CAIE M1 2024 March — Question 6 10 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionMarch
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNewton's laws and connected particles
TypeCar towing trailer, inclined road
DifficultyStandard +0.3 This is a standard connected particles problem with straightforward application of Newton's second law and energy methods. Part (a) requires resolving forces and applying F=ma to the system, then considering the trailer separately to find tension—routine mechanics. Part (b) uses work-energy theorem with given values. Both parts follow textbook procedures with no novel insight required, making it slightly easier than average.
Spec3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings
A car of mass 1800 kg is towing a trailer of mass 300 kg up a straight road inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = 0.05\). The car and trailer are connected by a tow-bar which is light and rigid and is parallel to the road. There is a resistance force of 800 N acting on the car and a resistance force of \(F\) N acting on the trailer. The driving force of the car's engine is 3000 N.
  1. It is given that \(F = 100\). Find the acceleration of the car and the tension in the tow-bar. [5]
  2. It is given instead that the total work done against \(F\) in moving a distance of 50 m up the road is 6000 J. The speed of the car at the start of the 50 m is \(20\) m s\(^{-1}\). Use an energy method to find the speed of the car at the end of the 50 m. [5]
Question 6:
AnswerMarks Guidance
6(a)Attempt at Newton’s second law on car or trailer or system *M1
sin/cos mix; sin or =2.865 needs to be
substituted; allow with =2.87 or 2.9 or better;
allow g missing.
Car: 3000−1800g0.05−800−T =1800a
 3000−900−800−T =1800a → 1 300−T =1800a 
Trailer: T −300g0.05−100=300a
 T −150−100=300a → T −250=300a 
System: 3000−1800g0.05−300g0.05−800−100 =(1800+300)a
AnswerMarks Guidance
 3000−900−150−800−100=2100a → 1 050=2100a A1 Any two equations correct.
Solving for a or TDM1 From equations with correct number of relevant
terms, allow g missing.
If no working seen, must be correct for their
equations.
AnswerMarks Guidance
Acceleration =0.5ms–2A1 Oe. Allow 0.499 from =2.87.
Tension =400NA1 Allow awrt 400 to 3sf, www.
Condone using car equation with 0.499 to get
T =400.5.
Using exact a from an inexact angle gives T =400.
5
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
6(b)Work done against resistance on car = 80050 =40 000 
or Work done by driving force = 300050 =150 000 B1
PE change =(1800+300)g500.05 =52 500 B1 Allow 2100g50sin2.9.
Or 2100g50sin2.87.
1
Initial KE = (1800+300)202  420 000 
AnswerMarks
2B1
1 1
(1800+300)v2 =300050−80050−6000−(1800+300)g500.05+ (1800+300)202
2 2
1050v2 =150 000−40 000−6000−52 500+420 000
 
1050v2 =471 500
AnswerMarks Guidance
 M1 Attempt at work-energy equation; correct number of
relevant terms; dimensionally correct; allow sign
errors; allow sin/cos mix in relevant resolved terms.
Only PE must be from a component.
Allow 2100g50sin2.9
or 2100g50sin2.87.
AnswerMarks Guidance
Speed = 21.2 ms−1 [21.1907…]A1 Awrt 21.2 to 3sf.
=2.87 gives 21.18909187.
=2.9 gives 21.17674855.
AnswerMarks Guidance
QuestionAnswer Marks
6(b)Alternative Method for Question 6(b): Using energy on Car or Trailer only (Must be finding Tension for this method)
Car: 3000−1800g0.05−800−T =1800a
6000
Trailer: T −300g0.05− =300a
50
2920
Solve to get T = =417.1428571
AnswerMarks
7B1
Work done against resistance on car = 80050 =40 000 
or Work done by driving force = 300050 =150 000 
2920  146000
or Work done against tension = 50 =
 
AnswerMarks
7  7 B1
PE change =1800g500.05 =45 000 
or
1
Initial KE = 1800202  360 000 
AnswerMarks Guidance
2B1 Or 300g500.05 =7500  .
Allow 1800g50sin2.9 or1800g50sin2.87.
1
Or 300202  60 000  .
2
1 2920 1
1800v2 =300050−80050− 50−1800g500.05+ 1800202
AnswerMarks Guidance
2 7 2M1 Attempt at work-energy equation; correct number of
relevant terms; dimensionally correct; allow sign
errors; allow sin/cos mix in relevant resolved terms.
Only PE must be from a component.
M0 if using T from part (a).
AnswerMarks Guidance
Speed = 21.2 ms−1 [21.1907…]A1 Awrt 21.2 to 3sf.
QuestionAnswer Marks
6(b)Special case: use of constant acceleration. Maximum 2 marks
6000 6000  103
F = =120, 2100a=3000− −800−2100g0.05  a= 
AnswerMarks Guidance
50 50  210B1 Oe.
6000
Allow 2100a=3000− −800−2100gsin2.9.
50
6000
Allow2100a=3000− −800−2100gsin2.87.
50
 103 
v2 =202 +250 → v=21.2 ms−1
 
