| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.8 This is a straightforward M1 equilibrium and resultant force question requiring only standard techniques: (a) sum forces to zero for equilibrium, (b) find resultant magnitude and use tan⁻¹ for angle, (c) apply F=ma. All steps are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 3.03d Newton's second law: 2D vectors3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{F}_3 + (3c\mathbf{i} + 4c\mathbf{j}) + (-14\mathbf{i} + 7\mathbf{j}) = \mathbf{0}\) | M1 | Uses the vector sum of all 3 forces being equal to zero. N.B. \(\mathbf{F}_3 = \mathbf{F}_1 + \mathbf{F}_2\) is M0 |
| \(\mathbf{F}_3 = (14 - 3c)\mathbf{i} + (-7 - 4c)\mathbf{j}\) | A1 | Cao. Must be in terms of \(c\), i and j but allow uncollected i's and j's |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{F}_1 + \mathbf{F}_2 = (6-14)\mathbf{i} + (8+7)\mathbf{j}\) \(= -8\mathbf{i} + 15\mathbf{j}\) | M1 | Finds the resultant using \(\mathbf{F}_1 + \mathbf{F}_2\) or \(-\) their \(\mathbf{F}_3\) |
| Find any relevant angle for their resultant | M1 | Uses trig to find a relevant angle for their resultant |
| Any of \(\tan^{-1}\left(\pm\frac{8}{15}\right)\), \(\tan^{-1}\left(\pm\frac{15}{8}\right)\), \(\sin^{-1}\left(\pm\frac{8}{17}\right)\), \(\cos^{-1}\left(\pm\frac{8}{17}\right)\),... | A1ft | Any correct relevant angle (does not need to be acute), ft on their resultant |
| \(120°\) or better (\(118.0724...\)) OR \(240°\) or better (\(241.9276...\)); In radians \(2.1\) or better (\(2.0607..\)) OR \(4.2\) or better (\(4.2224...\)) | A1 | Cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{(-8)^2 + 15^2}\) or their acceleration \(\sqrt{\left(\frac{-8}{m}\right)^2 + \left(\frac{15}{m}\right)^2}\) | M1 | Use of Pythagoras on their resultant force or their acceleration |
| \( | \text{their } \mathbf{R} | = 8.5m\) or their Resultant \(= m\mathbf{a}\) |
| A correct equation in \(m\) only e.g. \(17 = m \times 8.5\) | A1ft | ft on their resultant |
| \(m = 2\) | A1 | cso |
# Question 2:
## Part 2(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{F}_3 + (3c\mathbf{i} + 4c\mathbf{j}) + (-14\mathbf{i} + 7\mathbf{j}) = \mathbf{0}$ | M1 | Uses the vector **sum** of all 3 forces being equal to zero. **N.B.** $\mathbf{F}_3 = \mathbf{F}_1 + \mathbf{F}_2$ is M0 |
| $\mathbf{F}_3 = (14 - 3c)\mathbf{i} + (-7 - 4c)\mathbf{j}$ | A1 | Cao. Must be in terms of $c$, **i** and **j** but allow uncollected **i**'s and **j**'s |
## Part 2(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{F}_1 + \mathbf{F}_2 = (6-14)\mathbf{i} + (8+7)\mathbf{j}$ $= -8\mathbf{i} + 15\mathbf{j}$ | M1 | Finds the resultant using $\mathbf{F}_1 + \mathbf{F}_2$ or $-$ their $\mathbf{F}_3$ |
| Find any relevant angle for their resultant | M1 | Uses trig to find a relevant angle for their resultant |
| Any of $\tan^{-1}\left(\pm\frac{8}{15}\right)$, $\tan^{-1}\left(\pm\frac{15}{8}\right)$, $\sin^{-1}\left(\pm\frac{8}{17}\right)$, $\cos^{-1}\left(\pm\frac{8}{17}\right)$,... | A1ft | Any **correct** relevant angle (does not need to be acute), ft on their resultant |
| $120°$ or better ($118.0724...$) **OR** $240°$ or better ($241.9276...$); In radians $2.1$ or better ($2.0607..$) **OR** $4.2$ or better ($4.2224...$) | A1 | Cso |
## Part 2(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{(-8)^2 + 15^2}$ or their acceleration $\sqrt{\left(\frac{-8}{m}\right)^2 + \left(\frac{15}{m}\right)^2}$ | M1 | Use of Pythagoras on their resultant force **or** their acceleration |
| $|\text{their } \mathbf{R}| = 8.5m$ **or** their Resultant $= m\mathbf{a}$ | M1 | Allow their $\mathbf{R} = 8.5\,m$ |
| A correct equation in $m$ only e.g. $17 = m \times 8.5$ | A1ft | ft on their resultant |
| $m = 2$ | A1 | cso |
---\begin{enumerate}
\item A particle $P$ rests in equilibrium on a smooth horizontal plane.
\end{enumerate}
A system of three forces, $\mathbf { F } _ { 1 } \mathrm {~N} , \mathbf { F } _ { 2 } \mathrm {~N}$ and $\mathbf { F } _ { 3 } \mathrm {~N}$ where
$$\begin{aligned}
& \mathbf { F } _ { 1 } = ( 3 c \mathbf { i } + 4 c \mathbf { j } ) \\
& \mathbf { F } _ { 2 } = ( - 14 \mathbf { i } + 7 \mathbf { j } )
\end{aligned}$$
is applied to $P$.\\
Given that $P$ remains in equilibrium,\\
(a) find $\mathbf { F } _ { 3 }$ in terms of $c$, $\mathbf { i }$ and $\mathbf { j }$.
The force $\mathbf { F } _ { 3 }$ is removed from the system.\\
Given that $c = 2$\\
(b) find the size of the angle between the direction of $\mathbf { i }$ and the direction of the resultant force acting on $P$.
The mass of $P$ is $m \mathrm {~kg}$.\\
Given that the magnitude of the acceleration of $P$ is $8.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$\\
(c) find the value of $m$.
\hfill \mbox{\textit{Edexcel M1 2023 Q2 [10]}}