CAIE M1 2024 June — Question 4 7 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeVariable resistance: find k or constants
DifficultyStandard +0.3 This is a straightforward power-resistance question requiring standard formulas (P=Fv, resolving forces) with clear given values. Part (a)(i) is direct substitution into R=kv², part (a)(ii) requires resolving forces on an incline and using P=Fv, and part (b) involves solving a quadratic from P=kv³. All steps are routine M1 techniques with no novel insight required, making it slightly easier than average.
Spec6.02l Power and velocity: P = Fv
A car has mass 1400 kg. When the speed of the car is \(v\text{ ms}^{-1}\) the magnitude of the resistance to motion is \(kv^2\) N where \(k\) is a constant.
  1. The car moves at a constant speed of \(24\text{ ms}^{-1}\) up a hill inclined at an angle of \(\alpha\) to the horizontal where \(\sin \alpha = 0.12\). At this speed the magnitude of the resistance to motion is 480 N.
    1. Find the value of \(k\). [1]
    2. Find the power of the car's engine. [3]
  2. The car now moves at a constant speed on a straight level road. Given that its engine is working at 54 kW, find this speed. [3]
Question 4:
AnswerMarks
4(a)(i)5
k242 480   k 
 
AnswerMarks Guidance
6B1 480
OE, e.g. , 0.833 or better.
576
1
AnswerMarks Guidance
4(a)(ii)Attempt at Newton’s second law
DF 4801400g0.12 DF 2160*M1 3 terms; allow sign errors; allow sin/cos mix.
Allow DF 4801400gsin6.9 or better.
 5
May see DF   their  242 1400g0.12.
 6
AnswerMarks Guidance
Power their 216024DB1 For using P = DF x v, where DF is numerical.
51840WA1 Allow W missing, but if given in kW units must be
present.
Allow 51.84kW.
Allow 51800, 51.8kW.
3
AnswerMarks
4(b)54000  5
DF  and DF   their  v2
AnswerMarks Guidance
v  6*B1FT 5
FT their 0.
6
Get an equation of the form av3 b and attempt to solve for v to get a
AnswerMarks Guidance
positive valueDM1 a and b must both be positive or both negative.
Must get to a value for v; if cubic not seen, the cubic may
be implied by the correct answer for their equation.
AnswerMarks Guidance
Speed = 40.2ms–1A1 40.165977.
AWRT 40.2 from correct work.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
--- 4(a)(i) ---
4(a)(i) | 5
k242 480   k 
 
6 | B1 | 480
OE, e.g. , 0.833 or better.
576
1
--- 4(a)(ii) ---
4(a)(ii) | Attempt at Newton’s second law
DF 4801400g0.12 DF 2160 | *M1 | 3 terms; allow sign errors; allow sin/cos mix.
Allow DF 4801400gsin6.9 or better.
 5
May see DF   their  242 1400g0.12.
 6
Power their 216024 | DB1 | For using P = DF x v, where DF is numerical.
51840W | A1 | Allow W missing, but if given in kW units must be
present.
Allow 51.84kW.
Allow 51800, 51.8kW.
3
--- 4(b) ---
4(b) | 54000  5
DF  and DF   their  v2
v  6 | *B1FT | 5
FT their 0.
6
Get an equation of the form av3 b and attempt to solve for v to get a
positive value | DM1 | a and b must both be positive or both negative.
Must get to a value for v; if cubic not seen, the cubic may
be implied by the correct answer for their equation.
Speed = 40.2ms–1 | A1 | 40.165977.
AWRT 40.2 from correct work.
3
Question | Answer | Marks | Guidance
A car has mass 1400 kg. When the speed of the car is $v\text{ ms}^{-1}$ the magnitude of the resistance to motion is $kv^2$ N where $k$ is a constant.

\begin{enumerate}[label=(\alph*)]
\item The car moves at a constant speed of $24\text{ ms}^{-1}$ up a hill inclined at an angle of $\alpha$ to the horizontal where $\sin \alpha = 0.12$. At this speed the magnitude of the resistance to motion is 480 N.

\begin{enumerate}[label=(\roman*)]
\item Find the value of $k$. [1]

\item Find the power of the car's engine. [3]
\end{enumerate}

\item The car now moves at a constant speed on a straight level road.

Given that its engine is working at 54 kW, find this speed. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q4 [7]}}
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