| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Variable resistance or force differential equation |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question involving power-resistance-force relationships. Part (i) is a straightforward 'show that' using P=Fv with equilibrium forces. Parts (ii) and (iii) apply the same principles with minor variations (new equilibrium, then using F=ma). The question requires systematic application of well-practiced techniques rather than problem-solving insight, making it slightly easier than average for A-level. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
3 The resistance to the motion of a car of mass 600 kg is $k v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the car's speed and $k$ is a constant. The car ascends a hill of inclination $\alpha$, where $\sin \alpha = \frac { 1 } { 10 }$. The power exerted by the car's engine is 12000 W and the car has constant speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that $k = 0.6$.
The power exerted by the car's engine is increased to 16000 W .\\
(ii) Calculate the maximum speed of the car while ascending the hill.
The car now travels on horizontal ground and the power remains 16000 W .\\
(iii) Calculate the acceleration of the car at an instant when its speed is $32 \mathrm {~ms} ^ { - 1 }$.
\hfill \mbox{\textit{OCR M2 2008 Q3 [9]}}