| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Session | Specimen |
| Marks | 9 |
| 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 Further Mechanics power-force-acceleration question requiring P=Fv to find driving force, then F=ma with resistance forces. Part (a) is straightforward substitution; part (b) adds an incline component but follows the same method. Slightly above average difficulty due to being FM1 content and requiring careful force resolution, but it's a textbook application with no novel insight needed. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.03d Newton's second law: 2D vectors3.03f Weight: W=mg6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(P = Fv\): \(F = \frac{12000}{20}\) | B1 | 600 or equivalent |
| Equation of motion: \(F-(200+2v) = 600a\) | M1 | Use the model to form the equation of motion; must include all terms; condone sign errors |
| \(600 - 240 = 600a\) | A1ft | Correct for their \(F\) |
| \(360 = 600a,\ a = 0.6\ \text{(m s}^{-2}\text{)}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion: \(\frac{12000}{w}-(200+2w)-600g\sin\theta = -600\times 0.05\) | M1 | Use the model to form the equation of motion; all terms needed; condone sign errors and sin/cos confusion |
| (All correct A1A1; one error A1A0) | A1, A1 | |
| 3-term quadratic and solve: \(2w^2 + 590w - 12000 = 0\) | M1 | Dependent on preceding M1; use equation of motion to form 3-term quadratic in \(w\) only |
| \(w = \dfrac{-590 + \sqrt{590^2 + 96000}}{4} = 19.1\ \text{(m s}^{-1}\text{)}\) | A1 | Accept 19; do not accept more than 3 s.f. |
## Question 5:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $P = Fv$: $F = \frac{12000}{20}$ | B1 | 600 or equivalent |
| Equation of motion: $F-(200+2v) = 600a$ | M1 | Use the model to form the equation of motion; must include all terms; condone sign errors |
| $600 - 240 = 600a$ | A1ft | Correct for their $F$ |
| $360 = 600a,\ a = 0.6\ \text{(m s}^{-2}\text{)}$ | A1 | cao |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion: $\frac{12000}{w}-(200+2w)-600g\sin\theta = -600\times 0.05$ | M1 | Use the model to form the equation of motion; all terms needed; condone sign errors and sin/cos confusion |
| (All correct A1A1; one error A1A0) | A1, A1 | |
| 3-term quadratic and solve: $2w^2 + 590w - 12000 = 0$ | M1 | Dependent on preceding M1; use equation of motion to form 3-term quadratic in $w$ only |
| $w = \dfrac{-590 + \sqrt{590^2 + 96000}}{4} = 19.1\ \text{(m s}^{-1}\text{)}$ | A1 | Accept 19; do not accept more than 3 s.f. |
---\begin{enumerate}
\item A car of mass 600 kg is moving along a straight horizontal road.
\end{enumerate}
At the instant when the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the car is modelled as a force of magnitude $( 200 + 2 v ) \mathrm { N }$.
The engine of the car is working at a constant rate of 12 kW .\\
(a) Find the acceleration of the car at the instant when $v = 20$
Later on the car is moving up a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 14 }$
At the instant when the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the car from non-gravitational forces is modelled as a force of magnitude ( $200 + 2 v ) \mathrm { N }$.
The engine is again working at a constant rate of 12 kW .\\
At the instant when the car has speed $w \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the car is decelerating at $0.05 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(b) Find the value of $w$.
\hfill \mbox{\textit{Edexcel FM1 Q5 [9]}}