| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations with non-standard force laws |
| Difficulty | Standard +0.3 This is a relative velocity problem with bearings, requiring vector decomposition and basic trigonometry. While it involves multiple moving objects and bearing conversions, it's a standard M4 mechanics question with straightforward application of learned techniques—slightly easier than average due to the clear setup and routine methodology. |
| 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 |
| \(v^2 = 250^2 + 210^2 - 2\times250\times210\cos30°\) | M1 | Equation for \(v\) — Must be essentially correct; M0 for \(\cos150°\), \(\cos40°\) etc |
| Magnitude is \(125 \text{ ms}^{-1}\) (3 sf) | A1 | |
| \(\dfrac{\sin\theta}{210} = \dfrac{\sin30°}{125.2}\), \(\theta = 57.0°\) | M1 | Equation for a relevant angle — Use of sine rule and calculated side (less strict than previous M1) |
| Bearing is \(343°\) (3 sf) | A1 | |
| [4] | ||
| OR \(_{U}\mathbf{v}_P = \begin{pmatrix}250\sin40°\\250\cos40°\end{pmatrix} - \begin{pmatrix}210\sin70°\\210\cos70°\end{pmatrix}\) | M1 | Subtracting components |
| \(= \begin{pmatrix}-36.64\\119.7\end{pmatrix}\) | A1 | |
| Magnitude is \(125 \text{ ms}^{-1}\), Bearing is \(343°\) | M1 A1 | Finding magnitude or bearing; Both correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Shortest distance is \(15000\sin73°\) | M1 | Or other complete method for distance — M0 for \(15000\cos73°\) |
| \(= 14300 \text{ m}\) (3 sf) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Relative velocity perpendicular to \(\mathbf{v}_Q\) (diagram shown) | B1 | |
| \(\cos\phi = \dfrac{160}{250}\) | M1 | |
| \(\phi = 50.2°\) | A1 | Or \(\psi = 39.8°\) |
| Bearing is \(350°\) | A1 | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Shortest distance is \(15000\sin10.2°\) | M1 | Or other complete method for distance |
| \(= 2660 \text{ m}\) (3 sf) | A1 | |
| [2] |
## Question 4(i)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 250^2 + 210^2 - 2\times250\times210\cos30°$ | M1 | Equation for $v$ — Must be essentially correct; M0 for $\cos150°$, $\cos40°$ etc |
| Magnitude is $125 \text{ ms}^{-1}$ (3 sf) | A1 | |
| $\dfrac{\sin\theta}{210} = \dfrac{\sin30°}{125.2}$, $\theta = 57.0°$ | M1 | Equation for a relevant angle — Use of sine rule and calculated side (less strict than previous M1) |
| Bearing is $343°$ (3 sf) | A1 | |
| **[4]** | | |
| **OR** $_{U}\mathbf{v}_P = \begin{pmatrix}250\sin40°\\250\cos40°\end{pmatrix} - \begin{pmatrix}210\sin70°\\210\cos70°\end{pmatrix}$ | M1 | Subtracting components |
| $= \begin{pmatrix}-36.64\\119.7\end{pmatrix}$ | A1 | |
| Magnitude is $125 \text{ ms}^{-1}$, Bearing is $343°$ | M1 A1 | Finding magnitude or bearing; Both correct |
---
## Question 4(i)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Shortest distance is $15000\sin73°$ | M1 | Or other complete method for distance — M0 for $15000\cos73°$ |
| $= 14300 \text{ m}$ (3 sf) | A1 | |
| **[2]** | | |
---
## Question 4(ii)(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Relative velocity perpendicular to $\mathbf{v}_Q$ (diagram shown) | B1 | |
| $\cos\phi = \dfrac{160}{250}$ | M1 | |
| $\phi = 50.2°$ | A1 | Or $\psi = 39.8°$ |
| Bearing is $350°$ | A1 | |
| **[4]** | | |
---
## Question 4(ii)(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Shortest distance is $15000\sin10.2°$ | M1 | Or other complete method for distance |
| $= 2660 \text{ m}$ (3 sf) | A1 | |
| **[2]** | | |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{6e3d5f5e-7ffa-4111-903d-468fb4d20192-2_364_1313_1224_376}
An unidentified aircraft $U$ is flying horizontally with constant velocity $250 \mathrm {~ms} ^ { - 1 }$ in the direction with bearing $040 ^ { \circ }$. Two spotter planes $P$ and $Q$ are flying horizontally at the same height as $U$, and at one instant $P$ is 15000 m due west of $U$, and $Q$ is 15000 m due east of $U$ (see diagram).\\
(i) Plane $P$ is flying with constant velocity $210 \mathrm {~ms} ^ { - 1 }$ in the direction with bearing $070 ^ { \circ }$.\\
\hfill \mbox{\textit{OCR M4 2013 Q4 [12]}}