| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find k or constants |
| Difficulty | Standard +0.3 This is a standard FM1 work-energy-power question requiring force equilibrium at constant speed and the power equation P=Fv. The multi-part structure and need to find k implicitly through equilibrium conditions adds modest complexity, but the mathematical steps are routine for Further Maths students: setting up force equations with resistance kv², using sin values, and algebraic manipulation. Slightly easier than average FM1 questions. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Freewheeling down: equation of motion down the plane using the model | M1 | Dimensionally correct, correct no. of terms, condone sin/cos confusion |
| \(100g\sin\alpha - kV^2 = 0 \quad \left(kV^2 = \frac{100g}{35}\right)\) | A1 | Correct equation |
| Cycling up: equation of motion up the plane using the model | M1 | Dimensionally correct, correct no. of terms, condone sin/cos confusion |
| \(F - 100g\sin\beta - kV^2 = 0\) | A1 | Correct equation |
| Use of \(F = \frac{P}{V} \quad \left(\frac{P}{V} = \frac{100g}{70} + \frac{100g}{35}\right)\) | B1/M1 | Any equivalent form; use correct strategy to set up and solve the equations |
| \(\left(P = \frac{300gV}{70}\right) \quad P = 42V\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion horizontally using the model | M1 | Correct no. of terms. Allow \(F - kU^2 = 0\) but not \(F - kV^2 = 0\) |
| \(\frac{35V}{U} - kU^2 = 0\) | A1 | Correct equation |
| Solve for \(U\) in terms of \(V\) \(\left(\frac{35V}{U} - \frac{100g}{35V^2}U^2 = 0\right)\) | M1 | Use correct strategy to set up and solve the equations |
| \(U = 1.1V\) or \(U = 1.08V\) | A1 | Accept 2 sf or 3 sf. \(U = \sqrt[3]{\frac{5}{4}}V\) scores 3/4 (depends on use of \(g\)) |
## Question 3:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Freewheeling down: equation of motion down the plane using the model | M1 | Dimensionally correct, correct no. of terms, condone sin/cos confusion |
| $100g\sin\alpha - kV^2 = 0 \quad \left(kV^2 = \frac{100g}{35}\right)$ | A1 | Correct equation |
| Cycling up: equation of motion up the plane using the model | M1 | Dimensionally correct, correct no. of terms, condone sin/cos confusion |
| $F - 100g\sin\beta - kV^2 = 0$ | A1 | Correct equation |
| Use of $F = \frac{P}{V} \quad \left(\frac{P}{V} = \frac{100g}{70} + \frac{100g}{35}\right)$ | B1/M1 | Any equivalent form; use correct strategy to set up and solve the equations |
| $\left(P = \frac{300gV}{70}\right) \quad P = 42V$ | A1 | cao |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion horizontally using the model | M1 | Correct no. of terms. Allow $F - kU^2 = 0$ but not $F - kV^2 = 0$ |
| $\frac{35V}{U} - kU^2 = 0$ | A1 | Correct equation |
| Solve for $U$ in terms of $V$ $\left(\frac{35V}{U} - \frac{100g}{35V^2}U^2 = 0\right)$ | M1 | Use correct strategy to set up and solve the equations |
| $U = 1.1V$ or $U = 1.08V$ | A1 | Accept 2 sf or 3 sf. $U = \sqrt[3]{\frac{5}{4}}V$ scores 3/4 (depends on use of $g$) |
---\begin{enumerate}
\item The total mass of a cyclist and his bicycle is 100 kg .
\end{enumerate}
In all circumstances, the magnitude of the resistance to the motion of the cyclist from non-gravitational forces is modelled as being $k v ^ { 2 } \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the cyclist.
The cyclist can freewheel, without pedalling, down a slope that is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 1 } { 35 }$, at a constant speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$
When he is pedalling up a slope that is inclined to the horizontal at an angle $\beta$, where $\sin \beta = \frac { 1 } { 70 }$, and he is moving at the same constant speed $V \mathrm {~ms} ^ { - 1 }$, he is working at a constant rate of $P$ watts.\\
(a) Find $P$ in terms of $V$.
If he pedals and works at a rate of 35 V watts on a horizontal road, he moves at a constant speed of $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(b) Find $U$ in terms of $V$.
\hfill \mbox{\textit{Edexcel FM1 AS 2021 Q3 [11]}}