| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Instantaneous change in power or force |
| Difficulty | Moderate -0.3 This is a standard M2 power question requiring straightforward application of P=Fv and F=ma formulas across multiple parts. While it has four parts, each involves routine substitution into well-practiced equations with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(P = 13500\) W | B1 | 1 mark; or \(13.5\) kW |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(500 = 13500/v\) | M1 | |
| \(v = 27\) ms\(^{-1}\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(15000/25 - 500 = 950a\) | M1, A1 | 2 parts to F; A0 for 900a |
| \(a = 0.105\) or \(2/19\) | A1 | 3 marks total; or \(100/950\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(15000/26 - 500 -\) | M1 | 3 parts to F |
| \(950.9.8\sin5° = 950a\) | A1 | A0 for 900a |
| \(a = (-).773\) ms\(^{-2}\) | A1 | 3 marks total; s.c. accept \(0.77\); 9 marks total |
# Question 4:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $P = 13500$ W | B1 | 1 mark; or $13.5$ kW |
## Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $500 = 13500/v$ | M1 | |
| $v = 27$ ms$^{-1}$ | A1 | 2 marks total |
## Part (iii):
| Working | Mark | Guidance |
|---------|------|----------|
| $15000/25 - 500 = 950a$ | M1, A1 | 2 parts to F; A0 for 900a |
| $a = 0.105$ or $2/19$ | A1 | 3 marks total; or $100/950$ |
## Part (iv):
| Working | Mark | Guidance |
|---------|------|----------|
| $15000/26 - 500 -$ | M1 | 3 parts to F |
| $950.9.8\sin5° = 950a$ | A1 | A0 for 900a |
| $a = (-).773$ ms$^{-2}$ | A1 | 3 marks total; s.c. accept $0.77$; 9 marks total |
---4 A car of mass 900 kg is travelling at a constant speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ on a level road. The total resistance to motion is 450 N .\\
(i) Calculate the power output of the car's engine.
A roof box of mass 50 kg is mounted on the roof of the car. The total resistance to motion of the vehicle increases to 500 N .\\
(ii) The car's engine continues to work at the same rate. Calculate the maximum speed of the car on the level road.
The power output of the car's engine increases to 15000 W . The resistance to motion of the car, with roof box, remains 500 N .\\
(iii) Calculate the instantaneous acceleration of the car on the level road when its speed is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iv) The car climbs a hill which is at an angle of $5 ^ { \circ }$ to the horizontal. Calculate the instantaneous retardation of the car when its speed is $26 \mathrm {~ms} ^ { - 1 }$.
\hfill \mbox{\textit{OCR M2 2006 Q4 [9]}}