| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: find bearing/direction to intercept (exact intercept) |
| Difficulty | Challenging +1.2 This is a standard relative velocity interception problem requiring vector resolution, the cosine rule, and systematic application of mechanics formulae. While it involves multiple steps and careful geometric reasoning with bearings, it follows a well-established method taught in M4 with no novel insights required—moderately above average difficulty due to the multi-part nature and potential for errors in bearing conversions. |
| 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 multiplication1.10h Vectors in kinematics: uniform acceleration in vector form3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct velocity triangle | B1 | Or \((6\sin\beta)t = 4t + 50\sin 60\) and \((6\cos\beta)t = 50\cos 60\) (or in terms of \(\alpha\) only) |
| \(\frac{6}{\sin 150} = \frac{4}{\sin\alpha}\) | M1 | Implies previous B1 sine rule leading to \(\alpha = \ldots\) (allow this mark for \(\sin 30\)); Genuine attempt to solve for M1 |
| Bearing is \(\beta = \alpha + 60° = 079.5°\) | A1 | 79.471220... |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{6}{\sin 150} = \frac{w}{\sin(30-19.47...)}\) | M1* | Use of sine rule with \(30 -\) their \(\alpha\) (or \(150 -\) their \(\alpha\)); Or cosine rule \(w^2 = 4^2 + 6^2 - 2(4)(6)\cos(30-\alpha)\); Or \(\sqrt{(6\sin\beta - 4)^2 + (6\cos\beta)^2}\) |
| \(w = 2.19\) | A1 | 2.1927526... |
| \(t = \frac{50}{w} = 22.8\) | M1 dep*, A1 | M1 for use of \(s = ut\) with their \(w\); 22.802389... |
| [4] |
# Question 1:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct velocity triangle | B1 | Or $(6\sin\beta)t = 4t + 50\sin 60$ and $(6\cos\beta)t = 50\cos 60$ (or in terms of $\alpha$ only) |
| $\frac{6}{\sin 150} = \frac{4}{\sin\alpha}$ | M1 | Implies previous B1 sine rule leading to $\alpha = \ldots$ (allow this mark for $\sin 30$); Genuine attempt to solve for M1 |
| Bearing is $\beta = \alpha + 60° = 079.5°$ | A1 | 79.471220... |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{6}{\sin 150} = \frac{w}{\sin(30-19.47...)}$ | M1* | Use of sine rule with $30 -$ their $\alpha$ (or $150 -$ their $\alpha$); Or cosine rule $w^2 = 4^2 + 6^2 - 2(4)(6)\cos(30-\alpha)$; Or $\sqrt{(6\sin\beta - 4)^2 + (6\cos\beta)^2}$ |
| $w = 2.19$ | A1 | 2.1927526... |
| $t = \frac{50}{w} = 22.8$ | M1 dep*, A1 | M1 for use of $s = ut$ with their $w$; 22.802389... |
| **[4]** | | |
---1 Alan is running in a straight line on a bearing of $090 ^ { \circ }$ at a constant speed of $4 \mathrm {~ms} ^ { - 1 }$. Ben sees Alan when they are 50 m apart and Alan is on a bearing of $060 ^ { \circ }$ from Ben. Ben sets off immediately to intercept Alan by running at a constant speed of $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Calculate the bearing on which Ben should run to intercept Alan.\\
(ii) Calculate the magnitude of the velocity of Ben relative to Alan and find the time it takes, from the moment Ben sees Alan, for Ben to intercept Alan.
\hfill \mbox{\textit{OCR M4 2014 Q1 [7]}}