WJEC Further Unit 3 Specimen — Question 7 9 marks

Exam BoardWJEC
ModuleFurther Unit 3 (Further Unit 3)
SessionSpecimen
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeFind power and resistance simultaneously
DifficultyChallenging +1.2 This is a standard Further Maths mechanics problem requiring application of F=ma with power and resistance on an incline. It involves setting up two equations (one at acceleration, one at maximum velocity where a=0) and solving simultaneously. While it requires careful bookkeeping of forces and the P=Fv relationship, the method is routine for FM students with no novel insight needed. The 9 marks reflect multiple steps rather than exceptional difficulty.
Spec3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv

A vehicle of mass 6000 kg is moving up a slope inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac{6}{49}\). The vehicle's engine exerts a constant power of \(P\) W. The constant resistance to motion of the vehicle is \(R\) N. At the instant the vehicle is moving with velocity \(\frac{16}{5}\) ms\(^{-1}\), its acceleration is 2 ms\(^{-2}\). The maximum velocity of the vehicle is \(\frac{16}{3}\) ms\(^{-1}\). Determine the value of \(P\) and the value of \(R\). [9]

AnswerMarks Guidance
N2L: \(T - mg\sin\alpha - R = ma\)M1 A1 AO3 AO2
\(T = \frac{P}{v}\)B1 AO3
\(\frac{5P}{16} - 6000 \times 9.8 \times \frac{6}{49} - R = 6000 \times 2\)A1 AO1
\(\frac{5P}{16} - R = 19200\)A1 AO1
N2L with \(a = 0\): \(T - mg\sin\alpha - R = 0\)M1 A1 AO3 AO2
\(\frac{3P}{16} - 6000 \times 9.8 \times \frac{6}{49} - R = 0\)A1 AO1
\(\frac{3P}{16} - R = 7200\)A1 AO1
Solving simultaneously:m1 AO1
\(\frac{2P}{16} = 12000\)A1 AO1
\(P = 96000\); \(R = 10800\)A1 AO1
Total: [9]
N2L: $T - mg\sin\alpha - R = ma$ | M1 A1 | AO3 AO2 | dimensionally correct, all forces correct equation

$T = \frac{P}{v}$ | B1 | AO3 | used si

$\frac{5P}{16} - 6000 \times 9.8 \times \frac{6}{49} - R = 6000 \times 2$ | A1 | AO1 | correct equation in P & R

$\frac{5P}{16} - R = 19200$ | A1 | AO1

N2L with $a = 0$: $T - mg\sin\alpha - R = 0$ | M1 A1 | AO3 AO2 | dimensionally correct, all forces correct equation

$\frac{3P}{16} - 6000 \times 9.8 \times \frac{6}{49} - R = 0$ | A1 | AO1 | correct equation in P & R

$\frac{3P}{16} - R = 7200$ | A1 | AO1

Solving simultaneously: | m1 | AO1 | eliminating one variable, Dep. on both M's

$\frac{2P}{16} = 12000$ | A1 | AO1 | both answers cao

$P = 96000$; $R = 10800$ | A1 | AO1

**Total: [9]**
A vehicle of mass 6000 kg is moving up a slope inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac{6}{49}$. The vehicle's engine exerts a constant power of $P$ W. The constant resistance to motion of the vehicle is $R$ N. At the instant the vehicle is moving with velocity $\frac{16}{5}$ ms$^{-1}$, its acceleration is 2 ms$^{-2}$. The maximum velocity of the vehicle is $\frac{16}{3}$ ms$^{-1}$.

Determine the value of $P$ and the value of $R$. [9]

\hfill \mbox{\textit{WJEC Further Unit 3  Q7 [9]}}