CAIE M1 2021 June — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2021
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeCyclist or runner: find resistance or speed
DifficultyStandard +0.3 This is a straightforward two-part mechanics question requiring standard application of P=Fv and F=ma on horizontal ground, then resolving forces on an incline at constant speed. All steps are routine M1 techniques with no novel problem-solving required, making it slightly easier than average.
Spec6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product

2 A cyclist is travelling along a straight horizontal road. She is working at a constant rate of 150 W . At an instant when her speed is \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), her acceleration is \(0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). The resistance to motion is 20 N .
  1. Find the total mass of the cyclist and her bicycle.
    The cyclist comes to a straight hill inclined at an angle \(\theta\) above the horizontal. She ascends the hill at constant speed \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). She continues to work at the same rate as before and the resistance force is unchanged.
  2. Find the value of \(\theta\).

Question 2(a):
AnswerMarks Guidance
AnswerMarks Guidance
Forward force exerted by cyclist \(= \frac{150}{4}\ \text{N}\ [= 37.5\ \text{N}]\)B1 OE. \(P = Fv\) used correctly
\(\frac{150}{4} - 20 = m \times 0.25\)M1 Use of Newton's second law
\(m = 70\ \text{kg}\)A1
Question 2(b):
AnswerMarks Guidance
AnswerMarks Guidance
\(150/3 - 20 - 70g\sin\theta = 0\)M1 For resolving up the plane
\(\theta = 2.5°\) to 1 d.p.A1 FT From \(2.456\ldots\); FT \(\theta = \sin^{-1}\!\left(\dfrac{3}{m}\right)\) from (a)
**Question 2(a):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Forward force exerted by cyclist $= \frac{150}{4}\ \text{N}\ [= 37.5\ \text{N}]$ | B1 | OE. $P = Fv$ used correctly |
| $\frac{150}{4} - 20 = m \times 0.25$ | M1 | Use of Newton's second law |
| $m = 70\ \text{kg}$ | A1 | |

---

**Question 2(b):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $150/3 - 20 - 70g\sin\theta = 0$ | M1 | For resolving up the plane |
| $\theta = 2.5°$ to 1 d.p. | A1 FT | From $2.456\ldots$; FT $\theta = \sin^{-1}\!\left(\dfrac{3}{m}\right)$ from **(a)** |
2 A cyclist is travelling along a straight horizontal road. She is working at a constant rate of 150 W . At an instant when her speed is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, her acceleration is $0.25 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. The resistance to motion is 20 N .
\begin{enumerate}[label=(\alph*)]
\item Find the total mass of the cyclist and her bicycle.\\

The cyclist comes to a straight hill inclined at an angle $\theta$ above the horizontal. She ascends the hill at constant speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. She continues to work at the same rate as before and the resistance force is unchanged.
\item Find the value of $\theta$.
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2021 Q2 [5]}}