CAIE M1 2024 June — Question 1 3 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeEnergy method - driving force on horizontal road
DifficultyStandard +0.3 This is a straightforward work-energy problem requiring application of the work-energy principle: work done = change in KE + work against resistance. It involves standard formulas (KE = ½mv², work = force × distance) with clear given values and no conceptual complications, making it slightly easier than average but still requiring proper method across multiple steps.
Spec6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component
A cyclist and bicycle have a total mass of 72 kg. The cyclist rides along a horizontal road against a total resistance force of 28 N. Find the total work done by the cyclist to increase his speed from \(8\text{ ms}^{-1}\) to \(16\text{ ms}^{-1}\) while travelling a distance of 100 metres. [3]
Question 1:
AnswerMarks
11
Initial KE  72822304
2
1
OR Final KE  721629216
2
AnswerMarks Guidance
OR Work done against resistance 281002800B1 Correct expression for either KE or correct expression for
work done against resistance.
For reference, 1 72  162 82 6912.
2
Attempt at work-energy equation
1 1 
7282 WD 72162 28100
 
AnswerMarks Guidance
2 2 M1 4 terms; allow sign errors; dimensionally correct.
WD9712JA1 OE. Condone 9710J.
Do not ISW.
Alternative method for Question 1:
162 82
162 82 2100a  a 0.96
 
AnswerMarks Guidance
2100(B1) 192
OE, e.g. a .
200
Use of suvat in a complete method to find an expression
for a. Must be of the form 'a'.
Attempt at Newton’s second law
DF2872their 0.96
AnswerMarks Guidance
 (M1) Three terms; dimensionally correct; allow sign errors; must
be using their value of a.
AnswerMarks Guidance
WD97.121009712J(A1) OE. Condone 9710J.
Do not ISW.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 1:
1 | 1
Initial KE  72822304
2
1
OR Final KE  721629216
2
OR Work done against resistance 281002800 | B1 | Correct expression for either KE or correct expression for
work done against resistance.
For reference, 1 72  162 82 6912.
2
Attempt at work-energy equation
1 1 
7282 WD 72162 28100
 
2 2  | M1 | 4 terms; allow sign errors; dimensionally correct.
WD9712J | A1 | OE. Condone 9710J.
Do not ISW.
Alternative method for Question 1:
162 82
162 82 2100a  a 0.96
 
2100 | (B1) | 192
OE, e.g. a .
200
Use of suvat in a complete method to find an expression
for a. Must be of the form 'a'.
Attempt at Newton’s second law
DF2872their 0.96
  | (M1) | Three terms; dimensionally correct; allow sign errors; must
be using their value of a.
WD97.121009712J | (A1) | OE. Condone 9710J.
Do not ISW.
3
Question | Answer | Marks | Guidance
A cyclist and bicycle have a total mass of 72 kg. The cyclist rides along a horizontal road against a total resistance force of 28 N.

Find the total work done by the cyclist to increase his speed from $8\text{ ms}^{-1}$ to $16\text{ ms}^{-1}$ while travelling a distance of 100 metres. [3]

\hfill \mbox{\textit{CAIE M1 2024 Q1 [3]}}
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