| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Constant acceleration vector problems |
| Difficulty | Moderate -0.8 This is a straightforward vector kinematics question requiring standard integration and application of suvat-style equations with constant acceleration. Parts (a)-(c) are direct applications of v = u + at and r = ut + ½at², while parts (d)(i)-(ii) involve simple substitution and magnitude calculation. No problem-solving insight or novel reasoning required—purely procedural mechanics. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 20i + (-0.4i + 0.5j)t\) | M1, A1 | Use of constant acceleration equation to find expression for \(v\). Any correct expression. |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = (20 - 0.4t)i + 0.5tj\) | M1 | Simplifying \(v\). (May be implied.) (Missing brackets may be condoned if followed by correct working.) |
| \(20 - 0.4t = 0\) | m1 | Putting \(i\) component equal to zero. |
| \(t = \frac{20}{0.4} = 50\) seconds | A1 | Correct time. Candidates who are able to see the correct time without supporting working gain full marks. |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 20i \times t + \frac{1}{2}(-0.4i + 0.5j) \times t^2\) | M1, A1 | Use of constant acceleration equation to find expression for \(r\). Any correct expression. |
| Answer | Marks | Guidance |
|---|---|---|
| \(r = 20i \times 100 + \frac{1}{2}(-0.4i + 0.5j) \times 100^2\) | m1 | Substituting \(t = 100\) into their expression for \(r\) (dependent on M1 in part (c)) |
| \(= 2000i - 2000i + 2500j\) | A1 | Correct simplified position vector ie 2500j. |
| \(= 2500j\) | A1 | Conclusion that helicopter is due north provided their position vector is of the form \(kj\), where \(k>0\). Note if integration is used there is no need to prove that the constant is zero. Note marks for (d) (i) can be awarded if part c scores zero. |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = (20 - 0.4 \times 100)i + 0.5 \times 100j\) | m1 | Substituting \(t = 100\) into their expression for \(v\) (dependent on M1 in part (a)) or use of other constant acceleration equation and their position vector (dependent on M1 in part (c)). |
| \(= -20i + 50j\) | A1 | Correct simplified velocity. |
| \(v = \sqrt{20^2 + 50^2} = 53.9\) | A1 | Correct speed (accept \(10\sqrt{29}\)). Note marks for (d) (ii) can be awarded if part a scores zero. |
**5(a)**
| $v = 20i + (-0.4i + 0.5j)t$ | M1, A1 | Use of constant acceleration equation to find expression for $v$. Any correct expression. |
**5(b)**
| $v = (20 - 0.4t)i + 0.5tj$ | M1 | Simplifying $v$. (May be implied.) (Missing brackets may be condoned if followed by correct working.) |
| $20 - 0.4t = 0$ | m1 | Putting $i$ component equal to zero. |
| $t = \frac{20}{0.4} = 50$ seconds | A1 | Correct time. Candidates who are able to see the correct time without supporting working gain full marks. |
Condone $\frac{20i}{0.4i} = 50$
**5(c)**
| $r = 20i \times t + \frac{1}{2}(-0.4i + 0.5j) \times t^2$ | M1, A1 | Use of constant acceleration equation to find expression for $r$. Any correct expression. |
**5(d)(i)**
| $r = 20i \times 100 + \frac{1}{2}(-0.4i + 0.5j) \times 100^2$ | m1 | Substituting $t = 100$ into their expression for $r$ (dependent on M1 in part (c)) |
| $= 2000i - 2000i + 2500j$ | A1 | Correct simplified position vector ie 2500j. |
| $= 2500j$ | A1 | Conclusion that helicopter is due north provided their position vector is of the form $kj$, where $k>0$. Note if integration is used there is no need to prove that the constant is zero. Note marks for (d) (i) can be awarded if part c scores zero. |
| Total | 3 | |
**5(d)(ii)**
| $v = (20 - 0.4 \times 100)i + 0.5 \times 100j$ | m1 | Substituting $t = 100$ into their expression for $v$ (dependent on M1 in part (a)) or use of other constant acceleration equation and their position vector (dependent on M1 in part (c)). |
| $= -20i + 50j$ | A1 | Correct simplified velocity. |
| $v = \sqrt{20^2 + 50^2} = 53.9$ | A1 | Correct speed (accept $10\sqrt{29}$). Note marks for (d) (ii) can be awarded if part a scores zero. |5 The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively. A helicopter moves horizontally with a constant acceleration of $( - 0.4 \mathbf { i } + 0.5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$. At time $t = 0$, the helicopter is at the origin and has velocity $20 \mathrm { i } \mathrm { m } \mathrm { s } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for the velocity of the helicopter at time $t$ seconds.
\item Find the time when the helicopter is travelling due north.
\item Find an expression for the position vector of the helicopter at time $t$ seconds.
\item When $t = 100$ :
\begin{enumerate}[label=(\roman*)]
\item show that the helicopter is due north of the origin;
\item find the speed of the helicopter.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q5 [13]}}