| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Up and down hill: two equations |
| Difficulty | Moderate -0.3 This is a standard M2 work-energy-power question requiring the formula P=Fv and Newton's second law. Part (a) involves resolving forces at constant speed (equilibrium) and using P=Fv directly. Part (b) requires finding the driving force from power, then applying F=ma with component resolution. Both parts are routine applications of well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Motion down the plane: | M1 | Dimensionally correct. Condone sign errors and sin/cos confusion |
| \(F + 750g\sin\theta = 1200\) | A1 | \((F + 450 = 1200)\) |
| Use of \(P = Fv\): \(F = \frac{9000}{v}\) | B1 | Award in (b) if not seen in (a) |
| \(\frac{9000}{v} + 750g \times \frac{3}{49} = 1200\) | ||
| \(v = \frac{9000}{750} = 12\) | A1 | |
| (4 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = ma\): \(F - (750g\sin\theta + 1200) = 750a\) | M1 | Dimensionally correct. Condone sign errors and sin/cos confusion |
| \(\frac{9000}{4.5} - \left(750g \times \frac{3}{49} + 1200\right) = 750a\) | A1 | Unsimplified equation with at most one error |
| A1 | Correct unsimplified equation | |
| \(a = 0.47\ (0.467)\ (\text{m s}^{-2})\) | A1 | 2 or 3 sf only, not \(\frac{7}{15}\) |
| (4 marks total) |
# Question 1:
## Part 1a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Motion down the plane: | M1 | Dimensionally correct. Condone sign errors and sin/cos confusion |
| $F + 750g\sin\theta = 1200$ | A1 | $(F + 450 = 1200)$ |
| Use of $P = Fv$: $F = \frac{9000}{v}$ | B1 | Award in (b) if not seen in (a) |
| $\frac{9000}{v} + 750g \times \frac{3}{49} = 1200$ | | |
| $v = \frac{9000}{750} = 12$ | A1 | |
| **(4 marks total)** | | |
## Part 1b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = ma$: $F - (750g\sin\theta + 1200) = 750a$ | M1 | Dimensionally correct. Condone sign errors and sin/cos confusion |
| $\frac{9000}{4.5} - \left(750g \times \frac{3}{49} + 1200\right) = 750a$ | A1 | Unsimplified equation with at most one error |
| | A1 | Correct unsimplified equation |
| $a = 0.47\ (0.467)\ (\text{m s}^{-2})$ | A1 | 2 or 3 sf only, not $\frac{7}{15}$ |
| **(4 marks total)** | | |
---\begin{enumerate}
\item A truck of mass 750 kg is moving with constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ down a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 3 } { 49 }$. The resistance to motion of the truck is modelled as a constant force of magnitude 1200 N . The engine of the truck is working at a constant rate of 9 kW .\\
(a) Find the value of $v$.
\end{enumerate}
On another occasion the truck is moving up the same straight road. The resistance to motion of the truck from non-gravitational forces is modelled as a constant force of magnitude 1200 N . The engine of the truck is working at a constant rate of 9 kW .\\
(b) Find the acceleration of the truck at the instant when it is moving with speed $4.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
\hfill \mbox{\textit{Edexcel M2 2018 Q1 [8]}}