| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Standard +0.3 This is a straightforward statics problem requiring basic trigonometry to find angles in the triangle, then resolving forces vertically and horizontally. Part (a) involves simple ratio comparison after finding tensions via resolution, and part (b) is direct substitution. The geometry is given clearly and the problem follows standard textbook methods with no novel insight required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Obtains both \(\sin^{-1}\!\left(\frac{0.6}{0.8}\right)=48.59°\) and \(\sin^{-1}\!\left(\frac{0.6}{1.2}\right)=30°\) | B1 | AO 1.1b. Accept complementary angles or exact values \(\frac{\sqrt{3}}{2}\) and \(\frac{\sqrt{7}}{4}\) |
| Resolves forces horizontally to form equilibrium equation, one component correct. Or uses triangle of forces and sine rule: \(T_{OA}\cos A = T_{OB}\cos B\) | M1 | AO 3.3 |
| Correct equation with angles substituted: \(T_{OA}=T_{OB}\dfrac{\cos 30}{\cos 48.59}\) | A1 | AO 1.1b |
| Rearranges to show \(T_{OA}=kT_{OB}\) where \(1.305\leq k\leq 1.325\); completes argument to conclude tension in shorter string is over 30% more than tension in longer string: \(\therefore T_{OA}=1.309T_{OB}\), \(\therefore T_{OA}>1.3T_{OB}\) | R1 | AO 2.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_{OA} = 2g \times \text{their ratio}\) | B1 (1.1b) | Obtains \(T_{OA}\) using ratio from part (a) |
| \(mg = T_{OA}\sin A + T_{OB}\sin B\) | M1 (3.3) | Resolves forces vertically to form three-term equilibrium equation, at least two terms correct; or triangle of forces with sine rule |
| \(mg = T_{OA}\sin A + T_{OB}\sin B\) (fully correct) | A1 (1.1b) | Forms fully correct equation of forces in equilibrium |
| \(mg = 2.6g\sin 48.59 + 2g\sin 30\), \(m = 3.0\) | A1F (3.4) | Substitutes \(T_{OA}\), \(T_{OB}\) and correct angles; AWRT \(m = 3\); FT ratio from part (a) provided \(m =\) AWRT 3 |
# Question 18(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Obtains both $\sin^{-1}\!\left(\frac{0.6}{0.8}\right)=48.59°$ and $\sin^{-1}\!\left(\frac{0.6}{1.2}\right)=30°$ | B1 | AO 1.1b. Accept complementary angles or exact values $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{7}}{4}$ |
| Resolves forces horizontally to form equilibrium equation, one component correct. Or uses triangle of forces and sine rule: $T_{OA}\cos A = T_{OB}\cos B$ | M1 | AO 3.3 |
| Correct equation with angles substituted: $T_{OA}=T_{OB}\dfrac{\cos 30}{\cos 48.59}$ | A1 | AO 1.1b |
| Rearranges to show $T_{OA}=kT_{OB}$ where $1.305\leq k\leq 1.325$; completes argument to conclude tension in shorter string is **over** 30% more than tension in longer string: $\therefore T_{OA}=1.309T_{OB}$, $\therefore T_{OA}>1.3T_{OB}$ | R1 | AO 2.1 |
## Question 18(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_{OA} = 2g \times \text{their ratio}$ | B1 (1.1b) | Obtains $T_{OA}$ using ratio from part (a) |
| $mg = T_{OA}\sin A + T_{OB}\sin B$ | M1 (3.3) | Resolves forces vertically to form three-term equilibrium equation, at least two terms correct; or triangle of forces with sine rule |
| $mg = T_{OA}\sin A + T_{OB}\sin B$ (fully correct) | A1 (1.1b) | Forms fully correct equation of forces in equilibrium |
| $mg = 2.6g\sin 48.59 + 2g\sin 30$, $m = 3.0$ | A1F (3.4) | Substitutes $T_{OA}$, $T_{OB}$ and correct angles; AWRT $m = 3$; FT ratio from part (a) provided $m =$ AWRT 3 |
---18 An object, $O$, of mass $m$ kilograms is hanging from a ceiling by two light, inelastic strings of different lengths.
The shorter string, of length 0.8 metres, is fixed to the ceiling at $A$.\\
The longer string, of length 1.2 metres, is fixed to the ceiling at $B$.\\
This object hangs 0.6 metres directly below the ceiling as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-28_252_940_667_552}
18
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the shorter string is over $30 \%$ more than the tension in the longer string.\\
18
\item The tension in the longer string is known to be $2 g$ newtons.
Find the value of $m$.\\
A rough wooden ramp is 10 metres long and is inclined at an angle of $25 ^ { \circ }$ above the horizontal. The bottom of the ramp is at the point $O$.
A crate of mass 20 kg is at rest at the point $A$ on the ramp.\\
The crate is pulled up the ramp using a rope attached to the crate.\\
Once in motion, the rope remains taut and parallel to the line of greatest slope of the ramp.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-30_252_842_804_598}
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2022 Q18 [8]}}