| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring routine differentiation of vector functions and application of F=ma. All parts involve standard procedures (differentiate for velocity, differentiate again for acceleration, multiply by mass) with no problem-solving insight needed. The polynomial differentiation is elementary and the vector arithmetic is basic, making this easier than average A-level material. |
| Spec | 1.10c Magnitude and direction: of vectors3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(v = (6t^2 - 2t)i + (1 - 12t^2)j\) | M1, A1, A1 | 3 marks |
| (b)(i) \(v\left(\frac{1}{3}\right) = \left(\frac{6}{9} - \frac{2}{3}\right)i + \left(1 - \frac{12}{9}\right)j = -\frac{1}{3}j\) | M1, A1 | 2 marks |
| (ii) Travelling due south | A1 ft | 1 mark |
| (c) \(a = (12t - 2)i - 24tj\) and \(a(4) = 46i - 96j\) | M1, A1, A1 | 3 marks |
| (d) \(F = 6(46i - 96j) = 276i - 576j\) | M1 | — |
| \(F = \sqrt{276^2 + 576^2} = 639 \text{ N}\) | M1, A1 | 3 marks |
**(a)** $v = (6t^2 - 2t)i + (1 - 12t^2)j$ | M1, A1, A1 | 3 marks | Differentiating both components; one component correct; second component correct
**(b)(i)** $v\left(\frac{1}{3}\right) = \left(\frac{6}{9} - \frac{2}{3}\right)i + \left(1 - \frac{12}{9}\right)j = -\frac{1}{3}j$ | M1, A1 | 2 marks | Substituting the value for $t$ into their $v$; correct velocity
**(ii)** Travelling due south | A1 ft | 1 mark | Correct description (Follow through from $v = -\frac{1}{3}j$)
**(c)** $a = (12t - 2)i - 24tj$ and $a(4) = 46i - 96j$ | M1, A1, A1 | 3 marks | Differentiating their velocity; correct acceleration at time $t$; correct acceleration at $t = 4$
**(d)** $F = 6(46i - 96j) = 276i - 576j$ | M1 | — | Apply Newton's second law correctly
$F = \sqrt{276^2 + 576^2} = 639 \text{ N}$ | M1, A1 | 3 marks | Finding magnitude; correct magnitude
Or: $a = \sqrt{46^2 + 96^2} = 106.45$, then $F = 6 \times 106.45 = 639 \text{ N}$
**Total for Question 1: 12 marks**
---1 A particle moves in a horizontal plane, in which the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively. At time $t$ seconds, its position vector, $\mathbf { r }$ metres, is given by
$$\mathbf { r } = \left( 2 t ^ { 3 } - t ^ { 2 } + 6 \right) \mathbf { i } + \left( 8 - 4 t ^ { 3 } + t \right) \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the velocity of the particle at time $t$.
\item \begin{enumerate}[label=(\roman*)]
\item Find the velocity of the particle when $t = \frac { 1 } { 3 }$.
\item State the direction in which the particle is travelling at this time.
\end{enumerate}\item Find the acceleration of the particle when $t = 4$.
\item The mass of the particle is 6 kg . Find the magnitude of the resultant force on the particle when $t = 4$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2006 Q1 [12]}}