CAIE M1 2023 November — Question 6 9 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeTowing system: inclined road
DifficultyStandard +0.3 This is a standard A-level mechanics problem involving Newton's second law with inclined planes and power calculations. Part (a) requires resolving forces and applying F=ma to the system and individual components—routine but multi-step. Part (b) uses the power equation P=Fv at constant speed, requiring students to equate driving force with total resistance including the new component from gravity. While it involves several steps and careful bookkeeping of forces, it follows standard textbook methods without requiring novel insight or particularly challenging problem-solving.
Spec3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv
A railway engine of mass 120000 kg is towing a coach of mass 60000 kg up a straight track inclined at an angle of \(\alpha\) to the horizontal where \(\sin \alpha = 0.02\). There is a light rigid coupling, parallel to the track, connecting the engine and coach. The driving force produced by the engine is 125000 N and there are constant resistances to motion of 22000 N on the engine and 13000 N on the coach.
  1. Find the acceleration of the engine and find the tension in the coupling. [5]
At an instant when the engine is travelling at 30 m s\(^{-1}\), it comes to a section of track inclined upwards at an angle \(\beta\) to the horizontal. The power produced by the engine is now 4500000 W and, as a result, the engine maintains a constant speed.
  1. Assuming that the resistance forces remain unchanged, find the value of \(\beta\). [4]
Question 6:
AnswerMarks
6(a)Engine: 125000−120000g0.02−22000−T =120000a
125000−120000gsin(1.145..)−22000−T =120000a
 125000−24000−22000−T =120000a 79000−T =120000a
Coach: T −60000g0.02−13000=60000a
T −60000gsin(1.145..)−13000=60000a
T −12000−13000=60000a T −25000=60000a
System:
125000−120000g0.02−60000g0.02−22000−13000=(120000+60000)a
125000−120000gsin(1.145..)−60000gsin(1.145..)−22000−13000=(120000+60000)a
AnswerMarks Guidance
 1 25000−24000−12000−22000−13000=(120000+60000)a 54000=180000a *M1 Attempt at Newton’s second law at
least once; correct number of
relevant terms; allow sign errors;
allow sin/cos mix; allow g
missing; a value fororsinmust
be substituted.
Allow with =1.1 or better
=1.145991998 
.
AnswerMarks
A1Any equations correct.
A1Two equations correct.
If using separate equations for
engine and coach and different T ’s,
then allow M1A1A0 max.
AnswerMarks Guidance
Solve for T or aDM1 Using equations with the correct
number of relevant terms.
If no working seen, must be
solutions to their equation(s) to be
awarded M1.
Acceleration = 0.3 m s–2
and
AnswerMarks Guidance
Tension = 43 000 NA1 Allow 0.299 from use of =1.15.
Awrt 43000 to 3sf from correct
work.
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(b)4500000
Driving force, DF = =150000 
AnswerMarks Guidance
30B1 P
Use of F = ,
v
oe e.g. DF30=4500000.
Attempt to resolve parallel to the track once if using system equation, twice if using equations for
AnswerMarks Guidance
engine and coach separatelyM1 Correct number of relevant terms;
allow sign errors; allow sin/cos
mix; allow g missing.
Must be correct number of
equations depending on method.
System: 150000−120000gsin−60000gsin−22000−13000=0
or
for Engine: 150000−120000gsin−22000−T'=0
AnswerMarks Guidance
and Coach: T'−60000gsin−13000=0A1 Allow DF or their DF.
Must be using same T'.
AnswerMarks Guidance
Solve to get =3.7A1 3.663058552
awrt 3.7° www.
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | Engine: 125000−120000g0.02−22000−T =120000a
125000−120000gsin(1.145..)−22000−T =120000a
 125000−24000−22000−T =120000a 79000−T =120000a
Coach: T −60000g0.02−13000=60000a
T −60000gsin(1.145..)−13000=60000a
T −12000−13000=60000a T −25000=60000a
System:
125000−120000g0.02−60000g0.02−22000−13000=(120000+60000)a
125000−120000gsin(1.145..)−60000gsin(1.145..)−22000−13000=(120000+60000)a
 1 25000−24000−12000−22000−13000=(120000+60000)a 54000=180000a  | *M1 | Attempt at Newton’s second law at
least once; correct number of
relevant terms; allow sign errors;
allow sin/cos mix; allow g
missing; a value fororsinmust
be substituted.
Allow with =1.1 or better
=1.145991998 
.
A1 | Any equations correct.
A1 | Two equations correct.
If using separate equations for
engine and coach and different T ’s,
then allow M1A1A0 max.
Solve for T or a | DM1 | Using equations with the correct
number of relevant terms.
If no working seen, must be
solutions to their equation(s) to be
awarded M1.
Acceleration = 0.3 m s–2
and
Tension = 43 000 N | A1 | Allow 0.299 from use of =1.15.
Awrt 43000 to 3sf from correct
work.
5
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | 4500000
Driving force, DF = =150000 
30 | B1 | P
Use of F = ,
v
oe e.g. DF30=4500000.
Attempt to resolve parallel to the track once if using system equation, twice if using equations for
engine and coach separately | M1 | Correct number of relevant terms;
allow sign errors; allow sin/cos
mix; allow g missing.
Must be correct number of
equations depending on method.
System: 150000−120000gsin−60000gsin−22000−13000=0
or
for Engine: 150000−120000gsin−22000−T'=0
and Coach: T'−60000gsin−13000=0 | A1 | Allow DF or their DF.
Must be using same T'.
Solve to get =3.7 | A1 | 3.663058552
awrt 3.7° www.
4
Question | Answer | Marks | Guidance
A railway engine of mass 120000 kg is towing a coach of mass 60000 kg up a straight track inclined at an angle of $\alpha$ to the horizontal where $\sin \alpha = 0.02$. There is a light rigid coupling, parallel to the track, connecting the engine and coach. The driving force produced by the engine is 125000 N and there are constant resistances to motion of 22000 N on the engine and 13000 N on the coach.

\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the engine and find the tension in the coupling. [5]
\end{enumerate}

At an instant when the engine is travelling at 30 m s$^{-1}$, it comes to a section of track inclined upwards at an angle $\beta$ to the horizontal. The power produced by the engine is now 4500000 W and, as a result, the engine maintains a constant speed.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Assuming that the resistance forces remain unchanged, find the value of $\beta$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2023 Q6 [9]}}
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