| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Cyclist or runner: find resistance or speed |
| Difficulty | Standard +0.3 This is a straightforward two-part mechanics problem requiring standard application of P=Fv and F=ma with resolution of forces on an incline. Part (a) uses Newton's second law to find resistance from given power, speed, and acceleration. Part (b) applies equilibrium on an incline with the same power and resistance. The calculations are routine with no conceptual challenges beyond standard M1 content, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Peddling force \(=PF=\frac{250}{5}\) | B1 | For use of power \(=Fv\), e.g. \(5\times PF=250\). |
| \(\text{their }PF - R = 90\times 0.1\) | M1 | Using their \(PF\neq 250\). 3 terms; allow sign errors. Dimensionally correct. |
| Resistance \(=41\ \text{N}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{250}{v}-\text{their }41-90g\sin 2=0\) | M1 | For attempt at resolving up the hill; 3 terms; allow sign errors; must be a component of weight (NOT mass) but allow sin/cos mix; allow use of their 41. Dimensionally correct. OE e.g. \(v=\frac{250}{\text{their }41+90g\sin 2}\). |
| Steady speed \(=3.45\ \text{ms}^{-1}\) | A1 | 3.45258…. AWRT 3.45. |
## Question 3(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Peddling force $=PF=\frac{250}{5}$ | B1 | For use of power $=Fv$, e.g. $5\times PF=250$. |
| $\text{their }PF - R = 90\times 0.1$ | M1 | Using their $PF\neq 250$. 3 terms; allow sign errors. Dimensionally correct. |
| Resistance $=41\ \text{N}$ | A1 | |
## Question 3(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{250}{v}-\text{their }41-90g\sin 2=0$ | M1 | For attempt at resolving up the hill; 3 terms; allow sign errors; must be a component of weight (NOT mass) but allow sin/cos mix; allow use of their 41. Dimensionally correct. OE e.g. $v=\frac{250}{\text{their }41+90g\sin 2}$. |
| Steady speed $=3.45\ \text{ms}^{-1}$ | A1 | 3.45258…. AWRT 3.45. |
---3 A cyclist is riding along a straight horizontal road. The total mass of the cyclist and his bicycle is 90 kg . The power exerted by the cyclist is 250 W . At an instant when the cyclist's speed is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, his acceleration is $0.1 \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the constant resistance to motion acting on the cyclist.\\
The cyclist comes to the bottom of a hill inclined at $2 ^ { \circ }$ to the horizontal.
\item Given that the power and resistance to motion are unchanged, find the steady speed which the cyclist could maintain when riding up the hill.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q3 [5]}}