| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Constant acceleration vector (i and j) |
| Difficulty | Moderate -0.8 This is a straightforward M1 vector kinematics question requiring direct application of v = u + at in component form. Parts (a) and (b) involve routine calculations (finding magnitude and angle), while part (c) requires setting up a ratio condition for parallel vectors—all standard textbook exercises with no novel problem-solving required. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Velocity \(= (14\mathbf{i} - 5\mathbf{j}) + 2(-4\mathbf{i} + \mathbf{j})\) | M1 | Correct use of \(t=2\) to find the velocity (unsimplified) |
| Speed \(= \sqrt{6^2 + (-3)^2}\) | M1 | Use of Pythagoras to find speed when \(t=2\) with their velocity |
| Speed \(= \sqrt{45} = 3\sqrt{5} = 6.7\) (m s\(^{-1}\)) or better | A1 cso | \(\sqrt{45} = 3\sqrt{5} = 6.7\)(m s\(^{-1}\)) or better (6.70820...). Must come from correct velocity |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan^{-1}\left(\frac{3}{6}\right)\) with correct diagram showing components 6 and 3 | M1 A1ft | M1: Use trig to find equation in relevant angle e.g. \((90°-\theta)\) for their velocity. A1ft: Correct equation for relevant angle, ft on their v |
| \(27°\) or better OR \(333°\) or better; \(0.46\) rads or better OR \(5.8\) rads or better | A1 | Cao. No isw (A0 for negative answer) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{v} = (14\mathbf{i}-5\mathbf{j}) + (-4\mathbf{i}+\mathbf{j})T\) (allow \(t\)) | M1 | Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}T\) to obtain velocity vector in \(T\) |
| \(\frac{14-4T}{-5+T} = \frac{2}{-3}\) | M1 A1 | M1: Use ratios using their velocity to produce equation in \(T\) only (allow reciprocal and incorrect sign). A1: Correct equation in \(T\) only |
| \(T = 3.2\) | A1 | cao |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Velocity $= (14\mathbf{i} - 5\mathbf{j}) + 2(-4\mathbf{i} + \mathbf{j})$ | M1 | Correct use of $t=2$ to find the velocity (unsimplified) |
| Speed $= \sqrt{6^2 + (-3)^2}$ | M1 | Use of Pythagoras to find speed when $t=2$ with their velocity |
| Speed $= \sqrt{45} = 3\sqrt{5} = 6.7$ (m s$^{-1}$) or better | A1 cso | $\sqrt{45} = 3\sqrt{5} = 6.7$(m s$^{-1}$) or better (6.70820...). Must come from correct velocity |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan^{-1}\left(\frac{3}{6}\right)$ with correct diagram showing components 6 and 3 | M1 A1ft | M1: Use trig to find equation in relevant angle e.g. $(90°-\theta)$ for their velocity. A1ft: Correct equation for relevant angle, ft on their **v** |
| $27°$ or better OR $333°$ or better; $0.46$ rads or better OR $5.8$ rads or better | A1 | Cao. No isw (A0 for negative answer) |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = (14\mathbf{i}-5\mathbf{j}) + (-4\mathbf{i}+\mathbf{j})T$ (allow $t$) | M1 | Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}T$ to obtain velocity vector in $T$ |
| $\frac{14-4T}{-5+T} = \frac{2}{-3}$ | M1 A1 | M1: Use ratios using their velocity to produce equation in $T$ only (allow reciprocal and incorrect sign). A1: Correct equation in $T$ only |
| $T = 3.2$ | A1 | cao |\begin{enumerate}
\item A particle $P$ is moving with constant acceleration ( $- 4 \mathbf { i } + \mathbf { j }$ ) $\mathrm { ms } ^ { - 2 }$
\end{enumerate}
At time $t = 0 , P$ has velocity $( 14 \mathbf { i } - 5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$\\
(a) Find the speed of $P$ at time $t = 2$ seconds.\\
(b) Find the size of the angle between the direction of $\mathbf { i }$ and the direction of motion of $P$ at time $t = 2$ seconds.
At time $t = T$ seconds, $P$ is moving in the direction of vector ( $2 \mathbf { i } - 3 \mathbf { j }$ )\\
(c) Find the value of $T$
\hfill \mbox{\textit{Edexcel M1 2023 Q3 [10]}}