| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Instantaneous change in power or force |
| Difficulty | Standard +0.3 This is a straightforward two-part mechanics question applying the standard power-force-velocity relationship (P=Fv) at maximum speed, then using F=ma with the new power condition. Both parts use direct formula application with minimal problem-solving insight required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (a) at max. speed, \(a = 0\), \(\frac{P}{30} - R = 0 \therefore \frac{P}{30} - 2000 = 0\) | M1 A1 | |
| \(P = 60000\) W \(\therefore H = 60\) | A1 | |
| (b) \(1.2 \times 60 = 72\) | A1 | |
| \(F - R = ma \therefore \frac{72000}{30} - 2000 = m \times 0.32\) | M1 A1 | |
| \(400 = 0.32m \therefore m = 1250\) kg | A1 | (7) |
**(a)** at max. speed, $a = 0$, $\frac{P}{30} - R = 0 \therefore \frac{P}{30} - 2000 = 0$ | M1 A1 |
$P = 60000$ W $\therefore H = 60$ | A1 |
**(b)** $1.2 \times 60 = 72$ | A1 |
$F - R = ma \therefore \frac{72000}{30} - 2000 = m \times 0.32$ | M1 A1 |
$400 = 0.32m \therefore m = 1250$ kg | A1 | (7)2. A car is travelling along a straight horizontal road against resistances to motion which are constant and total 2000 N . When the engine of the car is working at a rate of $H$ kilowatts, the maximum speed of the car is $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $H$.
The car driver wishes to overtake another vehicle so she increases the rate of working of the engine by $20 \%$ and this results in an initial acceleration of $0.32 \mathrm {~ms} ^ { - 2 }$. Assuming that the resistances to motion remain constant,
\item find the mass of the car.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q2 [7]}}