| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Cyclist or runner: find resistance or speed |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem requiring standard application of P=Fv at terminal velocity where driving force equals resistance. Part (i) is a simple 'show that' calculation, and part (ii) adds a component of weight but follows the same method. The concepts are standard M1 material with no novel problem-solving required. |
| Spec | 3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (using \(F = P/v\) and Newton's 2nd law with \(a=0\)) | M1 | For using \(F = P/v\) and Newton's 2nd law with \(a = 0\) |
| \(100/4 - 4k = 0 \rightarrow k = 6.25\) | A1 | 2 marks total — AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (Newton's 2nd law, \(a=0\) uphill, 3-term equation) | M1 | For using Newton's 2nd law with \(a = 0\) uphill \(\rightarrow\) 3 term equation |
| \(100/v - 70g \times 0.05 - 6.25v = 0\) | A1 | |
| \([6.25v^2 + 35v - 100 = 0]\) or \([v^2 + 5.6v - 16 = 0]\) | M1 | For solving a 3-term quadratic for \(v\) |
| Maximum speed is \(2.08\ \text{ms}^{-1}\) | A1 | 4 marks total |
## Question 3:
**Part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| (using $F = P/v$ and Newton's 2nd law with $a=0$) | M1 | For using $F = P/v$ and Newton's 2nd law with $a = 0$ |
| $100/4 - 4k = 0 \rightarrow k = 6.25$ | A1 | **2 marks total** — AG |
**Part (ii):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| (Newton's 2nd law, $a=0$ uphill, 3-term equation) | M1 | For using Newton's 2nd law with $a = 0$ uphill $\rightarrow$ 3 term equation |
| $100/v - 70g \times 0.05 - 6.25v = 0$ | A1 | |
| $[6.25v^2 + 35v - 100 = 0]$ or $[v^2 + 5.6v - 16 = 0]$ | M1 | For solving a 3-term quadratic for $v$ |
| Maximum speed is $2.08\ \text{ms}^{-1}$ | A1 | **4 marks total** |
---3 The resistance to motion acting on a runner of mass 70 kg is $k v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the runner's speed and $k$ is a constant. The greatest power the runner can exert is 100 W . The runner's greatest steady speed on horizontal ground is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that $k = 6.25$.\\
(ii) Find the greatest steady speed of the runner while running uphill on a straight path inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.05$.
\hfill \mbox{\textit{CAIE M1 2013 Q3 [6]}}