| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Variable resistance or force differential equation |
| Difficulty | Standard +0.8 This is a multi-part Further Mechanics question requiring power-resistance relationships, force resolution on an incline, and work-energy principle application. Part (a) is routine (showing λ=8 using P=Fv at constant speed). Parts (b) and (c) require careful consideration of forces after the towbar breaks, with (c) involving work-energy principle with multiple resistance forces on an incline. The question demands systematic application of several mechanics principles but follows standard FM1 patterns without requiring novel insight. |
| 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 |
| Use of \(P = Fv : F = \frac{15000}{25} (= 600)\) | B1 | 600 or equivalent |
| Equation of motion | M1 | Use the model to form the equation of motion. If they start with two separate equations each one must be correct. |
| \(F - (200 + 200 + 25\lambda) = 0\) | A1 | Correct unsimplified equation |
| \(\lambda = 8\) * | A1* | Deduce given answer from correct working |
| Total: 4 marks | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation of motion | M1 | Use the model to form the equation of motion for the car (with \(v = 10\) used). All terms required. Dimensionally correct. Condone sign error and sin/cos confusion |
| \(\frac{15000}{10} - 280 - 600g\sin\theta = 600a\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation | |
| \(a = 1.38 \text{ m s}^{-2}\) (1.4) | A1 | 2 or 3 sf only – follows use of 9.8 |
| Total: 4 marks | (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Work energy equation | M1 | Complete strategy to form the work-energy equation. Condone sin/cos confusion and sign errors |
| \(\frac{1}{2} \times 150 \times 100 = 200d + 150gd\sin\theta\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation for \(d\) | |
| \(d = 25.2\) (m) (25) | A1 | Max 3 sf – follows use of 9.8 |
| Total: 4 marks | (4) | |
| Question Total: 12 marks | (12) |
## Question 4(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use of $P = Fv : F = \frac{15000}{25} (= 600)$ | B1 | 600 or equivalent |
| Equation of motion | M1 | Use the model to form the equation of motion. If they start with two separate equations each one must be correct. |
| $F - (200 + 200 + 25\lambda) = 0$ | A1 | Correct unsimplified equation |
| $\lambda = 8$ * | A1* | Deduce **given answer** from correct working |
| **Total: 4 marks** | **(4)** | |
---
## Question 4(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation of motion | M1 | Use the model to form the equation of motion for the car (with $v = 10$ used). All terms required. Dimensionally correct. Condone sign error and sin/cos confusion |
| $\frac{15000}{10} - 280 - 600g\sin\theta = 600a$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $a = 1.38 \text{ m s}^{-2}$ (1.4) | A1 | 2 or 3 sf only – follows use of 9.8 |
| **Total: 4 marks** | **(4)** | |
---
## Question 4(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Work energy equation | M1 | Complete strategy to form the work-energy equation. Condone sin/cos confusion and sign errors |
| $\frac{1}{2} \times 150 \times 100 = 200d + 150gd\sin\theta$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation for $d$ |
| $d = 25.2$ (m) (25) | A1 | Max 3 sf – follows use of 9.8 |
| **Total: 4 marks** | **(4)** | |
| **Question Total: 12 marks** | **(12)** | |
---\begin{enumerate}
\item A car of mass 600 kg pulls a trailer of mass 150 kg along a straight horizontal road. The trailer is connected to the car by a light inextensible towbar, which is parallel to the direction of motion of the car. The resistance to the motion of the trailer is modelled as a constant force of magnitude 200 N . At the instant when the speed of the car is $v \mathrm {~ms} ^ { - 1 }$, the resistance to the motion of the car is modelled as a force of magnitude $( 200 + \lambda v ) \mathrm { N }$, where $\lambda$ is a constant.
\end{enumerate}
When the engine of the car is working at a constant rate of 15 kW , the car is moving at a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$\\
(a) Show that $\lambda = 8$
Later on, the car is pulling the trailer up a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 15 }$\\
The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 200 N at all times. 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 $( 200 + 8 v ) \mathrm { N }$.
The engine of the car is again working at a constant rate of 15 kW .\\
When $v = 10$, the towbar breaks. The trailer comes to instantaneous rest after moving a distance $d$ metres up the road from the point where the towbar broke.\\
(b) Find the acceleration of the car immediately after the towbar breaks.\\
(c) Use the work-energy principle to find the value of $d$.\\
\hfill \mbox{\textit{Edexcel FM1 2019 Q4 [12]}}