| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| 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 straightforward multi-part mechanics question testing standard power-force-velocity relationships (P=Fv) and Newton's second law on an incline. Part (i) is direct substitution, part (ii) requires resolving forces on an incline but with given sin θ, and part (iii) combines these concepts with F=ma. All steps are routine applications of well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([20000/v = 650]\) | M1 | For using DF \(= P/v\) and for resolving forces along the direction of motion |
| Speed is 30.8 ms\(^{-1}\) | A1 | [2 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\text{DF} = 650 + 1400g \times \frac{1}{7}]\) | M1 | For resolving forces along the direction of motion |
| \(P/10 = 650 + 1400g \times \frac{1}{7}\) | M1 | For using DF \(= P/v\) |
| Power is 26500 W | A1 | [3 marks] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P = 0.8 \times 26500(21200)\) | B1\(\sqrt{}\) | ft \(0.8 \times P\) from (ii) |
| \([21200/20 + 1400g \times \frac{1}{7} - 650 = 1400a]\) | M1 | For using Newton's Second Law |
| Acceleration is 1.72 ms\(^{-2}\) | A1 | [3 marks] |
## Question 5:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[20000/v = 650]$ | M1 | For using DF $= P/v$ and for resolving forces along the direction of motion |
| Speed is 30.8 ms$^{-1}$ | A1 | **[2 marks]** |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{DF} = 650 + 1400g \times \frac{1}{7}]$ | M1 | For resolving forces along the direction of motion |
| $P/10 = 650 + 1400g \times \frac{1}{7}$ | M1 | For using DF $= P/v$ |
| Power is 26500 W | A1 | **[3 marks]** |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P = 0.8 \times 26500(21200)$ | B1$\sqrt{}$ | ft $0.8 \times P$ from (ii) |
| $[21200/20 + 1400g \times \frac{1}{7} - 650 = 1400a]$ | M1 | For using Newton's Second Law |
| Acceleration is 1.72 ms$^{-2}$ | A1 | **[3 marks]** |
---5 The motion of a car of mass 1400 kg is resisted by a constant force of magnitude 650 N .\\
(i) Find the constant speed of the car on a horizontal road, assuming that the engine works at a rate of 20 kW .\\
(ii) The car is travelling at a constant speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a hill inclined at an angle of $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 7 }$. Find the power of the car's engine.\\
(iii) The car descends the same hill with the engine working at $80 \%$ of the power found in part (ii). Find the acceleration of the car at an instant when the speed is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\hfill \mbox{\textit{CAIE M1 2016 Q5 [8]}}