| Exam Board | OCR MEI |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2011 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Topic | Variable Force |
| Type | Power-velocity relationship |
| Difficulty | Challenging +1.8 This M4 question requires multiple sophisticated techniques: deriving differential equations from power/force relationships, separating variables, integrating rational functions with partial fractions, and handling two distinct phases of motion. The power-velocity relationship P=8v⁴ giving F=8v³ is non-standard, and part (iii) requires setting up and solving a new DE after conditions change. While methodical, it demands strong integration skills and careful algebraic manipulation across extended multi-step reasoning, placing it well above average difficulty. |
| Spec | 6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
3 A car of mass 800 kg moves horizontally in a straight line with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t$ seconds. While $v \leqslant 20$, the power developed by the engine is $8 v ^ { 4 } \mathrm {~W}$. The total resistance force on the car is of magnitude $8 v ^ { 2 } \mathrm {~N}$. Initially $v = 2$ and the car is at a point O . At time $t$ s the displacement from O is $x \mathrm {~m}$.\\
(i) Find $v$ in terms of $x$ and show that when $v = 20 , x = 100 \ln 1.9$.\\
(ii) Find the relationship between $t$ and $x$, and show that when $v = 20 , t \approx 19.2$.
The driving force is removed at the instant when $v$ reaches 20 .\\
(iii) For the subsequent motion, find $v$ in terms of $t$. Calculate $t$ when $v = 2$.
\hfill \mbox{\textit{OCR MEI M4 2011 Q3 [24]}}