| Exam Board | OCR |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2013 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Angular kinematics – constant angular acceleration/deceleration |
| Difficulty | Moderate -0.8 This is a straightforward application of constant angular acceleration equations (rotational analogues of SUVAT). Part (i) requires finding angular acceleration from initial/final speeds and time using ω = ω₀ + αt. Part (ii) uses θ = ω₀t + ½αt² to find time. Both are direct substitutions into standard formulae with no conceptual challenges or problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(76 = 42 + \alpha \times 8\) | M1 | Using \(\omega_1 = \omega_0 + \alpha t\) |
| Angular acceleration is \(4.25 \text{ rad s}^{-2}\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\) | M1 | |
| \(810 = 42t + 2.125t^2\) | A1 | FT Quadratic equation for \(t\) |
| \(t = \dfrac{-42 \pm \sqrt{42^2 + 4 \times 2.125 \times 810}}{2 \times 2.125}\) | ||
| OR \(\omega_1^2 = 42^2 + 2 \times 4.25 \times 810\) | M1 | \(\omega_1^2 = \omega_0^2 + 2\alpha\theta\) and \(\omega_1 = \omega_0 + \alpha t\) — Or equivalent |
| \(\omega_1 = 93\) | A1 FT Equation for \(t\) | |
| \(93 = 42 + 4.25t\) | ||
| Time is \(12\) s | A1 | |
| [3] |
## Question 1(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $76 = 42 + \alpha \times 8$ | M1 | Using $\omega_1 = \omega_0 + \alpha t$ |
| Angular acceleration is $4.25 \text{ rad s}^{-2}$ | A1 | |
| **[2]** | | |
---
## Question 1(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ | M1 | |
| $810 = 42t + 2.125t^2$ | A1 | FT Quadratic equation for $t$ |
| $t = \dfrac{-42 \pm \sqrt{42^2 + 4 \times 2.125 \times 810}}{2 \times 2.125}$ | | |
| **OR** $\omega_1^2 = 42^2 + 2 \times 4.25 \times 810$ | M1 | $\omega_1^2 = \omega_0^2 + 2\alpha\theta$ and $\omega_1 = \omega_0 + \alpha t$ — Or equivalent |
| $\omega_1 = 93$ | | A1 FT Equation for $t$ |
| $93 = 42 + 4.25t$ | | |
| Time is $12$ s | A1 | |
| **[3]** | | |
---1 A camshaft inside an engine is rotating with angular speed $42 \mathrm { rads } ^ { - 1 }$. When the throttle is opened the camshaft speeds up with constant angular acceleration, and 8 seconds after the throttle was opened the angular speed is $76 \mathrm { rad } \mathrm { s } ^ { - 1 }$.\\
(i) Find the angular acceleration of the camshaft.\\
(ii) Find the time taken for the camshaft to turn through 810 radians from the moment that the throttle was opened.
\hfill \mbox{\textit{OCR M4 2013 Q1 [5]}}