WJEC Further Unit 3 2018 June — Question 2 10 marks

Exam BoardWJEC
ModuleFurther Unit 3 (Further Unit 3)
Year2018
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeUp and down hill: two equations
DifficultyChallenging +1.2 This is a mechanics problem requiring systematic application of power-force relationships and resistance modeling. Students must set up two equations from the maximum speed conditions (power = driving force × velocity, with driving force balancing resistance and weight component), recognize that resistance R = kv², solve simultaneously for P and k, then calculate resistance at the given speed. While it involves multiple steps and careful algebraic manipulation, the approach is methodical and follows standard A-level mechanics techniques without requiring novel insight—moderately harder than average due to the algebraic complexity and Further Maths context.
Spec3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv

A car of mass 750 kg is moving on a slope inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = 0.1\). When the car's engine is working at a constant power \(PW\), the car can travel at maximum speeds of \(14\text{ ms}^{-1}\) up the slope and \(28\text{ ms}^{-1}\) down the slope. In each case, the resistance to motion experienced by the car is proportional to the square of its speed. Find the value of \(P\) and determine the resistance to the motion of the car when its speed is \(10.5\text{ ms}^{-1}\). [10]

AnswerMarks Guidance
Resistance \(R = kv^2\)B1 si
Tractive force \(T = \frac{P}{v}\)M1 used
N2L up slopeM1 dim correct equation
\(\frac{P}{14} - 14^2k - 750g \times \frac{1}{10} = 0\)A1 correct equation, allow 750a RHS
N2L down slopeM1 dim correct equation
\(\frac{P}{28} - 28^2k + 750g \times \frac{1}{10} = 0\)A1 correct equation
\(\frac{4P}{14} - 4 \times 75g = 4 \times 14^2k\)
\(\frac{P}{28} + 75g = 4 \times 14^2k\)
\(\frac{7P}{28} = 5 \times 75g\)m1 one variable eliminated
\(P = 14700\)A1 cao \(P\) or \(k\)
\(\frac{14700}{14} - 75 \times 9.8 = 14^2k\)m1
\(k = \frac{45}{28}\)
Resistance \(R\) when \(v=10.5 = \frac{45}{28} \times 10.5^2\)
\(R = 177.(1875)\) (N)A1 cao
| Resistance $R = kv^2$ | B1 | si |
| Tractive force $T = \frac{P}{v}$ | M1 | used |
| **N2L up slope** | M1 | dim correct equation |
| $\frac{P}{14} - 14^2k - 750g \times \frac{1}{10} = 0$ | A1 | correct equation, allow 750a RHS |
| **N2L down slope** | M1 | dim correct equation |
| $\frac{P}{28} - 28^2k + 750g \times \frac{1}{10} = 0$ | A1 | correct equation |
| $\frac{4P}{14} - 4 \times 75g = 4 \times 14^2k$ | | |
| $\frac{P}{28} + 75g = 4 \times 14^2k$ | | |
| $\frac{7P}{28} = 5 \times 75g$ | m1 | one variable eliminated |
| $P = 14700$ | A1 | cao $P$ or $k$ |
| $\frac{14700}{14} - 75 \times 9.8 = 14^2k$ | m1 | |
| $k = \frac{45}{28}$ | | |
| Resistance $R$ when $v=10.5 = \frac{45}{28} \times 10.5^2$ | | |
| $R = 177.(1875)$ (N) | A1 | cao |
A car of mass 750 kg is moving on a slope inclined at an angle $\theta$ to the horizontal, where $\sin\theta = 0.1$. When the car's engine is working at a constant power $PW$, the car can travel at maximum speeds of $14\text{ ms}^{-1}$ up the slope and $28\text{ ms}^{-1}$ down the slope. In each case, the resistance to motion experienced by the car is proportional to the square of its speed. Find the value of $P$ and determine the resistance to the motion of the car when its speed is $10.5\text{ ms}^{-1}$. [10]

\hfill \mbox{\textit{WJEC Further Unit 3 2018 Q2 [10]}}