| Exam Board | Edexcel |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2018 |
| 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 work-energy-power question requiring application of P=Fv and F=ma with speed-dependent resistance. Part (a) involves straightforward substitution into Newton's second law to find λ, while part (b) requires equilibrium conditions on an incline. Both parts follow routine procedures with no novel insight required, making it slightly easier than average for FM1. |
| Spec | 6.02l Power and velocity: P = Fv |
| V349 SIHI NI IMIMM ION OC | VJYV SIHIL NI LIIIM ION OO | VJYV SIHIL NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of \(P = Fv\) | B1 | Seen or implied; allow in (b) if not seen in (a) |
| Equation of motion: \(F - \lambda v = 750 \times 0.6\) | M1 | Requires all three terms; dimensionally correct; condone sign errors; need not substitute for \(F\) |
| \(\frac{18000}{15} - \lambda \times 15 = 750 \times 0.6\) | A1 | Correct unsimplified equation |
| \(1200 - 15\lambda = 450 \Rightarrow \lambda = 50\) | A1* | Obtain given answer correctly; sufficient working required |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Overall strategy: form quadratic in \(V\) and solve | M1 | Complete strategy |
| Equation of motion (all terms required) | M1 | Condone sign errors and sin/cos confusion; need not substitute for \(F\) |
| \(\frac{12000}{V} - 50V - 750g\sin\alpha = 0\) | A1 | Substituted equation with at most one error; allow in \(F\) or \(V\) |
| \(\frac{12000}{V} - 50V - 490 = 0 \Rightarrow 5V^2 + 49V - 1200 = 0\) | A1 | Correct quadratic; e.g. \(5V^2 + 49V - 1200 = 0\); allow in \(F\) or \(V\) |
| \(V = \frac{-49 + \sqrt{49^2 + 20 \times 1200}}{10} = 11.3\) only | A1 | Accept 11 or 11.3; negative root should be rejected if seen |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $P = Fv$ | B1 | Seen or implied; allow in (b) if not seen in (a) |
| Equation of motion: $F - \lambda v = 750 \times 0.6$ | M1 | Requires all three terms; dimensionally correct; condone sign errors; need not substitute for $F$ |
| $\frac{18000}{15} - \lambda \times 15 = 750 \times 0.6$ | A1 | Correct unsimplified equation |
| $1200 - 15\lambda = 450 \Rightarrow \lambda = 50$ | A1* | Obtain given answer correctly; sufficient working required |
**Total: 4 marks**
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## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Overall strategy: form quadratic in $V$ and solve | M1 | Complete strategy |
| Equation of motion (all terms required) | M1 | Condone sign errors and sin/cos confusion; need not substitute for $F$ |
| $\frac{12000}{V} - 50V - 750g\sin\alpha = 0$ | A1 | Substituted equation with at most one error; allow in $F$ or $V$ |
| $\frac{12000}{V} - 50V - 490 = 0 \Rightarrow 5V^2 + 49V - 1200 = 0$ | A1 | Correct quadratic; e.g. $5V^2 + 49V - 1200 = 0$; allow in $F$ or $V$ |
| $V = \frac{-49 + \sqrt{49^2 + 20 \times 1200}}{10} = 11.3$ only | A1 | Accept 11 or 11.3; negative root should be rejected if seen |
**Total: 5 marks**
---\begin{enumerate}
\item A van of mass 750 kg is moving along a straight horizontal road. At the instant when the van is moving at $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the van is modelled as a force of magnitude $\lambda \nu \mathrm { N }$, where $\lambda$ is a constant.
\end{enumerate}
The engine of the van is working at a constant rate of 18 kW .\\
At the instant when $v = 15$, the acceleration of the van is $0.6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$\\
(a) Show that $\lambda = 50$
The van now moves up a straight road inclined at an angle to the horizontal, where $\sin \alpha = \frac { 1 } { 15 }$\\
At the instant when the van is moving at $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the van from non-gravitational forces is modelled as a force of magnitude 50 v . When the engine of the van is working at a constant rate of 12 kW , the van is moving at a constant speed $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(b) Find the value of $V$.
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\hfill \mbox{\textit{Edexcel FM1 AS 2018 Q3 [9]}}