| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| 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 M4 interception problem requiring vector resolution, relative velocity concepts, and solving simultaneous equations with bearings. While it involves multiple steps and careful angle work, it follows a well-established method taught explicitly in mechanics modules. The calculations are straightforward once the vector equations are set up correctly, making it moderately above average difficulty but not requiring novel insight. |
| 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 |
| \(180 - 60 - 35 = 85\) | B1 | Correct angle (soi) |
| \(\frac{\sin\theta}{5} = \frac{\sin 85}{8.5}\) | M1 | Sine rule with cv(85); \(\theta = 35.873446\ldots\) |
| bearing \(= (180 - 85 - \theta) - 35\) | M1 | |
| \(= 24.1\) (3 sf) | A1 [4] | \(24.12655\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(w^2 = 5^2 + 8.5^2 - 2(5)(8.5)\cos(180-85-\theta)\) | *M1 | \(\frac{w}{\sin(180-85-\theta)} = \frac{8.5}{\sin 85}\) |
| \(w = 7.32\) (3sf) | A1 | \(7.32344\ldots\) |
| \(t = \frac{9500}{w}\) | M1dep* | Dependent on both previous M marks; \(t = 1297.20427\) |
| \(s = 5t\) | M1 | |
| \(s = 6486\) (4sf) | A1 [5] | \(6486.021\ldots\); Accept 3sf or better |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $180 - 60 - 35 = 85$ | B1 | Correct angle (soi) |
| $\frac{\sin\theta}{5} = \frac{\sin 85}{8.5}$ | M1 | Sine rule with cv(85); $\theta = 35.873446\ldots$ |
| bearing $= (180 - 85 - \theta) - 35$ | M1 | |
| $= 24.1$ (3 sf) | A1 **[4]** | $24.12655\ldots$ |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $w^2 = 5^2 + 8.5^2 - 2(5)(8.5)\cos(180-85-\theta)$ | *M1 | $\frac{w}{\sin(180-85-\theta)} = \frac{8.5}{\sin 85}$ |
| $w = 7.32$ (3sf) | A1 | $7.32344\ldots$ |
| $t = \frac{9500}{w}$ | M1dep* | Dependent on both previous M marks; $t = 1297.20427$ |
| $s = 5t$ | M1 | |
| $s = 6486$ (4sf) | A1 **[5]** | $6486.021\ldots$; Accept 3sf or better |
---2 A ship $S$ is travelling with constant speed $5 \mathrm {~ms} ^ { - 1 }$ on a course with bearing $325 ^ { \circ }$. A second ship $T$ observes $S$ when $S$ is 9500 m from $T$ on a bearing of $060 ^ { \circ }$ from $T$. Ship $T$ sets off in pursuit, travelling with constant speed $8.5 \mathrm {~ms} ^ { - 1 }$ in a straight line.\\
(i) Find the bearing of the course which $T$ should take in order to intercept $S$.\\
(ii) Find the distance travelled by $S$ from the moment that $T$ sets off in pursuit until the point of interception.
\hfill \mbox{\textit{OCR M4 2017 Q2 [9]}}