| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration on incline given power |
| Difficulty | Standard +0.3 This is a standard FM1 power-force-velocity question with straightforward application of P=Fv and F=ma. Part (a) uses constant resistance at constant speed (equilibrium), part (b) applies Newton's second law with variable resistance and component of weight, and part (c) finds terminal velocity. All parts follow routine procedures with no novel insight required, making it slightly easier than average for Further Maths. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(F = \frac{1000P}{v}\) where \(v = \frac{72000}{3600} (= 20)\) | M1 | Use of \(P = Fv\), condone \(\frac{P}{72}\) or \(\frac{P}{20}\) |
| Use equation of motion: \(F - 900 = 0\), e.g. \(\frac{1000P}{20} = 900\) | M1 | Condone use of \(P\) instead of \(1000P\) and condone 72 instead of 20 for method mark |
| \(P = 18\) | A1 | Allow \(P = 18000\) leading to final answer \(P = 18\). Ignore units. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use equation of motion for car and power equation to give equation in \(a\) only | M1 | All required terms present and no extras. Condone \(\pm\) sign errors and cos/sin confusion. M0 if a term is missing or weight not resolved. |
| \(\frac{30000}{10} - 1000g\sin\alpha - 20 \times 10 = 1000a\) | A1 | Correct equation with at most one error |
| (above equation fully correct) | A1 | Correct equation in \(a\) only |
| \(a = 2.4\) | A1 | Answer of 2.4 only. A0 for \(\frac{12}{5}\) when using \(g=9.8\). A0 for \(g=9.81\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{30000}{U}\) | M1 | Use of \(P = Fv\), condone incorrect number of zeros. M0 if using speed of 10 from part (b) |
| Equation of motion for car | M1 | All required terms present and no extras. Condone \(\pm\) sign errors and cos/sin confusion. M0 if term missing or weight not resolved. M0 if using resistance of 200 from part (b) |
| \(F - 1000g\sin\alpha - 20U = 0\) | A1 | Correct unsimplified equation. \(F\) does not need to be substituted, sin/cos need not be substituted |
| \(\frac{30000}{U} - 1000g\left(\frac{2}{49}\right) - 20U = 0\) or \(U^2 + 20U - 1500 = 0\) | A1 | Correct equation in \(U\) only |
| \(U = 30\) | A1 | Answer of 30 only |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $F = \frac{1000P}{v}$ where $v = \frac{72000}{3600} (= 20)$ | M1 | Use of $P = Fv$, condone $\frac{P}{72}$ or $\frac{P}{20}$ |
| Use equation of motion: $F - 900 = 0$, e.g. $\frac{1000P}{20} = 900$ | M1 | Condone use of $P$ instead of $1000P$ and condone 72 instead of 20 for method mark |
| $P = 18$ | A1 | Allow $P = 18000$ leading to final answer $P = 18$. Ignore units. |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use equation of motion for car **and** power equation to give equation in $a$ only | M1 | All required terms present and no extras. Condone $\pm$ sign errors and cos/sin confusion. M0 if a term is missing or weight not resolved. |
| $\frac{30000}{10} - 1000g\sin\alpha - 20 \times 10 = 1000a$ | A1 | Correct equation with at most one error |
| (above equation fully correct) | A1 | Correct equation in $a$ only |
| $a = 2.4$ | A1 | Answer of 2.4 only. A0 for $\frac{12}{5}$ when using $g=9.8$. A0 for $g=9.81$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{30000}{U}$ | M1 | Use of $P = Fv$, condone incorrect number of zeros. M0 if using speed of 10 from part (b) |
| Equation of motion for car | M1 | All required terms present and no extras. Condone $\pm$ sign errors and cos/sin confusion. M0 if term missing or weight not resolved. M0 if using resistance of 200 from part (b) |
| $F - 1000g\sin\alpha - 20U = 0$ | A1 | Correct unsimplified equation. $F$ does not need to be substituted, sin/cos need not be substituted |
| $\frac{30000}{U} - 1000g\left(\frac{2}{49}\right) - 20U = 0$ or $U^2 + 20U - 1500 = 0$ | A1 | Correct equation in $U$ only |
| $U = 30$ | A1 | Answer of 30 only |
---\begin{enumerate}
\item A car of mass 1000 kg moves in a straight line along a horizontal road at a constant speed of $72 \mathrm {~km} \mathrm {~h} ^ { - 1 }$
\end{enumerate}
\begin{itemize}
\item The resistance to the motion of the car is modelled as a constant force of magnitude 900 N
\end{itemize}
The engine of the car is working at a constant rate of $P \mathrm {~kW}$.\\
Using the model,\\
(a) find the value of $P$.
The car now travels in a straight line up a road which is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 2 } { 49 }$
\begin{itemize}
\item In a refined model, the resistance to the motion of the car from non-gravitational forces is now modelled as a force of magnitude $20 v$ newtons, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the car
\end{itemize}
At the instant when the engine of the car is working at a constant rate of 30 kW and the car is moving up the road at $10 \mathrm {~ms} ^ { - 1 }$, the acceleration of the car is $a \mathrm {~ms} ^ { - 2 }$
Using the refined model,\\
(b) find the value of $a$.
Later on, when the engine of the car is again working at a constant rate of 30 kW , the car is moving up the road at a constant speed $U \mathrm {~m} \mathrm {~s} ^ { - 1 }$
Using the refined model,\\
(c) find the value of $U$.
\hfill \mbox{\textit{Edexcel FM1 2024 Q3 [12]}}