| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Particle motion: 2D direction/bearing |
| Difficulty | Moderate -0.3 This is a straightforward 2D kinematics question requiring standard application of v = u + at in vector form. Part (a) is direct substitution, part (b) requires setting the j-component to zero (routine), and part (c) is a simple verification using Pythagoras. Slightly easier than average due to the mechanical nature of the calculations with no conceptual challenges. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{v}=4\mathbf{j}+(3\mathbf{i}-5\mathbf{j})t\) | M1A1 | Use of \(\mathbf{v}=\mathbf{u}+\mathbf{a}t\) and \(\mathbf{u}\neq0\) or integration; correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{v}=3t\mathbf{i}+(4-5t)\mathbf{j}\); \(4-5t=0\) | M1 | \(\mathbf{j}\) component of velocity equal to zero |
| \(t=0.8\text{ s}\) | A1 | Correct \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{v}=12\mathbf{i}-16\mathbf{j}\) | M1 | Finding velocity when \(t=4\) |
| A1 | Correct velocity | |
| \(v=\sqrt{12^2+16^2}=20\) | dM1 | Finding the magnitude |
| A1 | Correct speed from correct working |
## Question 7:
**Part (a)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{v}=4\mathbf{j}+(3\mathbf{i}-5\mathbf{j})t$ | M1A1 | Use of $\mathbf{v}=\mathbf{u}+\mathbf{a}t$ and $\mathbf{u}\neq0$ or integration; correct expression |
**Part (b)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{v}=3t\mathbf{i}+(4-5t)\mathbf{j}$; $4-5t=0$ | M1 | $\mathbf{j}$ component of velocity equal to zero |
| $t=0.8\text{ s}$ | A1 | Correct $t$ |
**Part (c)**
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{v}=12\mathbf{i}-16\mathbf{j}$ | M1 | Finding velocity when $t=4$ |
| | A1 | Correct velocity |
| $v=\sqrt{12^2+16^2}=20$ | dM1 | Finding the magnitude |
| | A1 | Correct speed from correct working |
---7 A particle moves on a smooth horizontal surface with acceleration $( 3 \mathbf { i } - 5 \mathbf { j } ) \mathrm { ms } ^ { - 2 }$. Initially the velocity of the particle is $4 \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the velocity of the particle at time $t$ seconds.
\item Find the time when the particle is travelling in the $\mathbf { i }$ direction.
\item Show that when $t = 4$ the speed of the particle is $20 \mathrm {~ms} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2005 Q7 [8]}}