| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Year | 2020 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Constant acceleration vector problems |
| Difficulty | Moderate -0.8 This is a straightforward kinematics question requiring standard integration of constant acceleration to find velocity and position vectors. Part (a) uses v = u + at directly, part (b) solves a simple quadratic from the j-component, and part (c) substitutes to find λ. All steps are routine A-level mechanics with no problem-solving insight required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) or integrate to give: \(\mathbf{v} = (-2\mathbf{i} + 2\mathbf{j}) + 2(4\mathbf{i} - 5\mathbf{j})\) | M1 | For any complete method to give a v expression with correct no. of terms with \(t=2\) used; if integrating, must see initial velocity as constant. Allow sign errors. |
| \((6\mathbf{i} - 8\mathbf{j})\) (m s\(^{-1}\)) | A1 | cao; isw if they go on to find the speed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Solve using \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) or integration (M0 if \(\mathbf{u} = \mathbf{0}\)) | M1 | Complete method for j component of displacement in \(t\) (or \(T\)) only, using \(\mathbf{a} = (4\mathbf{i} - 5\mathbf{j})\); allow sign errors |
| \(-4.5\mathbf{j} = 2t\mathbf{j} - \frac{1}{2}t^2 5\mathbf{j}\) (j terms only) | A1 | Correct j vector equation in \(t\) or \(T\); ignore i terms |
| Equate j components: \(-4.5 = 2T - \frac{5}{2}T^2\) | M1 | Must have earned 1st M mark; equation in \(T\) only; must be 3-term quadratic in \(T\) |
| \(T = 1.8\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Substitute \(T\) value into i component equation: \(\lambda = -2T + \frac{1}{2}T^2 \times 4\) | M1 | Must have earned 1st M mark in (b); complete method with equation in \(\lambda\) only (no i's) from appropriate displacement; M0 if \(\mathbf{a} = \mathbf{0}\) used; expression for \(\lambda\) must be quadratic in \(T\) |
| \(\lambda = 2.9\) or \(2.88\) or \(\frac{72}{25}\) oe | A1 | cao |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ or integrate to give: $\mathbf{v} = (-2\mathbf{i} + 2\mathbf{j}) + 2(4\mathbf{i} - 5\mathbf{j})$ | M1 | For any complete method to give a **v** expression with correct no. of terms with $t=2$ used; if integrating, must see initial velocity as constant. Allow sign errors. |
| $(6\mathbf{i} - 8\mathbf{j})$ (m s$^{-1}$) | A1 | cao; isw if they go on to find the speed |
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Solve using $\mathbf{r} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ or integration (M0 if $\mathbf{u} = \mathbf{0}$) | M1 | Complete method for **j** component of displacement in $t$ (or $T$) only, using $\mathbf{a} = (4\mathbf{i} - 5\mathbf{j})$; allow sign errors |
| $-4.5\mathbf{j} = 2t\mathbf{j} - \frac{1}{2}t^2 5\mathbf{j}$ (**j** terms only) | A1 | Correct **j** vector equation in $t$ or $T$; ignore **i** terms |
| Equate **j** components: $-4.5 = 2T - \frac{5}{2}T^2$ | M1 | Must have earned 1st M mark; equation in $T$ only; must be 3-term quadratic in $T$ |
| $T = 1.8$ | A1 | cao |
## Question 2(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitute $T$ value into **i** component equation: $\lambda = -2T + \frac{1}{2}T^2 \times 4$ | M1 | Must have earned 1st M mark in (b); complete method with equation in $\lambda$ only (no **i**'s) from appropriate displacement; M0 if $\mathbf{a} = \mathbf{0}$ used; expression for $\lambda$ must be quadratic in $T$ |
| $\lambda = 2.9$ or $2.88$ or $\frac{72}{25}$ oe | A1 | cao |
---\begin{enumerate}
\item A particle $P$ moves with acceleration $( 4 \mathbf { i } - 5 \mathbf { j } ) \mathrm { ms } ^ { - 2 }$
\end{enumerate}
At time $t = 0 , P$ is moving with velocity $( - 2 \mathbf { i } + 2 \mathbf { j } ) \mathrm { ms } ^ { - 1 }$\\
(a) Find the velocity of $P$ at time $t = 2$ seconds.
At time $t = 0 , P$ passes through the origin $O$.\\
At time $t = T$ seconds, where $T > 0$, the particle $P$ passes through the point $A$.\\
The position vector of $A$ is ( $\lambda \mathbf { i } - 4.5 \mathbf { j }$ )m relative to $O$, where $\lambda$ is a constant.\\
(b) Find the value of $T$.\\
(c) Hence find the value of $\lambda$
\hfill \mbox{\textit{Edexcel Paper 3 2020 Q2 [8]}}