| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2015 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach when exact intercept not possible |
| Difficulty | Standard +0.8 This is a relative velocity problem requiring vector resolution, bearing calculations, and optimization to find closest approach. It involves multiple steps: setting up position vectors, finding the velocity of B to minimize distance, calculating the shortest distance using perpendicular distance from relative motion, and computing total distances traveled. While the concepts are standard M4 material, the multi-part nature, bearing conversions, and geometric reasoning required place it moderately above average difficulty. |
| 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 |
| \(\cos\theta = 40/75 \Rightarrow \theta = 57.769\ldots\) | M1 A1 | A1 maybe implied (or A1 for 32.23095…) |
| Bearing is \(\theta + 180° = 237.8°\) | A1 | |
| Shortest distance \(d = 400\cos(180 - 50 - \theta)\) | M1 | M1 for \(d = 400\cos(180 - 50 - \text{cv}(\theta))\) |
| \(d = 122\) m (3sf) | A1 [5] | 122.07235… |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Time to closest approach \(= \dfrac{400\sin(180-50-\theta)}{\sqrt{75^2 - 40^2}}\) | *M1 B1 B1 | M1 for use of \(t = \frac{s}{u}\), B1 for numerator (380.91775…), B1 for denominator (63.44288…) |
| \(t = 6.004\ldots\) | A1 | 6.0041050… |
| Total distance \(= 75t + 40t\) | M1 dep* | \(115(\text{cv}(t))\) |
| 690 m (3 sf) | A1 [6] | 690.47207… if M0 then SC B1 for 450.307… or 240.164… |
# Question 3:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta = 40/75 \Rightarrow \theta = 57.769\ldots$ | M1 A1 | A1 maybe implied (or A1 for 32.23095…) |
| Bearing is $\theta + 180° = 237.8°$ | A1 | |
| Shortest distance $d = 400\cos(180 - 50 - \theta)$ | M1 | M1 for $d = 400\cos(180 - 50 - \text{cv}(\theta))$ |
| $d = 122$ m (3sf) | A1 **[5]** | 122.07235… |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Time to closest approach $= \dfrac{400\sin(180-50-\theta)}{\sqrt{75^2 - 40^2}}$ | *M1 B1 B1 | M1 for use of $t = \frac{s}{u}$, B1 for numerator (380.91775…), B1 for denominator (63.44288…) |
| $t = 6.004\ldots$ | A1 | 6.0041050… |
| Total distance $= 75t + 40t$ | M1 dep* | $115(\text{cv}(t))$ |
| 690 m (3 sf) | A1 **[6]** | 690.47207… if M0 then SC B1 for 450.307… or 240.164… |
---3 Two planes, $A$ and $B$, flying at the same altitude, are participating in an air show. Initially the planes are 400 m apart and plane $B$ is on a bearing of $130 ^ { \circ }$ from plane $A$. Plane $A$ is moving due south with a constant speed of $75 \mathrm {~ms} ^ { - 1 }$. Plane $B$ is moving at a constant speed of $40 \mathrm {~ms} ^ { - 1 }$ and has set a course to get as close as possible to $A$.\\
(i) Find the bearing of the course set by $B$ and the shortest distance between the two planes in the subsequent motion.\\
(ii) Find the total distance travelled by $A$ and $B$ from the instant when they are initially 400 m apart to the point of their closest approach.
\hfill \mbox{\textit{OCR M4 2015 Q3 [11]}}