| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Velocity from acceleration and initial conditions |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question testing basic vector calculus concepts: finding magnitude (speed), parallel vectors (equal components), differentiation for acceleration, and perpendicular vectors (zero i-component). All parts require routine application of standard techniques with no problem-solving insight needed, making it easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(7\mathbf{i} - 3\mathbf{j}\) seen or implied | B1 | cao |
| \(\sqrt{7^2 + (-3)^2}\) | M1 | Use of Pythagoras, including square root, on a velocity vector at \(t = 0\) |
| \(\sqrt{58}\), 7.6 or better (m s\(^{-1}\)) | A1 | cao. Must come from a correct v. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(t^2 - 3t + 7 = 2t^2 - 3\) OR \(\frac{t^2 - 3t + 7}{2t^2 - 3} = \frac{1}{1} = 1\) | M1 | Equating i and j components of v or ratio of 1:1 to obtain a quadratic in \(t\) only. If using a constant \(k\), it must be eliminated. N.B. M0 if they write \(\mathbf{i} + \mathbf{j} = (t^2 - 3t + 7)\mathbf{i} + (2t^2 - 3)\mathbf{j}\) |
| \(t = 2\) only | A1 | N.B. Allow M1A1 for correct trial and error giving \(\mathbf{v} = 5\mathbf{i} + 5\mathbf{j}\) when \(t=2\), but M0 if they don't get \(t=2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate v wrt \(t\) to give a vector | M1 | At least one power decreasing by 1 in each component. M0 if clearly dividing by \(t\). Both i and j needed. Allow recovery if i and j disappear then reappear. |
| \((2t - 3)\mathbf{i} + 4t\mathbf{j}\) | A1 | cao (must be a vector). N.B. Allow M1A0 for \(2t - 3\mathbf{i} + 4t\mathbf{j}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(2t - 3 = 0\) | M1 | \(2t - 3 = 0\) or derivative of i-component of v \(= 0\). N.B. M0 if they equate derivative of both components to zero. |
| \(t = 1.5\) | A1 | cao. Correct answer with no working can score both marks. |
## Question 3:
**Part (a)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $7\mathbf{i} - 3\mathbf{j}$ seen or implied | B1 | cao |
| $\sqrt{7^2 + (-3)^2}$ | M1 | Use of Pythagoras, including square root, on a **velocity** vector at $t = 0$ |
| $\sqrt{58}$, 7.6 or better (m s$^{-1}$) | A1 | cao. Must come from a correct **v**. |
**Part (b)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $t^2 - 3t + 7 = 2t^2 - 3$ **OR** $\frac{t^2 - 3t + 7}{2t^2 - 3} = \frac{1}{1} = 1$ | M1 | Equating **i** and **j** components of **v** or ratio of 1:1 to obtain a quadratic in $t$ only. If using a constant $k$, it must be eliminated. N.B. M0 if they write $\mathbf{i} + \mathbf{j} = (t^2 - 3t + 7)\mathbf{i} + (2t^2 - 3)\mathbf{j}$ |
| $t = 2$ only | A1 | N.B. Allow M1A1 for correct trial and error giving $\mathbf{v} = 5\mathbf{i} + 5\mathbf{j}$ when $t=2$, but M0 if they don't get $t=2$ |
**Part (c)**
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate **v** wrt $t$ to give a vector | M1 | At least one power decreasing by 1 in **each** component. M0 if clearly dividing by $t$. Both **i** and **j** needed. Allow recovery if **i** and **j** disappear then reappear. |
| $(2t - 3)\mathbf{i} + 4t\mathbf{j}$ | A1 | cao (must be a vector). N.B. Allow M1A0 for $2t - 3\mathbf{i} + 4t\mathbf{j}$ |
**Part (d)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $2t - 3 = 0$ | M1 | $2t - 3 = 0$ or derivative of **i**-component of **v** $= 0$. N.B. M0 if they equate derivative of both components to zero. |
| $t = 1.5$ | A1 | cao. Correct answer with no working can score both marks. |
---\begin{enumerate}
\item At time $t$ seconds, where $t \geqslant 0$, a particle $P$ has velocity $\mathbf { v } \mathrm { ms } ^ { - 1 }$ where
\end{enumerate}
$$\mathbf { v } = \left( t ^ { 2 } - 3 t + 7 \right) \mathbf { i } + \left( 2 t ^ { 2 } - 3 \right) \mathbf { j }$$
Find\\
(a) the speed of $P$ at time $t = 0$\\
(b) the value of $t$ when $P$ is moving parallel to $( \mathbf { i } + \mathbf { j } )$\\
(c) the acceleration of $P$ at time $t$ seconds\\
(d) the value of $t$ when the direction of the acceleration of $P$ is perpendicular to $\mathbf { i }$
\hfill \mbox{\textit{Edexcel Paper 3 2023 Q3 [9]}}