| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | January |
| Marks | 10 |
| 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 M1 equilibrium problem requiring resolution of forces in two directions and basic trigonometry to find angles. While it involves multiple steps (finding angles, resolving horizontally and vertically, solving simultaneous equations), these are routine techniques practiced extensively at this level 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 |
| \(PM = 3.5 - 2\tan 45° = 1.5\) OR \(PB = \sqrt{3.5^2 + \left(\frac{2}{\sin 45°}\right)^2 - 2\times3.5\times\left(\frac{2}{\sin 45°}\right)\cos 45°} = 2.5\) | M1 | Finding length of \(PM\) or \(PB\) |
| \(\tan\alpha = \frac{1.5}{2};\ \cos\alpha = \frac{4}{5};\ \sin\alpha = \frac{3}{5}\) OR \(\alpha = 37°\) or \((90°-\alpha) = 53°\) (at least 2SF) | A1 | Correct trig ratio for \(\alpha\) or \((90°-\alpha)\) or correct value |
| \(T_p\cos\alpha + T_Q\cos 45° = 6g\) | M1 A2 | Resolving vertically; \(-1\) each error |
| \(T_p\sin\alpha = T_Q\cos 45°\) | M1 A1 | Resolving horizontally |
| \(T_p = \frac{30g}{7} = 42\text{ N}\);\ \(T_Q = 36\) or \(35.6\text{ N}\) | DM1 A1; A1 | DM1 dependent on all THREE previous M marks. Units not needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{T_p}{\sin 45°} = \frac{6g}{\sin(45°+\alpha)}\) OR \(\frac{T_Q}{\sin(180°-\alpha)} = \frac{6g}{\sin(45°+\alpha)}\) | M1 A2 | \(-1\) ee |
| \(\frac{T_Q}{\sin(180°-\alpha)} = \frac{6g}{\sin(45°+\alpha)}\) OR \(\frac{T_p}{\sin 45°} = \frac{6g}{\sin(45°+\alpha)}\) OR \(\frac{T_p}{\sin 45°} = \frac{T_Q}{\sin(180°-\alpha)}\) | M1 A1 | Treat omission of \(g\) as one error |
# Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $PM = 3.5 - 2\tan 45° = 1.5$ OR $PB = \sqrt{3.5^2 + \left(\frac{2}{\sin 45°}\right)^2 - 2\times3.5\times\left(\frac{2}{\sin 45°}\right)\cos 45°} = 2.5$ | M1 | Finding length of $PM$ or $PB$ |
| $\tan\alpha = \frac{1.5}{2};\ \cos\alpha = \frac{4}{5};\ \sin\alpha = \frac{3}{5}$ OR $\alpha = 37°$ or $(90°-\alpha) = 53°$ (at least 2SF) | A1 | Correct trig ratio for $\alpha$ or $(90°-\alpha)$ or correct value |
| $T_p\cos\alpha + T_Q\cos 45° = 6g$ | M1 A2 | Resolving vertically; $-1$ each error |
| $T_p\sin\alpha = T_Q\cos 45°$ | M1 A1 | Resolving horizontally |
| $T_p = \frac{30g}{7} = 42\text{ N}$;\ $T_Q = 36$ or $35.6\text{ N}$ | DM1 A1; A1 | DM1 dependent on all THREE previous M marks. Units not needed |
**Alternative (Lami's Theorem):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{T_p}{\sin 45°} = \frac{6g}{\sin(45°+\alpha)}$ OR $\frac{T_Q}{\sin(180°-\alpha)} = \frac{6g}{\sin(45°+\alpha)}$ | M1 A2 | $-1$ ee |
| $\frac{T_Q}{\sin(180°-\alpha)} = \frac{6g}{\sin(45°+\alpha)}$ OR $\frac{T_p}{\sin 45°} = \frac{6g}{\sin(45°+\alpha)}$ OR $\frac{T_p}{\sin 45°} = \frac{T_Q}{\sin(180°-\alpha)}$ | M1 A1 | Treat omission of $g$ as one error |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0d5a56ba-6a33-4dc8-b612-d2957211124f-14_451_551_255_699}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A small metal box of mass 6 kg is attached at $B$ to two ropes $B P$ and $B Q$. The fixed points $P$ and $Q$ are on a horizontal ceiling and $P Q = 3.5 \mathrm {~m}$. The box hangs in equilibrium at a vertical distance of 2 m below the line $P Q$, with the ropes in a vertical plane and with angle $B Q P = 45 ^ { \circ }$, as shown in Figure 3. The box is modelled as a particle and the ropes are modelled as light inextensible strings. Find\\
(i) the tension in $B P$,\\
(ii) the tension in $B Q$.
\hfill \mbox{\textit{Edexcel M1 2019 Q5 [10]}}