| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Engine power on road constant/variable speed |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question involving power-force-velocity relationships and work-energy principle. Part (a) requires applying P=Fv with Newton's second law to find forces, while part (b) uses work-energy principle with given resistance and component of weight down the slope. Both parts follow routine M2 procedures with straightforward arithmetic, making it slightly easier than average A-level difficulty. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \frac{1000P}{10}\) | M1 | Use of \(P = Fv\) |
| Equation of motion | M1 | All terms required. Dimensionally correct. Accept separate equations for the car and trailer. |
| \(F - 300g\sin\alpha - 800g\sin\alpha - 200 - 600 = 1100 \times 0.5\) | A1 | Unsimplified equation(s) with at most one error |
| A1 | Correct unsimplified equation in \(F\) (or \(P\)) | |
| \(P = 21.2\) | A1 | Accept 21. Max. 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Work-energy equation for the trailer | M1 | The Q requires work-energy. Equation for the trailer, with all terms dimensionally correct. Condone sign errors |
| \(200d = \frac{1}{2} \cdot 300 \cdot 12^2 - 300dg\sin\alpha\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation | |
| Solve for \(d\) | DM1 | Dependent on the previous M1 |
| \(d = 52.7\ \text{(m)}\) or \(53\ \text{(m)}\) | A1 | Max 3 sf |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \frac{1000P}{10}$ | M1 | Use of $P = Fv$ |
| Equation of motion | M1 | All terms required. Dimensionally correct. Accept separate equations for the car and trailer. |
| $F - 300g\sin\alpha - 800g\sin\alpha - 200 - 600 = 1100 \times 0.5$ | A1 | Unsimplified equation(s) with at most one error |
| | A1 | Correct unsimplified equation in $F$ (or $P$) |
| $P = 21.2$ | A1 | Accept 21. Max. 3 sf |
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Work-energy equation for the trailer | M1 | The Q requires work-energy. Equation for the trailer, with all terms dimensionally correct. Condone sign errors |
| $200d = \frac{1}{2} \cdot 300 \cdot 12^2 - 300dg\sin\alpha$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| Solve for $d$ | DM1 | Dependent on the previous M1 |
| $d = 52.7\ \text{(m)}$ or $53\ \text{(m)}$ | A1 | Max 3 sf |6. A car of mass 800 kg pulls a trailer of mass 300 kg up a straight road which is inclined to the horizontal at an angle $\alpha$, where $\sin \alpha = \frac { 1 } { 14 }$. The trailer is attached to the car by a light inextensible towbar which is parallel to the direction of motion of the car. The car's engine works at a constant rate of $P \mathrm {~kW}$. The non-gravitational resistances to motion are constant and of magnitude 600 N on the car and 200 N on the trailer.
At a given instant the car is moving at $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is accelerating at $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
\begin{enumerate}[label=(\alph*)]
\item Find the value of $P$.
When the car is moving up the road at $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, 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. The non-gravitational resistance to the motion of the trailer remains constant and of magnitude 200 N .
\item Find, using the work-energy principle, the value of $d$.\\
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\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q6 [10]}}