| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Constant power on horizontal road |
| Difficulty | Moderate -0.3 This is a straightforward application of the power-force-velocity relationship (P=Fv) and Newton's second law. Part (i) requires combining F=ma with P=Fv to find velocity; part (ii) is a simple show-that using P=Fv at maximum speed (a=0); part (iii) uses work=power×time. All steps are standard M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(DF - 400 = 1200\times0.5\) | M1, A1 | For using Newton's 2nd law (3 terms needed) |
| \(20000 = 1000v\) | M1 | For using \(P = Fv\) |
| Speed is 20 ms⁻¹ | A1 (×4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(20000/v - 400 = 0\) | M1 | For using \(P = Fv\) and Newton's 2nd law with \(a = 0\) and \(F = 400\) |
| \(v_{max} = 50\) ms⁻¹ | A1 (×2) | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(20000 = \frac{1500000}{\Delta T}\) or distance \(= 1500000/400 = 3750\) and time \(= 3750/50\) | M1 | For using \(P = \frac{\Delta W}{\Delta T}\) or for using 'distance = work done/400' and 'time = distance/50' |
| Time taken is 75 s | A1 (×2) |
# Question 6:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $DF - 400 = 1200\times0.5$ | M1, A1 | For using Newton's 2nd law (3 terms needed) |
| $20000 = 1000v$ | M1 | For using $P = Fv$ |
| Speed is 20 ms⁻¹ | A1 (×4) | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $20000/v - 400 = 0$ | M1 | For using $P = Fv$ and Newton's 2nd law with $a = 0$ and $F = 400$ |
| $v_{max} = 50$ ms⁻¹ | A1 (×2) | AG |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $20000 = \frac{1500000}{\Delta T}$ or distance $= 1500000/400 = 3750$ and time $= 3750/50$ | M1 | For using $P = \frac{\Delta W}{\Delta T}$ or for using 'distance = work done/400' and 'time = distance/50' |
| Time taken is 75 s | A1 (×2) | |
---6 A car of mass 1200 kg travels along a horizontal straight road. The power of the car's engine is 20 kW . The resistance to the car's motion is 400 N .\\
(i) Find the speed of the car at an instant when its acceleration is $0.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) Show that the maximum possible speed of the car is $50 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
The work done by the car's engine as the car travels from a point $A$ to a point $B$ is 1500 kJ .\\
(iii) Given that the car is travelling at its maximum possible speed between $A$ and $B$, find the time taken to travel from $A$ to $B$.
\hfill \mbox{\textit{CAIE M1 2004 Q6 [8]}}