| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Standard +0.3 This is a standard FM1 power-speed-force question requiring the formula P=Fv and Newton's second law. Part (a) involves substituting values and solving F=ma with given power and resistance. Part (b) adds an incline component but uses equilibrium (constant speed), making it straightforward. Both parts are routine applications of well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation of motion: \(F-(900+9\times25)=1200a\) | M1 | Dimensionally correct. Condone sign errors |
| Use of \(25000=F\times25\) | M1 | Correct use of \(P=Fv\). Allow in (b) if not seen in (a). |
| \(\frac{25000}{25}-(900+225)=1200a\) | A1 | Correct unsimplified equation |
| \(a=-\frac{5}{48}\), deceleration \(=\frac{5}{48}\) \((=0.10416..)\) \((\text{ms}^{-2})\) | A1 | 0.10 or better. Final answer must be positive. |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation of motion: \(\frac{25000}{V}-1200g\sin\theta-(900+9V)=0\) | M1, A1, A1 | Need all terms. Dimensionally correct. Condone sign errors; unsimplified equation with at most one error; correct unsimplified equation |
| Form quadratic and solve for \(V\) | M1 | Complete method to solve for \(V\) |
| \((9V^2+1488V-25000=0)\), \(V=15.4\ (15)\) | A1 | Correct to 2 sf or 3 sf |
| Total: (5) |
## Question 2:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion: $F-(900+9\times25)=1200a$ | M1 | Dimensionally correct. Condone sign errors |
| Use of $25000=F\times25$ | M1 | Correct use of $P=Fv$. Allow in (b) if not seen in (a). |
| $\frac{25000}{25}-(900+225)=1200a$ | A1 | Correct unsimplified equation |
| $a=-\frac{5}{48}$, deceleration $=\frac{5}{48}$ $(=0.10416..)$ $(\text{ms}^{-2})$ | A1 | 0.10 or better. Final answer must be positive. |
| **Total: (4)** | | |
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion: $\frac{25000}{V}-1200g\sin\theta-(900+9V)=0$ | M1, A1, A1 | Need all terms. Dimensionally correct. Condone sign errors; unsimplified equation with at most one error; correct unsimplified equation |
| Form quadratic and solve for $V$ | M1 | Complete method to solve for $V$ |
| $(9V^2+1488V-25000=0)$, $V=15.4\ (15)$ | A1 | Correct to 2 sf or 3 sf |
| **Total: (5)** | | |
---\begin{enumerate}
\item A truck of mass 1200 kg is moving along a straight horizontal road.
\end{enumerate}
At the instant when the speed of the truck is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the truck is modelled as a force of magnitude $( 900 + 9 v ) \mathrm { N }$.
The engine of the truck is working at a constant rate of 25 kW .\\
(a) Find the deceleration of the truck at the instant when $v = 25$
Later on, the truck is moving up a straight road that is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 20 }$
At the instant when the speed of the truck is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the truck from non-gravitational forces is modelled as a force of magnitude ( $900 + 9 v$ ) N.
When the engine of the truck is working at a constant rate of 25 kW the truck is moving up the road at a constant speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(b) Find the value of $V$.
\hfill \mbox{\textit{Edexcel FM1 2020 Q2 [9]}}