| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Standard +0.3 This is a straightforward vector mechanics problem requiring resolution of forces and use of F=ma. Part (a) uses the condition that acceleration is parallel to motion direction (standard technique), and part (b) involves basic kinematics with constant acceleration. The multi-step nature and vector notation add slight complexity, but the methods are standard A-level mechanics procedures with no novel insight required. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((4\mathbf{i}-\mathbf{j})+(\lambda\mathbf{i}+\mu\mathbf{j}) = (4+\lambda)\mathbf{i}+(-1+\mu)\mathbf{j}\) | M1 | Adding forces, \(\mathbf{i}\)'s and \(\mathbf{j}\)'s collected or single column vector |
| Use ratios to obtain equation in \(\lambda\) and \(\mu\) only | M1 | Must use ratios; ignore equation if they go on to use ratios |
| \(\dfrac{(4+\lambda)}{(-1+\mu)} = \dfrac{3}{1}\) or \(\dfrac{\frac{1}{4}(4+\lambda)}{\frac{1}{4}(-1+\mu)} = \dfrac{3}{1}\) | A1 | Correct equation |
| \(\lambda - 3\mu + 7 = 0\) | A1* | Allow \(0 = \lambda - 3\mu + 7\) but nothing else |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\lambda=2 \Rightarrow \mu=3\); Resultant force \(= (6\mathbf{i}+2\mathbf{j})\) N | M1 | \(\lambda\) and \(\mu\) must be substituted to find resultant |
| \((6\mathbf{i}+2\mathbf{j}) = 4\mathbf{a}\) OR \( | (6\mathbf{i}+2\mathbf{j}) | = 4a\) |
| Use \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(\mathbf{u}=\mathbf{0}\), their \(\mathbf{a}\) and \(t=4\) | DM1 | Dependent on previous M; or integrate twice with \(\mathbf{u}=\mathbf{0}\), \(t=4\) |
| \(\mathbf{r} = \frac{1}{2}\times\frac{(6\mathbf{i}+2\mathbf{j})}{4}\times 4^2 = (12\mathbf{i}+4\mathbf{j})\) | ||
| \(\sqrt{12^2+4^2}\) | M1 | Use Pythagoras with square root to find magnitude |
| \(\sqrt{160},\ 2\sqrt{40},\ 4\sqrt{10}\) or \(13\) or better (m) | A1 | cao |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(4\mathbf{i}-\mathbf{j})+(\lambda\mathbf{i}+\mu\mathbf{j}) = (4+\lambda)\mathbf{i}+(-1+\mu)\mathbf{j}$ | M1 | Adding forces, $\mathbf{i}$'s and $\mathbf{j}$'s collected or single column vector |
| Use ratios to obtain equation in $\lambda$ and $\mu$ only | M1 | Must use ratios; ignore equation if they go on to use ratios |
| $\dfrac{(4+\lambda)}{(-1+\mu)} = \dfrac{3}{1}$ or $\dfrac{\frac{1}{4}(4+\lambda)}{\frac{1}{4}(-1+\mu)} = \dfrac{3}{1}$ | A1 | Correct equation |
| $\lambda - 3\mu + 7 = 0$ | A1* | Allow $0 = \lambda - 3\mu + 7$ but nothing else |
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\lambda=2 \Rightarrow \mu=3$; Resultant force $= (6\mathbf{i}+2\mathbf{j})$ N | M1 | $\lambda$ and $\mu$ must be substituted to find resultant |
| $(6\mathbf{i}+2\mathbf{j}) = 4\mathbf{a}$ OR $|(6\mathbf{i}+2\mathbf{j})| = 4a$ | M1 | Use $\mathbf{F}=4\mathbf{a}$ or $|\mathbf{F}|=4a$; independent mark |
| Use $\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ with $\mathbf{u}=\mathbf{0}$, their $\mathbf{a}$ and $t=4$ | DM1 | Dependent on previous M; or integrate twice with $\mathbf{u}=\mathbf{0}$, $t=4$ |
| $\mathbf{r} = \frac{1}{2}\times\frac{(6\mathbf{i}+2\mathbf{j})}{4}\times 4^2 = (12\mathbf{i}+4\mathbf{j})$ | | |
| $\sqrt{12^2+4^2}$ | M1 | Use Pythagoras with square root to find magnitude |
| $\sqrt{160},\ 2\sqrt{40},\ 4\sqrt{10}$ or $13$ or better (m) | A1 | cao |
---\begin{enumerate}
\item \hspace{0pt} [In this question, $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors.]
\end{enumerate}
A particle $P$ of mass 4 kg is at rest at the point $A$ on a smooth horizontal plane.\\
At time $t = 0$, two forces, $\mathbf { F } _ { 1 } = ( 4 \mathbf { i } - \mathbf { j } ) \mathrm { N }$ and $\mathbf { F } _ { 2 } = ( \lambda \mathbf { i } + \mu \mathbf { j } ) \mathrm { N }$, where $\lambda$ and $\mu$ are constants, are applied to $P$
Given that $P$ moves in the direction of the vector ( $3 \mathbf { i } + \mathbf { j }$ )\\
(a) show that
$$\lambda - 3 \mu + 7 = 0$$
At time $t = 4$ seconds, $P$ passes through the point $B$.\\
Given that $\lambda = 2$\\
(b) find the length of $A B$.
\hfill \mbox{\textit{Edexcel Paper 3 2022 Q3 [9]}}