| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find k or constants |
| Difficulty | Standard +0.3 This is a standard M2 power-force-velocity question requiring the formula P=Fv and F=ma. Part (a) uses maximum speed where acceleration is zero (driving force equals resistance), giving a straightforward algebraic 'show that'. Part (b) applies Newton's second law with the resistance formula from part (a). Both parts follow textbook methods with no novel problem-solving required, making this slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(25920 = k(36^2)(36)\) | \(k = 25920 \div 36^3 = \frac{4}{9}\) | M1 A1 M1 A1 |
| (b) \(25920 = 25\left(\frac{4}{9}\right)(25)^2 + 460a\) | \(a = 1.50 \text{ ms}^{-2}\) | M1 A1 A1 M1 A1 |
**(a)** $25920 = k(36^2)(36)$ | $k = 25920 \div 36^3 = \frac{4}{9}$ | M1 A1 M1 A1 |
**(b)** $25920 = 25\left(\frac{4}{9}\right)(25)^2 + 460a$ | $a = 1.50 \text{ ms}^{-2}$ | M1 A1 A1 M1 A1 | 9 marksA motor-cycle and its rider have a total mass of 460 kg. The maximum rate at which the cycle's engine can work is 25 920 W and the maximum speed of the cycle on a horizontal road is 36 ms$^{-1}$. A variable resisting force acts on the cycle and has magnitude $kv^2$, where $v$ is the speed of the cycle in ms$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{5}{8}$. [4 marks]
\item Find the acceleration of the cycle when it is moving at 25 ms$^{-1}$ on the horizontal road, with its engine working at full power. [5 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q5 [9]}}