| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: verify/find meeting point (position vector method) |
| Difficulty | Moderate -0.3 This is a standard M3 relative velocity problem with straightforward vector arithmetic. Part (a) requires simple vector subtraction, part (b) applies the standard position formula, and part (c) involves setting components equal to check for collision. While it requires understanding of relative motion concepts, the execution is routine with no novel problem-solving required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(_A v_B = v_B - v_A = (3\mathbf{i}+4\mathbf{j})-(5\mathbf{i}-\mathbf{j})\) | M1 | |
| \(= -2\mathbf{i}+5\mathbf{j}\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(_A r_{0B} = (40\mathbf{i}-90\mathbf{j})-(-60\mathbf{i}+160\mathbf{j}) = 100\mathbf{i}-250\mathbf{j}\) | M1 | Simplification not necessary |
| \(_A r_B = (100\mathbf{i}-250\mathbf{j})+(-2\mathbf{i}+5\mathbf{j})t\) | m1 A1F | Total: 3 |
| Alternative: \(r_A = (60\mathbf{i}+160\mathbf{j})+(5\mathbf{i}-\mathbf{j})t\) | M1 | |
| \(r_B = (40\mathbf{i}-90\mathbf{j})+(3\mathbf{i}+4\mathbf{j})t\) | ||
| \(_A r_B = [(40\mathbf{i}-90\mathbf{j})+(3\mathbf{i}+4\mathbf{j})t]-[(60\mathbf{i}+160\mathbf{j})+(5\mathbf{i}-\mathbf{j})t]\) | m1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(_A r_B = (100-2t)\mathbf{i}+(-250+5t)\mathbf{j}\) | M1 | Collecting \(\mathbf{i}\) and \(\mathbf{j}\) terms |
| \((100-2t)=0 \Leftrightarrow t=50\) | A1F | |
| \((-250+5t)=0 \Leftrightarrow t=50\) | ||
| \(\therefore A\) and \(B\) would collide | E1 | Total: 3 |
| Alternative: \([(100-2t)\mathbf{i}+(-250+5t)\mathbf{j}]\cdot(-2\mathbf{i}+5\mathbf{j})=0\) | M1 | |
| \(-200+4t-1250+25t=0 \Rightarrow t=50\) | A1 | |
| \(\ | {_A r_B}\ | \sqrt{(100-2\times50)^2+(-250+5\times50)^2}=0\) |
| \(\therefore A\) and \(B\) would collide | E1 |
## Question 2:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $_A v_B = v_B - v_A = (3\mathbf{i}+4\mathbf{j})-(5\mathbf{i}-\mathbf{j})$ | M1 | |
| $= -2\mathbf{i}+5\mathbf{j}$ | A1 | **Total: 2** |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $_A r_{0B} = (40\mathbf{i}-90\mathbf{j})-(-60\mathbf{i}+160\mathbf{j}) = 100\mathbf{i}-250\mathbf{j}$ | M1 | Simplification not necessary |
| $_A r_B = (100\mathbf{i}-250\mathbf{j})+(-2\mathbf{i}+5\mathbf{j})t$ | m1 A1F | **Total: 3** |
| **Alternative:** $r_A = (60\mathbf{i}+160\mathbf{j})+(5\mathbf{i}-\mathbf{j})t$ | M1 | |
| $r_B = (40\mathbf{i}-90\mathbf{j})+(3\mathbf{i}+4\mathbf{j})t$ | | |
| $_A r_B = [(40\mathbf{i}-90\mathbf{j})+(3\mathbf{i}+4\mathbf{j})t]-[(60\mathbf{i}+160\mathbf{j})+(5\mathbf{i}-\mathbf{j})t]$ | m1A1 | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $_A r_B = (100-2t)\mathbf{i}+(-250+5t)\mathbf{j}$ | M1 | Collecting $\mathbf{i}$ and $\mathbf{j}$ terms |
| $(100-2t)=0 \Leftrightarrow t=50$ | A1F | |
| $(-250+5t)=0 \Leftrightarrow t=50$ | | |
| $\therefore A$ and $B$ would collide | E1 | **Total: 3** |
| **Alternative:** $[(100-2t)\mathbf{i}+(-250+5t)\mathbf{j}]\cdot(-2\mathbf{i}+5\mathbf{j})=0$ | M1 | |
| $-200+4t-1250+25t=0 \Rightarrow t=50$ | A1 | |
| $\|{_A r_B}\|\sqrt{(100-2\times50)^2+(-250+5\times50)^2}=0$ | | |
| $\therefore A$ and $B$ would collide | E1 | |
---2 The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed due east and due north respectively.\\
Two runners, Albina and Brian, are running on level parkland with constant velocities of $( 5 \mathbf { i } - \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ and $( 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ respectively. Initially, the position vectors of Albina and Brian are $( - 60 \mathbf { i } + 160 \mathbf { j } ) \mathrm { m }$ and $( 40 \mathbf { i } - 90 \mathbf { j } ) \mathrm { m }$ respectively, relative to a fixed origin in the parkland.
\begin{enumerate}[label=(\alph*)]
\item Write down the velocity of Brian relative to Albina.
\item Find the position vector of Brian relative to Albina $t$ seconds after they leave their initial positions.
\item Hence determine whether Albina and Brian will collide if they continue running with the same velocities.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2008 Q2 [8]}}