| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| 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. Students apply Newton's first law (constant velocity → zero resultant force) and solve two simultaneous equations using basic trigonometry. The setup is clear, the method is routine, and it's slightly easier than average due to straightforward arithmetic and no conceptual surprises. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Perpendicular to direction of motion: \(500\sin 40° = P\sin\alpha°\) | M1 A1 | Correct number of terms; condone sign errors and sin/cos confusion |
| Parallel to direction of motion: \(500\cos 40° + P\cos\alpha° = 900\) | M1 A1 | If using \(F=ma\) must have \(a=0\); condone sign errors and sin/cos confusion |
| (i) \(\tan\alpha° = \dfrac{500\sin 40°}{900 - 500\cos 40°}\) | M1 | Form and solve equation in \(\alpha\) (do not penalise accuracy) |
| \(\alpha = 32\) or better \((31.8683...)\) | A1 | Accept 32, 31.9, 31.87,… as final answer |
| (ii) \(P = \dfrac{500\sin 40°}{\sin 31.868...°}\) | M1 | Form and solve equation in \(P\) (do not penalise accuracy) |
| \(P = 610\) or better \((608.736...)\) | A1 | Accept 610, 609, 608.7,… as final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Cosine Rule: \(P^2 = 500^2 + 900^2 - 2\times500\times900\cos 40°\) | M1 A1 | |
| Sine Rule: \(\dfrac{P}{\sin 40°} = \dfrac{500}{\sin\alpha°} = \dfrac{900}{\sin(140°-\alpha°)}\) | M1 A1 | Any two |
| (i) \(\alpha = 32\) or better | M1 A1 | |
| (ii) \(P = 610\) or better | M1 A1 |
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Perpendicular to direction of motion: $500\sin 40° = P\sin\alpha°$ | M1 A1 | Correct number of terms; condone sign errors and sin/cos confusion |
| Parallel to direction of motion: $500\cos 40° + P\cos\alpha° = 900$ | M1 A1 | If using $F=ma$ must have $a=0$; condone sign errors and sin/cos confusion |
| **(i)** $\tan\alpha° = \dfrac{500\sin 40°}{900 - 500\cos 40°}$ | M1 | Form and solve equation in $\alpha$ (do not penalise accuracy) |
| $\alpha = 32$ or better $(31.8683...)$ | A1 | Accept 32, 31.9, 31.87,… as **final** answer |
| **(ii)** $P = \dfrac{500\sin 40°}{\sin 31.868...°}$ | M1 | Form and solve equation in $P$ (do not penalise accuracy) |
| $P = 610$ or better $(608.736...)$ | A1 | Accept 610, 609, 608.7,… as **final** answer |
**Alternative (Triangle of Forces):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Cosine Rule: $P^2 = 500^2 + 900^2 - 2\times500\times900\cos 40°$ | M1 A1 | |
| Sine Rule: $\dfrac{P}{\sin 40°} = \dfrac{500}{\sin\alpha°} = \dfrac{900}{\sin(140°-\alpha°)}$ | M1 A1 | Any two |
| **(i)** $\alpha = 32$ or better | M1 A1 | |
| **(ii)** $P = 610$ or better | M1 A1 | |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{84c0eead-0a87-4d87-b33d-794a94bb466c-18_502_1429_280_319}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A boat is pulled along a river at a constant speed by two ropes.\\
The banks of the river are parallel and the boat travels horizontally in a straight line, parallel to the riverbanks.
\begin{itemize}
\item The tension in the first rope is 500 N acting at an angle of $40 ^ { \circ }$ to the direction of motion, as shown in Figure 3.
\item The tension in the second rope is $P$ newtons, acting at an angle of $\alpha ^ { \circ }$ to the direction of motion, also shown in Figure 3.
\item The resistance to motion of the boat as it moves through the water is a constant force of magnitude 900 N
\end{itemize}
The boat is modelled as a particle. The ropes are modelled as being light and lying in a horizontal plane.
Use the model to find\\
(i) the value of $\alpha$\\
(ii) the value of $P$
\hfill \mbox{\textit{Edexcel M1 2023 Q6 [8]}}