| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Engine switched off: find distance or speed |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard application of power = force × velocity and work-energy principle. Part (a) involves resolving forces on an incline at constant speed (equilibrium) then calculating power, while part (b) is a direct application of work-energy with given resistance force. Both parts follow textbook methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Driving force \(= 150 + 500g\sin\theta\) | M1 | Requires both terms; condone sign and sin/cos confusion |
| \(= 150 + 500\times9.8\times\frac{1}{20}\ (= 395)\) (N) | A2 | \(-1\) each error; use of 9.81 is an error |
| Rate of work \(= 20\times(395)\) | M1 | Use of \(P = Fv\) |
| \(= 7.9\) kW | A1 | or 7900 (W) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(150d + 500g\times\sin\theta\times d\ (= \text{their } 395\cdot d) = \frac{1}{2}\times500\times20^2\) | M1 | Energy equation; requires all 3 terms of correct form, no duplication; condone sign errors and sin/cos confusion |
| \(150d + 500g\times\frac{1}{20}\times d\ (= \text{their } 395\cdot d) = \frac{1}{2}\times500\times20^2\) | A2 | Correct unsimplified equation in \(d\); \(-1\) each error; M1A2 available for correct work leading to \(-ve\ d\) |
| \(d = 250\) (m) | A1 | Accept 253; answer must be \(+ve\) |
| [4] |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Driving force $= 150 + 500g\sin\theta$ | M1 | Requires both terms; condone sign and sin/cos confusion |
| $= 150 + 500\times9.8\times\frac{1}{20}\ (= 395)$ (N) | A2 | $-1$ each error; use of 9.81 is an error |
| Rate of work $= 20\times(395)$ | M1 | Use of $P = Fv$ |
| $= 7.9$ kW | A1 | or 7900 (W) |
| **[5]** | | |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $150d + 500g\times\sin\theta\times d\ (= \text{their } 395\cdot d) = \frac{1}{2}\times500\times20^2$ | M1 | Energy equation; requires all 3 terms of correct form, no duplication; condone sign errors and sin/cos confusion |
| $150d + 500g\times\frac{1}{20}\times d\ (= \text{their } 395\cdot d) = \frac{1}{2}\times500\times20^2$ | A2 | Correct unsimplified equation in $d$; $-1$ each error; M1A2 available for correct work leading to $-ve\ d$ |
| $d = 250$ (m) | A1 | Accept 253; answer must be $+ve$ |
| **[4]** | | |
---2. A car of mass 500 kg is moving at a constant speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 20 }$. The resistance to motion from non-gravitational forces is modelled as a constant force of magnitude 150 N .
\begin{enumerate}[label=(\alph*)]
\item Find the rate of working of the engine of the car.
When the car is travelling up the road at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the engine is switched off. The car then comes to instantaneous rest, without braking, having moved a distance $d$ metres up the road from the point where the engine was switched off. The resistance to motion from non-gravitational forces is again modelled as a constant force of magnitude 150 N .
\item Use the work-energy principle to find the value of $d$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2015 Q2 [9]}}