left(\frac{28.8}{P+30}\right)^2=1\) | A1 | AG. Must have evidence of where 28.8 and 14.4 come from. A0 for any error seen. A0 if use of inexact angles seen. Any inexact decimals seen for force components scores M1A1A1A0 max 3/4. If exact values of \(\sin\alpha\) and \(\cos\alpha\) not shown, \(28.8=(P+30)\cos\theta\) or \(14.4=(30-P)\sin\theta\) from no working scores M1A1A1A0 max 3/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sub \(P=6\) into \(\left(\frac{14.4}{30-P}\right)^2+\left(\frac{28.8}{P+30}\right)^2\) to get \(\left[\left(\frac{14.4}{24}\right)^2+\left(\frac{28.8}{36}\right)^2\right]=\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=0.36+0.64=1\) | B1 | Must see either \(\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=1\) or \(0.36+0.64=1\) as minimum working |
| \(\theta = 36.9\) | B1 | AWRT 36.9 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolving either direction | M1 | Correct number of terms, allow sign errors, allow sin/cos mix. Do not allow with just $\sin\alpha$ and $\cos\alpha$ |
| $(33+15)\times\frac{3}{5} = P\cos\theta + 30\cos\theta$ OR $(33+15)\cos\left(\tan^{-1}\frac{4}{3}\right) = P\cos\theta + 30\cos\theta$ OR $19.8+9 = P\cos\theta + 30\cos\theta$ | A1 | OE, but see note for final A1. Allow: $28.8=(P+30)\cos\theta$; $(33+15)\cos53(.1)=P\cos\theta+30\cos\theta$; $19.81+9.01=P\cos\theta+30\cos\theta$; $19.86+9.03=P\cos\theta+30\cos\theta$ |
| $15\times\frac{4}{5}+30\sin\theta = 33\times\frac{4}{5}+P\sin\theta$ OR $15\sin\left(\tan^{-1}\frac{4}{3}\right)+30\sin\theta = 33\sin\left(\tan^{-1}\frac{4}{3}\right)+P\sin\theta$ OR $12+30\sin\theta = 26.4+P\sin\theta$ | A1 | OE, but see note for final A1. Allow: $14.4=(30-P)\sin\theta$; $15\sin53(.1)+30\sin\theta=33\sin53(.1)+P\sin\theta$; $12.00+30\sin\theta=26.39+P\sin\theta$; $11.98+30\sin\theta=26.35+P\sin\theta$ |
| $[\text{Use } \cos^2\theta+\sin^2\theta=1 \text{ with}]\ \cos\theta=\frac{28.8}{P+30}$ and $\sin\theta=\frac{14.4}{30-P}$ to get $\left(\frac{14.4}{30-P}\right)^2+\left(\frac{28.8}{P+30}\right)^2=1$ | A1 | AG. Must have evidence of where 28.8 and 14.4 come from. A0 for any error seen. A0 if use of inexact angles seen. Any inexact decimals seen for force components scores M1A1A1A0 max 3/4. If exact values of $\sin\alpha$ and $\cos\alpha$ not shown, $28.8=(P+30)\cos\theta$ or $14.4=(30-P)\sin\theta$ from no working scores M1A1A1A0 max 3/4 |
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## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sub $P=6$ into $\left(\frac{14.4}{30-P}\right)^2+\left(\frac{28.8}{P+30}\right)^2$ to get $\left[\left(\frac{14.4}{24}\right)^2+\left(\frac{28.8}{36}\right)^2\right]=\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=0.36+0.64=1$ | B1 | Must see either $\left(\frac{3}{5}\right)^2+\left(\frac{4}{5}\right)^2=1$ or $0.36+0.64=1$ as minimum working |
| $\theta = 36.9$ | B1 | AWRT 36.9 |
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\includegraphics[max width=\textwidth, alt={}, center]{e5ee28f2-5876-4149-9a77-18c5792c1bd8-04_442_636_264_758}
Coplanar forces of magnitudes $30 \mathrm {~N} , 15 \mathrm {~N} , 33 \mathrm {~N}$ and $P \mathrm {~N}$ act at a point in the directions shown in the diagram, where $\tan \alpha = \frac { 4 } { 3 }$. The system is in equilibrium.
\begin{enumerate}[label=(\alph*)]
\item Show that $\left( \frac { 14.4 } { 30 - P } \right) ^ { 2 } + \left( \frac { 28.8 } { P + 30 } \right) ^ { 2 } = 1$.
\item Verify that $P = 6$ satisfies this equation and find the value of $\theta$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q3 [6]}}