AnswerMarks
 210 B1
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | Attempt at Newton’s second law on car or trailer or system | *M1 | Correct number of terms; allow sign errors; allow
sin/cos mix; sin or =2.865 needs to be
substituted; allow with =2.87 or 2.9 or better;
allow g missing.
Car: 3000−1800g0.05−800−T =1800a
 3000−900−800−T =1800a → 1 300−T =1800a 
Trailer: T −300g0.05−100=300a
 T −150−100=300a → T −250=300a 
System: 3000−1800g0.05−300g0.05−800−100 =(1800+300)a
 3000−900−150−800−100=2100a → 1 050=2100a  | A1 | Any two equations correct.
Solving for a or T | DM1 | From equations with correct number of relevant
terms, allow g missing.
If no working seen, must be correct for their
equations.
Acceleration =0.5ms–2 | A1 | Oe. Allow 0.499 from =2.87.
Tension =400N | A1 | Allow awrt 400 to 3sf, www.
Condone using car equation with 0.499 to get
T =400.5.
Using exact a from an inexact angle gives T =400.
5
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | Work done against resistance on car = 80050 =40 000 
or Work done by driving force = 300050 =150 000  | B1
PE change =(1800+300)g500.05 =52 500  | B1 | Allow 2100g50sin2.9.
Or 2100g50sin2.87.
1
Initial KE = (1800+300)202  420 000 
2 | B1
1 1
(1800+300)v2 =300050−80050−6000−(1800+300)g500.05+ (1800+300)202
2 2
1050v2 =150 000−40 000−6000−52 500+420 000
 
1050v2 =471 500
  | M1 | Attempt at work-energy equation; correct number of
relevant terms; dimensionally correct; allow sign
errors; allow sin/cos mix in relevant resolved terms.
Only PE must be from a component.
Allow 2100g50sin2.9
or 2100g50sin2.87.
Speed = 21.2 ms−1 [21.1907…] | A1 | Awrt 21.2 to 3sf.
=2.87 gives 21.18909187.
=2.9 gives 21.17674855.
Question | Answer | Marks | Guidance
6(b) | Alternative Method for Question 6(b): Using energy on Car or Trailer only (Must be finding Tension for this method)
Car: 3000−1800g0.05−800−T =1800a
6000
Trailer: T −300g0.05− =300a
50
2920
Solve to get T = =417.1428571
7 | B1
Work done against resistance on car = 80050 =40 000 
or Work done by driving force = 300050 =150 000 
2920  146000
or Work done against tension = 50 =
 
7  7  | B1
PE change =1800g500.05 =45 000 
or
1
Initial KE = 1800202  360 000 
2 | B1 | Or 300g500.05 =7500  .
Allow 1800g50sin2.9 or1800g50sin2.87.
1
Or 300202  60 000  .
2
1 2920 1
1800v2 =300050−80050− 50−1800g500.05+ 1800202
2 7 2 | M1 | Attempt at work-energy equation; correct number of
relevant terms; dimensionally correct; allow sign
errors; allow sin/cos mix in relevant resolved terms.
Only PE must be from a component.
M0 if using T from part (a).
Speed = 21.2 ms−1 [21.1907…] | A1 | Awrt 21.2 to 3sf.
Question | Answer | Marks | Guidance
6(b) | Special case: use of constant acceleration. Maximum 2 marks
6000 6000  103
F = =120, 2100a=3000− −800−2100g0.05  a= 
50 50  210 | B1 | Oe.
6000
Allow 2100a=3000− −800−2100gsin2.9.
50
6000
Allow2100a=3000− −800−2100gsin2.87.
50
 103 
v2 =202 +250 → v=21.2 ms−1
 
 210  | B1
5
Question | Answer | Marks | Guidance
A car of mass 1800 kg is towing a trailer of mass 300 kg up a straight road inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.05$. The car and trailer are connected by a tow-bar which is light and rigid and is parallel to the road. There is a resistance force of 800 N acting on the car and a resistance force of $F$ N acting on the trailer. The driving force of the car's engine is 3000 N.

\begin{enumerate}[label=(\alph*)]
\item It is given that $F = 100$.

Find the acceleration of the car and the tension in the tow-bar. [5]

\item It is given instead that the total work done against $F$ in moving a distance of 50 m up the road is 6000 J. The speed of the car at the start of the 50 m is $20$ m s$^{-1}$.

Use an energy method to find the speed of the car at the end of the 50 m. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2024 Q6 [10]}}
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