CAIE M1 2022 November — Question 4 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypePower from force and derived speed (non-equilibrium)
DifficultyStandard +0.3 This is a straightforward power and energy question requiring standard M1 techniques: F=ma to find driving force, P=Fv for power, then work-energy principle. All steps are routine applications of formulas with no conceptual challenges or novel problem-solving required, making it slightly easier than average.
Spec1.08d Evaluate definite integrals: between limits6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product
A car of mass 1200 kg is travelling along a straight horizontal road \(AB\). There is a constant resistance force of magnitude 500 N. When the car passes point \(A\), it has a speed of \(15 \text{ m s}^{-1}\) and an acceleration of \(0.8 \text{ m s}^{-2}\).
  1. Find the power of the car's engine at the point \(A\). [3]
The car continues to work with this power as it travels from \(A\) to \(B\). The car takes 53 seconds to travel from \(A\) to \(B\) and the speed of the car at \(B\) is \(32 \text{ m s}^{-1}\).
  1. Show that the distance \(AB\) is 1362.6 m. [3]
Question 4:
AnswerMarks
4Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
AnswerMarks Guidance
4(a)P = D × 15 B1
For any D. OE including .
15
AnswerMarks Guidance
D – 500 = 1200 × 0.8 (⇒ D = 1460)M1 Attempt at Newton’s second law with three terms.
Allow sign errors.
AnswerMarks Guidance
Power = 21900 WA1 Allow 21900 without units or 21.9 kW, but not simply
21.9 without units or with wrong units.
3
AnswerMarks
4(b)1 1
[Change in KE =] 1200322 − 1200152
2 2
=614400−135000= 
AnswerMarks Guidance
479400B1 Sight of both KEs.
Work done by engine = 21900 × 53 ( = 1160700)B1ft WD
OE e.g. 21900=
53
FT their 21900.
AnswerMarks Guidance
Distance AB = 1362.6 mB1 AG Must come from 1160700 – 500d = 479400 OE
e.g. 500d=681300.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | P = D × 15 | B1 | P
For any D. OE including .
15
D – 500 = 1200 × 0.8 (⇒ D = 1460) | M1 | Attempt at Newton’s second law with three terms.
Allow sign errors.
Power = 21900 W | A1 | Allow 21900 without units or 21.9 kW, but not simply
21.9 without units or with wrong units.
3
--- 4(b) ---
4(b) | 1 1
[Change in KE =] 1200322 − 1200152
2 2
=614400−135000= 
479400 | B1 | Sight of both KEs.
Work done by engine = 21900 × 53 ( = 1160700) | B1ft | WD
OE e.g. 21900=
53
FT their 21900.
Distance AB = 1362.6 m | B1 | AG Must come from 1160700 – 500d = 479400 OE
e.g. 500d=681300.
3
Question | Answer | Marks | Guidance
A car of mass 1200 kg is travelling along a straight horizontal road $AB$. There is a constant resistance force of magnitude 500 N. When the car passes point $A$, it has a speed of $15 \text{ m s}^{-1}$ and an acceleration of $0.8 \text{ m s}^{-2}$.

\begin{enumerate}[label=(\alph*)]
\item Find the power of the car's engine at the point $A$. [3]
\end{enumerate}

The car continues to work with this power as it travels from $A$ to $B$. The car takes 53 seconds to travel from $A$ to $B$ and the speed of the car at $B$ is $32 \text{ m s}^{-1}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that the distance $AB$ is 1362.6 m. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2022 Q4 [6]}}
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