| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Maximum speed on incline vs horizontal |
| Difficulty | Standard +0.3 This is a standard M2 power-force-motion question requiring the formula P=Fv and Newton's second law on an incline. Part (a) involves calculating driving force from power, then applying F=ma with resistance and weight component. Part (b) uses equilibrium at constant speed. Both parts follow routine procedures with straightforward arithmetic, making it slightly easier than average for M2. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.03d Newton's second law: 2D vectors3.03f Weight: W=mg6.02l Power and velocity: P = Fv |
A van of mass 1500 kg is driving up a straight road inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac{1}{12}$. The resistance to motion due to non-gravitational forces is modelled as a constant force of magnitude 1000 N.
Given that initially the speed of the van is 30 m s$^{-1}$ and that the van's engine is working at a rate of 60 kW,
\begin{enumerate}[label=(\alph*)]
\item calculate the magnitude of the initial deceleration of the van.
[4]
\end{enumerate}
When travelling up the same hill, the rate of working of the van's engine is increased to 80 kW. Using the same model for the resistance due to non-gravitational forces,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item calculate in m s$^{-1}$ the constant speed which can be sustained by the van at this rate of working.
[4]
\item Give one reason why the use of this model for resistance may mean that your answer to part (b) is too high.
[1]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2002 Q2 [9]}}