| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.3 This is a straightforward M1 vector mechanics question requiring standard techniques: vector addition, magnitude calculation using Pythagoras, F=ma, and SUVAT with vectors. All parts follow routine procedures with no problem-solving insight needed, making it slightly easier than average but not trivial due to the multi-part nature and need for careful algebraic manipulation. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.02e Two-dimensional constant acceleration: with vectors3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Resultant \(= (6\mathbf{i} - 3\mathbf{j}) + (3\mathbf{i} + 15\mathbf{j}) = 9\mathbf{i} + 12\mathbf{j}\) | M1 | Summing the two vectors |
| Correct resultant | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Magnitude \(= \sqrt{9^2 + 12^2} = 15 \text{ N}\) | M1 | Finding magnitude with an addition sign |
| Correct magnitude based on their answer to (a) | A1F |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(1.5m = 9\) or \(2m = 12\), \(m = 6 \text{ kg}\) | M1 | Applying Newton's second law to one or both components |
| Correct mass, follow through their answer to (a) | A1F | Do not award if vector division with 2 components has been used, e.g. \(\frac{9\mathbf{i}+12\mathbf{j}}{1.5\mathbf{i}+2\mathbf{j}} = 6\) or \(6\mathbf{i}+6\mathbf{j}\) etc without a correct previous statement gives M0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{r} = \frac{1}{2}(1.5\mathbf{i} + 2\mathbf{j})t^2\) | M1 | Using a constant acceleration equation to find position vector with \(\mathbf{u} = 0\mathbf{i} + 0\mathbf{j}\) |
| Correct position vector | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathbf{r} = \frac{1}{2}(1.5\mathbf{i} + 2\mathbf{j}) \times 2^2 = 3\mathbf{i} + 4\mathbf{j}\) | M1 | Finding position vector when \(t = 2\); \((\mathbf{r} = (1.5\mathbf{i}+2\mathbf{j}) \times 2 = 3\mathbf{i}+4\mathbf{j}\) scores M0 unless clear how 2 was obtained) |
| \(d = \sqrt{(3)^2 + (4)^2} = \sqrt{25} = 5\) | A1 | Correct distance |
# Question 6:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Resultant $= (6\mathbf{i} - 3\mathbf{j}) + (3\mathbf{i} + 15\mathbf{j}) = 9\mathbf{i} + 12\mathbf{j}$ | M1 | Summing the two vectors |
| Correct resultant | A1 | |
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Magnitude $= \sqrt{9^2 + 12^2} = 15 \text{ N}$ | M1 | Finding magnitude with an addition sign |
| Correct magnitude based on their answer to (a) | A1F | |
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $1.5m = 9$ or $2m = 12$, $m = 6 \text{ kg}$ | M1 | Applying Newton's second law to one or both components |
| Correct mass, follow through their answer to (a) | A1F | Do not award if vector division with 2 components has been used, e.g. $\frac{9\mathbf{i}+12\mathbf{j}}{1.5\mathbf{i}+2\mathbf{j}} = 6$ or $6\mathbf{i}+6\mathbf{j}$ etc without a correct previous statement gives M0A0 |
## Part (d)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{r} = \frac{1}{2}(1.5\mathbf{i} + 2\mathbf{j})t^2$ | M1 | Using a constant acceleration equation to find position vector with $\mathbf{u} = 0\mathbf{i} + 0\mathbf{j}$ |
| Correct position vector | A1 | |
## Part (d)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathbf{r} = \frac{1}{2}(1.5\mathbf{i} + 2\mathbf{j}) \times 2^2 = 3\mathbf{i} + 4\mathbf{j}$ | M1 | Finding position vector when $t = 2$; $(\mathbf{r} = (1.5\mathbf{i}+2\mathbf{j}) \times 2 = 3\mathbf{i}+4\mathbf{j}$ scores M0 unless clear how 2 was obtained) |
| $d = \sqrt{(3)^2 + (4)^2} = \sqrt{25} = 5$ | A1 | Correct distance |
---6 Two forces, $\mathbf { P } = ( 6 \mathbf { i } - 3 \mathbf { j } )$ newtons and $\mathbf { Q } = ( 3 \mathbf { i } + 15 \mathbf { j } )$ newtons, act on a particle. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are perpendicular.
\begin{enumerate}[label=(\alph*)]
\item Find the resultant of $\mathbf { P }$ and $\mathbf { Q }$.
\item Calculate the magnitude of the resultant of $\mathbf { P }$ and $\mathbf { Q }$.
\item When these two forces act on the particle, it has an acceleration of $( 1.5 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$. Find the mass of the particle.
\item The particle was initially at rest at the origin.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for the position vector of the particle when the forces have been applied to the particle for $t$ seconds.
\item Find the distance of the particle from the origin when the forces have been applied to the particle for 2 seconds.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q6 [10]}}