| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Instantaneous change in power or force |
| Difficulty | Moderate -0.3 This is a straightforward M2 mechanics question requiring standard application of P=Fv and F=ma with resolution of forces on an incline. Part (a) uses equilibrium at constant speed (a standard 'show that' requiring one equation), and part (b) applies Newton's second law with a new power value. Both parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03a Force: vector nature and diagrams3.03c Newton's second law: F=ma one dimension6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(T = \frac{15000}{20} = 750\) | M1 | |
| R(parallel to road): \(T = R + 750g\sin\theta\) | M1 A1 | |
| \(R = 750 - 750 \times 9.8 \times \frac{1}{15}\) | ||
| \(R = 260*\) | A1 | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(T' = \frac{18000}{20} = 900\) | M1 | |
| \(T' - 260 - 750g \times \sin\theta = 750a\) | M1 A1 | |
| \(a = \frac{900 - 260 - 750 \times 9.8 \times \frac{1}{15}}{750}\) | ||
| \(a = 0.2\) | A1 | (4 marks) |
# Question 4:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| $T = \frac{15000}{20} = 750$ | M1 | |
| R(parallel to road): $T = R + 750g\sin\theta$ | M1 A1 | |
| $R = 750 - 750 \times 9.8 \times \frac{1}{15}$ | | |
| $R = 260*$ | A1 | (4 marks) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| $T' = \frac{18000}{20} = 900$ | M1 | |
| $T' - 260 - 750g \times \sin\theta = 750a$ | M1 A1 | |
| $a = \frac{900 - 260 - 750 \times 9.8 \times \frac{1}{15}}{750}$ | | |
| $a = 0.2$ | A1 | (4 marks) |
---4. A car of mass 750 kg is moving up a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 15 }$. The resistance to motion of the car from non-gravitational forces has constant magnitude R newtons. The power developed by the car's engine is 15 kW and the car is moving at a constant speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { R } = 260$.
The power developed by the car's engine is now increased to 18 kW . The magnitude of the resistance to motion from non-gravitational forces remains at 260 N . At the instant when the car is moving up the road at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the car's acceleration is a $\mathrm { m } \mathrm { s } ^ { - 2 }$.
\item Find the value of a.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q4 [8]}}