Moderate -0.5 This is a straightforward work-energy theorem application requiring only one equation: work done by driving force minus work against resistance equals kinetic energy gained. The calculation is direct with all values given, making it easier than average despite being a mechanics problem.
A cyclist is riding a bicycle along a straight horizontal road \(AB\) of length 50 m. The cyclist starts from rest at \(A\) and reaches a speed of \(6 \text{ m s}^{-1}\) at \(B\). The cyclist produces a constant driving force of magnitude 100 N. There is a resistance force, and the work done against the resistance force from \(A\) to \(B\) is 3560 J.
Find the total mass of the cyclist and bicycle. [3]
Work energy equation. Three terms. Allow sign errors.
Dimensionally correct.
Answer
Marks
mass = 80 kg
A1
3
SC: Acceleration considered as a constant
62 =0+2a50 [⇒a=0.36]
3560
100− =mtheir 0.36
Answer
Marks
Guidance
50
M1
From use of v=6, u=0, s=50. Must be using
correct suvat formulae.
For equation involving mass using N2L with three
terms.
Allow sign errors in N2L.
Answer
Marks
Guidance
mass = 80 kg
A1
Question
Answer
Marks
Question 1:
1 | Work done by cyclist = 50 × 100 (= 5000 J) | B1
1
Their 5000 – 3560 = m62
2 | M1 | Work energy equation. Three terms. Allow sign errors.
Dimensionally correct.
mass = 80 kg | A1
3
SC: Acceleration considered as a constant
62 =0+2a50 [⇒a=0.36]
3560
100− =mtheir 0.36
50 | M1 | From use of v=6, u=0, s=50. Must be using
correct suvat formulae.
For equation involving mass using N2L with three
terms.
Allow sign errors in N2L.
mass = 80 kg | A1
Question | Answer | Marks | Guidance
A cyclist is riding a bicycle along a straight horizontal road $AB$ of length 50 m. The cyclist starts from rest at $A$ and reaches a speed of $6 \text{ m s}^{-1}$ at $B$. The cyclist produces a constant driving force of magnitude 100 N. There is a resistance force, and the work done against the resistance force from $A$ to $B$ is 3560 J.
Find the total mass of the cyclist and bicycle. [3]
\hfill \mbox{\textit{CAIE M1 2022 Q1 [3]}}