| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration on incline given power |
| Difficulty | Standard +0.8 This is a standard Further Maths mechanics problem involving power, resistance forces, and motion on an incline. Part (a) requires applying Newton's second law with variable resistance and calculating power using P=Fv, while part (b) involves recognizing that maximum speed occurs at equilibrium (a=0) and solving for the angle. The multi-step nature, variable resistance force, and need to coordinate multiple mechanics principles make it moderately challenging but still follows standard FM1/FM2 patterns without requiring novel insight. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv |
| Answer | Marks |
|---|---|
| B1 | \(3500g\left(\frac{3}{40}\right) = 2100\) |
| \(F = \frac{P}{25}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F - R - mg \sin \alpha = ma\) | M1, A1 | All forces, dim. correct; M1: Allow \(mg \cos \alpha\) or sign errors, but not both |
| \(\frac{P}{25} - 40(25) - 3500g\left(\frac{3}{40}\right) = 3500(0 \cdot 2)\) | A1 | Correct equation; FT their \(F\) |
| \(P = 60\,000\) (W) or 60 (kW) | A1[5] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = \frac{40 \times 1000}{20}\) (= 2000) | B1 | si |
| Using N2L with \(a = 0\) | M1 | All forces, dim. correct; M1: Allow \(mg \cos \alpha\) or sign errors, but not both |
| Answer | Marks | Guidance |
|---|---|---|
| \(2000 - 40(20) - 3500g \sin \alpha = 0\) | A1 | Correct equation; FT their \(F\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha = 2^0\) | A1[5] | cao |
## Part (a)
| B1 | $3500g\left(\frac{3}{40}\right) = 2100$
$F = \frac{P}{25}$ | B1 |
Using N2L up slope
$F - R - mg \sin \alpha = ma$ | M1, A1 | All forces, dim. correct; M1: Allow $mg \cos \alpha$ or sign errors, but not both
$\frac{P}{25} - 40(25) - 3500g\left(\frac{3}{40}\right) = 3500(0 \cdot 2)$ | A1 | Correct equation; FT their $F$
$P = 60\,000$ (W) or 60 (kW) | A1[5] | cao
## Part (b)
$F = \frac{40 \times 1000}{20}$ (= 2000) | B1 | si
Using N2L with $a = 0$ | M1 | All forces, dim. correct; M1: Allow $mg \cos \alpha$ or sign errors, but not both
$F - R - mg \sin \alpha = 0$
$2000 - 40(20) - 3500g \sin \alpha = 0$ | A1 | Correct equation; FT their $F$
$\sin \alpha = \frac{12}{343} = 0 \cdot 03498 \ldots$
$\alpha = 2^0$ | A1[5] | cao
**Total for Question 6: 10**
---A vehicle of mass 3500 kg is moving up a slope inclined at an angle $\alpha$ to the horizontal. When the vehicle is travelling at a velocity of $v\text{ ms}^{-1}$, the resistance to motion can be modelled by a variable force of magnitude $40v$ N.
\begin{enumerate}[label=(\alph*)]
\item Given that $\sin\alpha = \frac{3}{49}$, calculate the power developed by the engine at the instant when the speed of the vehicle is $25\text{ ms}^{-1}$ and its deceleration is $0.2\text{ ms}^{-2}$. [5]
\item When the vehicle's engine is working at a constant rate of 40 kW, the maximum speed that can be maintained up the slope is $20\text{ ms}^{-1}$. Find the value of $\alpha$. Give your answer in degrees, correct to one significant figure. [5]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2022 Q6 [10]}}