| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Moderate -0.3 This is a standard two-string equilibrium problem requiring resolution of forces in two directions and solving simultaneous equations. While it involves trigonometry and careful angle work, it's a textbook M1 exercise with no novel insight required—slightly easier than average due to being a routine application of equilibrium principles. |
| Spec | 3.03e Resolve forces: two dimensions3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T_P \cos 55 = T_Q \cos 35\) | M1 A1 | M1 for resolving horizontally, correct no. of terms, both \(T_P\) and \(T_Q\) resolved. M0 if \(T_P = T_Q\) assumed |
| \(T_P \sin 55 + T_Q \sin 35 = 2g\) | M1 A1 | M1 for resolving vertically, correct no. of terms, both resolved. M0 if \(T_P = T_Q\) assumed |
| Eliminating \(T_P\) or \(T_Q\) | M1 | Third M1 (independent) for eliminating either tension |
| \(T_P = 16\text{N}\) or \(16.1\text{N}\); \(T_Q = 11\text{N}\) or \(11.2\text{N}\) | A1 A1 | N.B. If both given to more than 3SF, deduct the third A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Along \(RP\): \(T_P = 2g\cos 35° = 16\text{N}\) or \(16.1\text{N}\) | M1 M1 A1 A1 | First M2 for resolving along string. A1 correct equation (\(T_P = 2g\sin 35°\) scores M2A0A0) |
| Along \(RQ\): \(T_Q = 2g\cos 55° = 11\text{N}\) or \(11.2\text{N}\) | M1 A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T_P = 2g\cos 35°= 16\text{N}\) or \(16.1\text{N}\) | M2, A1(third), A1(third) | |
| \(T_Q = T_P \tan 35°\) or \(\sqrt{(2g)^2 - T_P^2} = 11\text{N}\) or \(11.2\text{N}\) | M1, A1(second), A1(fourth) | N.B. If Sine Rule used with 35°, 55°, 80° in triangle: 3 M marks available, at most 1 A mark |
## Question 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T_P \cos 55 = T_Q \cos 35$ | M1 A1 | M1 for resolving horizontally, correct no. of terms, both $T_P$ and $T_Q$ resolved. M0 if $T_P = T_Q$ assumed |
| $T_P \sin 55 + T_Q \sin 35 = 2g$ | M1 A1 | M1 for resolving vertically, correct no. of terms, both resolved. M0 if $T_P = T_Q$ assumed |
| Eliminating $T_P$ or $T_Q$ | M1 | Third M1 (independent) for eliminating either tension |
| $T_P = 16\text{N}$ or $16.1\text{N}$; $T_Q = 11\text{N}$ or $11.2\text{N}$ | A1 A1 | N.B. If both given to more than 3SF, deduct the third A1 |
**ALT 1 (resolving along each string):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Along $RP$: $T_P = 2g\cos 35° = 16\text{N}$ or $16.1\text{N}$ | M1 M1 A1 A1 | First M2 for resolving along string. A1 correct equation ($T_P = 2g\sin 35°$ scores M2A0A0) |
| Along $RQ$: $T_Q = 2g\cos 55° = 11\text{N}$ or $11.2\text{N}$ | M1 A1 A1 | |
**ALT 2 (Triangle of Forces):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T_P = 2g\cos 35°= 16\text{N}$ or $16.1\text{N}$ | M2, A1(third), A1(third) | |
| $T_Q = T_P \tan 35°$ or $\sqrt{(2g)^2 - T_P^2} = 11\text{N}$ or $11.2\text{N}$ | M1, A1(second), A1(fourth) | N.B. If Sine Rule used with 35°, 55°, 80° in triangle: 3 M marks available, at most 1 A mark |3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{de3245a7-cf6e-423e-8689-9a074bdbc23b-04_540_958_116_482}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle of mass 2 kg is suspended from a horizontal ceiling by two light inextensible strings, $P R$ and $Q R$. The particle hangs at $R$ in equilibrium, with the strings in a vertical plane. The string $P R$ is inclined at $55 ^ { \circ }$ to the horizontal and the string $Q R$ is inclined at $35 ^ { \circ }$ to the horizontal, as shown in Figure 1.
\section*{Find}
(i) the tension in the string $P R$,\\
(ii) the tension in the string $Q R$.
\hfill \mbox{\textit{Edexcel M1 2015 Q3 [7]}}