| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| 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 power-resistance problem requiring two simultaneous equations from Newton's second law (P=Fv) in different scenarios. While it involves multiple steps and careful equation manipulation, it follows a well-established template that M2 students practice extensively. The 'show that' part provides scaffolding, and the techniques (resolving forces, using P=Fv, solving simultaneous equations) are routine for this module. |
| Spec | 3.03d Newton's second law: 2D vectors3.03v Motion on rough surface: including inclined planes6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
4 A car of mass 800 kg experiences a resistance of magnitude $k v ^ { 2 } \mathrm {~N}$, where $k$ is a constant and $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the car's speed. The car's engine is working at a constant rate of $P \mathrm {~W}$. At an instant when the car is travelling on a horizontal road with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ its acceleration is $0.75 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. At an instant when the car is ascending a hill of constant slope $12 ^ { \circ }$ to the horizontal with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ its acceleration is $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Show that $k = 0.900$, correct to 3 decimal places, and find $P$.
The power is increased to $1.5 P \mathrm {~W}$.\\
(ii) Calculate the maximum steady speed of the car on a horizontal road.
\hfill \mbox{\textit{OCR M2 2009 Q4 [10]}}