| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find steady/maximum speed given power |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard application of P=Fv and F=ma. Part (i) involves finding driving force from power equation then applying Newton's second law. Part (ii) requires recognizing that at steady speed, acceleration is zero and resolving forces parallel to the slope. Both parts use routine techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Driving Force \(= 20000/20\ (= 1000)\) | B1 | |
| M1 | Attempt at N2L with 3 terms. Signs may not be correct at this stage | |
| \(20000/20 - 800 = 1600a\) | A1 | Using their \(20000/20\), but not \(20000\) |
| \(a = 0.125 \text{ ms}^{-2}\) | A1 | Allow \(\frac{1}{8}\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(20000/v\) | B1 | |
| M1 | 3 terms with attempt at resolving weight; \(g\) can be omitted at this stage; if \(F = \ldots\) then \(F = 0\) somewhere to award M | |
| \(DF - 800 - 1600g\sin 4 = 0\) | A1 | aef |
| \(v = 10.6 \text{ ms}^{-1}\) | A1 | |
| [4] |
## Question 2:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Driving Force $= 20000/20\ (= 1000)$ | B1 | |
| | M1 | Attempt at N2L with 3 terms. Signs may not be correct at this stage |
| $20000/20 - 800 = 1600a$ | A1 | Using their $20000/20$, but not $20000$ |
| $a = 0.125 \text{ ms}^{-2}$ | A1 | Allow $\frac{1}{8}$ |
| **[4]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $20000/v$ | B1 | |
| | M1 | 3 terms with attempt at resolving weight; $g$ can be omitted at this stage; if $F = \ldots$ then $F = 0$ somewhere to award M |
| $DF - 800 - 1600g\sin 4 = 0$ | A1 | aef |
| $v = 10.6 \text{ ms}^{-1}$ | A1 | |
| **[4]** | | |
---2 A car of mass 1600 kg moves along a straight horizontal road. The resistance to the motion of the car has constant magnitude 800 N and the car's engine is working at a constant rate of 20 kW .\\
(i) Find the acceleration of the car at an instant when the car's speed is $20 \mathrm {~ms} ^ { - 1 }$.
The car now moves up a hill inclined at $4 ^ { \circ }$ to the horizontal. The car's engine continues to work at 20 kW and the magnitude of the resistance to motion remains at 800 N .\\
(ii) Find the greatest steady speed at which the car can move up the hill.
\hfill \mbox{\textit{OCR M2 2012 Q2 [8]}}