| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Standard +0.8 This FM1 question requires applying P=Fv at constant speed (part a), then using Newton's second law with power, resistance, and gravitational components on an incline (part b). While the techniques are standard for Further Maths, the multi-step nature, unit conversion, and combining power/resistance/incline dynamics makes it moderately challenging—harder than typical A-level mechanics but routine for FM1 students. |
| Spec | 3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{60000}{v}\) | M1 | Use of \(P = Fv\) with \(P = 60\) or \(60000\). Allow \(U\) instead of \(v\). Allow if seen in (b) |
| Equation of motion parallel to road: \(F - 37.5v = 0\) | M1 | Correct no. of terms, condone sign errors, allow \(U\) instead of \(v\). Neither \(v\) nor \(F\) need to be substituted. N.B. M0 for \(60000 - 37.5v = 0\) |
| \(\frac{60000}{v} - 37.5v = 0\) (Allow \(v\) replaced by \(U\)) | A1 | Correct unsimplified equation in \(v\) only, seen or implied. A0 if they replace only one of the \(v\)'s by \(U\) |
| \(U = 144\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion parallel to the slope | M1 | Correct no. of terms, condone sign errors and sin/cos confusion. M0 if they use \(v = 40\) |
| \(F - 37.5V - 750g\sin\alpha = 750 \times 2\) | A1 | Correct equation with at most 1 error, \(\sin\alpha\) does not need to be substituted |
| \(\frac{60000}{V} - 37.5V - 750g\sin\alpha = 750 \times 2\) | A1 | Correct equation in \(V\) only, \(\sin\alpha\) does not need to be substituted |
| \(V = 20\) | A1 | cao |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{60000}{v}$ | M1 | Use of $P = Fv$ with $P = 60$ or $60000$. Allow $U$ instead of $v$. Allow if seen in (b) |
| Equation of motion parallel to road: $F - 37.5v = 0$ | M1 | Correct no. of terms, condone sign errors, allow $U$ instead of $v$. Neither $v$ nor $F$ need to be substituted. N.B. M0 for $60000 - 37.5v = 0$ |
| $\frac{60000}{v} - 37.5v = 0$ (Allow $v$ replaced by $U$) | A1 | Correct unsimplified equation in $v$ only, seen or implied. A0 if they replace only one of the $v$'s by $U$ |
| $U = 144$ | A1 | cao |
**(4 marks)**
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## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion parallel to the slope | M1 | Correct no. of terms, condone sign errors and sin/cos confusion. M0 if they use $v = 40$ |
| $F - 37.5V - 750g\sin\alpha = 750 \times 2$ | A1 | Correct equation with at most 1 error, $\sin\alpha$ does not need to be substituted |
| $\frac{60000}{V} - 37.5V - 750g\sin\alpha = 750 \times 2$ | A1 | Correct equation in $V$ only, $\sin\alpha$ does not need to be substituted |
| $V = 20$ | A1 | cao |
**(4 marks)**
---\begin{enumerate}
\item A racing car of mass 750 kg is moving along a straight horizontal road at a constant speed of $U \mathbf { k m ~ h } ^ { - \mathbf { 1 } }$. The engine of the racing car is working at a constant rate of 60 kW .
\end{enumerate}
The resistance to the motion of the racing car is modelled as a force of magnitude $37.5 v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the racing car.
Using the model,\\
(a) find the value of $U$
Later on, the racing car is accelerating up a straight road which is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 5 } { 49 }$. The engine of the racing car is working at a constant rate of 60 kW .
The total resistance to the motion of the racing car from non-gravitational forces is modelled as a force of magnitude $37.5 v \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the racing car. At the instant when the acceleration of the racing car is $2 \mathrm {~ms} ^ { - 2 }$, the speed of the racing car is $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$
Using the model,\\
(b) find the value of $V$
\hfill \mbox{\textit{Edexcel FM1 AS 2023 Q2 [8]}}