AQA Further AS Paper 2 Mechanics 2022 June — Question 5 5 marks

Exam BoardAQA
ModuleFurther AS Paper 2 Mechanics (Further AS Paper 2 Mechanics)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeVariable resistance: find constant speed
DifficultyModerate -0.3 This is a straightforward maximum speed problem requiring students to recognize that at maximum speed, driving force equals resistance force, then apply P=Fv. The conversion from km/h to m/s and single calculation make it slightly easier than average, though the 'fully justify' requirement adds minimal complexity.
Spec6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

5 A car, of mass 1000 kg , is travelling on a straight horizontal road. When the car travels at a speed of \(v \mathrm {~ms} ^ { - 1 }\), it experiences a resistance force of magnitude \(25 v\) newtons. The car has a maximum speed of \(72 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) on the straight road.
Find the maximum power output of the car.
Fully justify your answer.

Question 5:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
At maximum speed driving force equals resistance, \(R = F\)E1 Explains that at maximum speed the driving force equals the resistance or that acceleration is zero
\(v = \frac{72 \times 1000}{60 \times 60} = 20 \text{ ms}^{-1}\)B1 Converts km h\(^{-1}\) to m s\(^{-1}\) to obtain \(20 \text{ ms}^{-1}\)
\(R = 25v = 25(20) = 500\)M1 Obtains correct driving force or resistive force using their value for speed
\(P = Fv\), \(P = 500 \times 20 = 10000\)M1 Forms an equation by modelling power as \(Fv\) with both \(F\) and \(v\) substituted
Max \(P = 10\) kWA1 Must state units
## Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| At maximum speed driving force equals resistance, $R = F$ | E1 | Explains that at maximum speed the driving force equals the resistance or that acceleration is zero |
| $v = \frac{72 \times 1000}{60 \times 60} = 20 \text{ ms}^{-1}$ | B1 | Converts km h$^{-1}$ to m s$^{-1}$ to obtain $20 \text{ ms}^{-1}$ |
| $R = 25v = 25(20) = 500$ | M1 | Obtains correct driving force or resistive force using their value for speed |
| $P = Fv$, $P = 500 \times 20 = 10000$ | M1 | Forms an equation by modelling power as $Fv$ with both $F$ and $v$ substituted |
| Max $P = 10$ kW | A1 | Must state units |
5 A car, of mass 1000 kg , is travelling on a straight horizontal road.

When the car travels at a speed of $v \mathrm {~ms} ^ { - 1 }$, it experiences a resistance force of magnitude $25 v$ newtons.

The car has a maximum speed of $72 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ on the straight road.\\
Find the maximum power output of the car.\\
Fully justify your answer.\\

\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2022 Q5 [5]}}