CAIE M1 2022 June — Question 6 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeVariable resistance: find constant speed
DifficultyStandard +0.3 Part (a) requires applying the work-energy principle with power, gravitational PE, and KE changes—a standard multi-step mechanics problem with 5 marks. Part (b) is a routine P=Fv calculation at constant speed. Both parts use familiar M1 techniques without requiring novel insight, making this slightly easier than average.
Spec6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02j Conservation with elastics: springs and strings6.02l Power and velocity: P = Fv
A car of mass \(900\text{kg}\) is moving up a hill inclined at \(\sin^{-1} 0.12\) to the horizontal. The initial speed of the car is \(11\text{ms}^{-1}\). After \(12\text{s}\), the car has travelled \(150\text{m}\) up the hill and has speed \(16\text{ms}^{-1}\). The engine of the car is working at a constant rate of \(24\text{kW}\).
  1. Find the work done against the resistive forces during the \(12\text{s}\). [5]
  2. The car then travels along a straight horizontal road. There is a resistance to the motion of the car of \((1520 + 4v)\text{N}\) when the speed of the car is \(v\text{ms}^{-1}\). The car travels at a constant speed with the engine working at a constant rate of \(32\text{kW}\). Find this speed. [3]
Question 6:
AnswerMarks
6Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9709/42 Cambridge International AS & A Level – Mark Scheme May/June 2022
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2022 Page 5 of 18
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(a)KE change  0.5900162 0.5900112
 1152005445060750 B1
PE900g1500.12 162000B1 Allow 900g150sin6.89 or 900g150sin6.9
Not from use of constant acceleration/Newton’s second law.
AnswerMarks Guidance
[Work done by car’s engine =] 2400012 288000B1 WD
OE e.g. 24000
12
Work done against resistive forces
2400012900g1500.120.5900162 0.5900112
AnswerMarks Guidance
28800016200011520054450M1 Use of work-energy 5 terms; dimensionally correct.
Work done by car’s engine not from using one of the given
speeds.
Allow sign errors.
AnswerMarks Guidance
Work done = 65250 JA1 or 65.25 kJ
Allow AWRT 65300 J or 65.3 kJ from correct work
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(b)32000
Driving Force
AnswerMarks Guidance
vB1 OE e.g. 32000 DFv
32000
15204v
AnswerMarks Guidance
vM1 Apply N2L to the car with a0 (3 terms) and attempt to
solve a 3-term quadratic in v.
For reference 4v2 1520v320000
Allow if no working seen and have correct real solution(s) to
their 3-term quadratic. If working shown and if using the
formula, it must be using the correct formula. If factorising
must have 3 of the 4 terms correct of v20v400
AnswerMarks Guidance
Speed 20 ms1A1 Only.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9709/42 Cambridge International AS & A Level – Mark Scheme May/June 2022
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2022 Page 5 of 18
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | KE change  0.5900162 0.5900112
 1152005445060750  | B1
PE900g1500.12 162000 | B1 | Allow 900g150sin6.89 or 900g150sin6.9
Not from use of constant acceleration/Newton’s second law.
[Work done by car’s engine =] 2400012 288000 | B1 | WD
OE e.g. 24000
12
Work done against resistive forces
2400012900g1500.120.5900162 0.5900112
28800016200011520054450 | M1 | Use of work-energy 5 terms; dimensionally correct.
Work done by car’s engine not from using one of the given
speeds.
Allow sign errors.
Work done = 65250 J | A1 | or 65.25 kJ
Allow AWRT 65300 J or 65.3 kJ from correct work
5
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | 32000
Driving Force
v | B1 | OE e.g. 32000 DFv
32000
15204v
v | M1 | Apply N2L to the car with a0 (3 terms) and attempt to
solve a 3-term quadratic in v.
For reference 4v2 1520v320000
Allow if no working seen and have correct real solution(s) to
their 3-term quadratic. If working shown and if using the
formula, it must be using the correct formula. If factorising
must have 3 of the 4 terms correct of v20v400
Speed 20 ms1 | A1 | Only.
3
Question | Answer | Marks | Guidance
A car of mass $900\text{kg}$ is moving up a hill inclined at $\sin^{-1} 0.12$ to the horizontal. The initial speed of the car is $11\text{ms}^{-1}$. After $12\text{s}$, the car has travelled $150\text{m}$ up the hill and has speed $16\text{ms}^{-1}$. The engine of the car is working at a constant rate of $24\text{kW}$.

\begin{enumerate}[label=(\alph*)]
\item Find the work done against the resistive forces during the $12\text{s}$. [5]

\item The car then travels along a straight horizontal road. There is a resistance to the motion of the car of $(1520 + 4v)\text{N}$ when the speed of the car is $v\text{ms}^{-1}$. The car travels at a constant speed with the engine working at a constant rate of $32\text{kW}$.

Find this speed. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q6 [8]}}
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