| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a straightforward M2 work-energy question requiring standard techniques: part (a) uses Power = Force × velocity to find the driving force, then resolves forces parallel to the slope at constant speed; part (b) applies work-energy principle with resistance and gravitational component. Both parts follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02a Work done: concept and definition6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Leading to \(\sin \theta = \frac{1}{14}\) | M1 A1 M1 A1 ft cso A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y \approx 102\) accept 102.4, 100 | M1 A1 M1 A1 | (4 marks) [9 marks total] |
| Answer | Marks | Guidance |
|---|---|---|
| \(y \approx 102\) accept 102.4, 100 | M1 M1 A1 A1 | (4 marks) |
**Part (a)**
$20000 = 16F$ ($F = 1250$)
$F = 550 + 1000 \times 9.8 \sin \theta$
Leading to $\sin \theta = \frac{1}{14}$ | M1 A1 M1 A1 ft cso A1 | (5 marks)
Fits their $F$
**Part (b)**
N2L: $550 + 1000 \times 9.8 \times \sin \theta = 1000a$
$(550 + 1000 \times 9.8 \times \frac{1}{14} = 1000a)$
or $1250 = 1000a$
$(a = (-)1.25)$
$v^2 = u^2 + 2as \Rightarrow 16^2 = 2 \times 1.25 \times y$
$y \approx 102$ accept 102.4, 100 | M1 A1 M1 A1 | (4 marks) [9 marks total]
**Alternative to (b)**
Work-Energy: $\frac{1}{2} \times 1000 \times 16^2 - 1000 \times 9.8 \times \frac{1}{14} \times y = 550y$
$y \approx 102$ accept 102.4, 100 | M1 M1 A1 A1 | (4 marks)
---A car of mass 1000 kg is moving at a constant speed of 16 m s$^{-1}$ up a straight road inclined at an angle $\theta$ to the horizontal. The rate of working of the engine of the car is 20 kW and the resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 550 N.
\begin{enumerate}[label=(\alph*)]
\item Show that $\sin \theta = \frac{1}{14}$. [5]
\end{enumerate}
When the car is travelling up the road at 16 m s$^{-1}$, the engine is switched off. The car comes to rest, without braking, having moved a distance $y$ metres from the point where the engine was switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 550 N.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $y$. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2008 Q3 [9]}}