| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Constant acceleration vector (i and j) |
| Difficulty | Standard +0.3 This is a standard M1 vector kinematics question requiring application of v = u + at, use of speed formula |v| = √(i² + j²), and bearing calculation. Part (b) involves algebraic manipulation to reach a given quadratic, which is routine. The multi-part structure and bearing calculation add slight complexity, but all techniques are standard M1 fare with no novel problem-solving required. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((5\mathbf{i}-8\mathbf{j})+5(-\lambda\mathbf{i}+2\lambda\mathbf{j})\) (m s\(^{-1}\)) | M1 A1 | Use of \(\mathbf{v}=\mathbf{u}+\mathbf{a}t\) to form vector expression in \(\lambda\) and \(t\). A1 correct unsimplified with \(t=5\). Column vectors allowed for M but not A |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(13 = \sqrt{(5-5\lambda)^2+(-8+10\lambda)^2}\) | M1 A1 | Collect \(\mathbf{i}\)'s and \(\mathbf{j}\)'s, correct use of Pythagoras to form equation in \(\lambda\) |
| \(25\lambda^2 - 42\lambda - 16 = 0\)* | A1* | cso. Expand brackets and correctly reach given answer. Allow \(0=25\lambda^2-42\lambda-16\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((-2\mathbf{i}+4\mathbf{j})\) seen or implied | B1 | Or column vector |
| \((5\mathbf{i}-8\mathbf{j})+(-2\mathbf{i}+4\mathbf{j})4\) | M1 A1 | Complete method for velocity at \(t=4\). A1 correct unsimplified; correct velocity is \(\mathbf{v}=-3\mathbf{i}+8\mathbf{j}\) |
| Using trig with velocity vector at \(t=4\), e.g. \(\tan^{-1}\left(\pm\frac{8}{3}\right)\), \(\tan^{-1}\left(\pm\frac{3}{8}\right)\), \(\sin^{-1}\left(\pm\frac{8}{\sqrt{73}}\right)\) | M1 | Use their velocity vector at \(t=4\) with trig to find relevant angle |
| \(339°\) | A1 | cao. Degrees sign not required. N.B. if both values of \(\lambda\) used, can score all marks except last one |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(5\mathbf{i}-8\mathbf{j})+5(-\lambda\mathbf{i}+2\lambda\mathbf{j})$ (m s$^{-1}$) | M1 A1 | Use of $\mathbf{v}=\mathbf{u}+\mathbf{a}t$ to form vector expression in $\lambda$ and $t$. A1 correct unsimplified with $t=5$. Column vectors allowed for M but not A |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $13 = \sqrt{(5-5\lambda)^2+(-8+10\lambda)^2}$ | M1 A1 | Collect $\mathbf{i}$'s and $\mathbf{j}$'s, correct use of Pythagoras to form equation in $\lambda$ |
| $25\lambda^2 - 42\lambda - 16 = 0$* | A1* | cso. Expand brackets and correctly reach given answer. Allow $0=25\lambda^2-42\lambda-16$ |
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(-2\mathbf{i}+4\mathbf{j})$ seen or implied | B1 | Or column vector |
| $(5\mathbf{i}-8\mathbf{j})+(-2\mathbf{i}+4\mathbf{j})4$ | M1 A1 | Complete method for velocity at $t=4$. A1 correct unsimplified; correct velocity is $\mathbf{v}=-3\mathbf{i}+8\mathbf{j}$ |
| Using trig with velocity vector at $t=4$, e.g. $\tan^{-1}\left(\pm\frac{8}{3}\right)$, $\tan^{-1}\left(\pm\frac{3}{8}\right)$, $\sin^{-1}\left(\pm\frac{8}{\sqrt{73}}\right)$ | M1 | Use their velocity vector at $t=4$ with trig to find relevant angle |
| $339°$ | A1 | cao. Degrees sign not required. N.B. if both values of $\lambda$ used, can score all marks except last one |\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors directed due east and due north respectively.]
\end{enumerate}
A particle $P$ moves with constant acceleration $( - \lambda \mathbf { i } + 2 \lambda \mathbf { j } ) \mathrm { ms } ^ { - 2 }$, where $\lambda$ is a positive constant.
At time $t = 0$, the velocity of $P$ is $( 5 \mathbf { i } - 8 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$\\
(a) Find the velocity of $P$ when $t = 5 \mathrm {~s}$, giving your answer in terms of $\mathbf { i } , \mathbf { j }$ and $\lambda$.
The speed of $P$ when $t = 5 \mathrm {~s}$ is $13 \mathrm {~ms} ^ { - 1 }$\\
(b) Show that
$$25 \lambda ^ { 2 } - 42 \lambda - 16 = 0$$
(c) Find the direction of motion of $P$ when $t = 4 \mathrm {~s}$, giving your answer as a bearing to the nearest degree.
\hfill \mbox{\textit{Edexcel M1 2023 Q4 [10]}}