| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 10 |
| 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 standard three-part mechanics question requiring routine application of P=Fv and F=ma on horizontal and inclined planes. Part (i) uses P=Fv to find driving force then applies Newton's second law; part (ii) adds a component for gravity on the incline; part (iii) uses work-energy or SUVAT with deceleration. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03c Newton's second law: F=ma one dimension6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Driving force \(= 36000/20\) | B1 | For use of power \(= Fv\) |
| \([36000/20 - R = 3200 \times 0.2]\) | M1 | Use of Newton's Second Law |
| \(R = 1160\) N | A1 | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Driving force \(F = 3200g\sin1.5 + 1160\) | M1 | Resolving along plane |
| \([\text{Power} = (3200g\sin1.5 + 1160) \times 30]\) | M1 | Use of \(P = Fv\) |
| Power \(= 59900\) W (59929.87...) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([-(3200g\sin1.5 + 1160) = 3200a]\) | M1 | Use of Newton's Second Law |
| \((a = -0.62426\ldots)\) | A1 | |
| \([0^2 = 30^2 + 2as]\) | M1 | Use of \(v^2 = u^2 + 2as\) to find \(s\) |
| Distance \(s = 721\) m (720.84...) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([3200g\sin1.5s]\) or \([\frac{1}{2} \times 3200 \times 900]\) | M1 | For PE gain or KE loss |
| \(3200g\sin1.5s\) and \(\frac{1}{2} \times 3200 \times 900\) | A1 | For PE gain and KE loss |
| \([\frac{1}{2} \times 3200 \times 900 = 1160s + 3200g\sin1.5s]\) | M1 | For work/energy equation |
| Distance \(s = 721\) m (720.84...) | A1 | |
| 4 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Driving force $= 36000/20$ | B1 | For use of power $= Fv$ |
| $[36000/20 - R = 3200 \times 0.2]$ | M1 | Use of Newton's Second Law |
| $R = 1160$ N | A1 | |
| | [3] | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Driving force $F = 3200g\sin1.5 + 1160$ | M1 | Resolving along plane |
| $[\text{Power} = (3200g\sin1.5 + 1160) \times 30]$ | M1 | Use of $P = Fv$ |
| Power $= 59900$ W (59929.87...) | A1 | |
| | 3 | |
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[-(3200g\sin1.5 + 1160) = 3200a]$ | M1 | Use of Newton's Second Law |
| $(a = -0.62426\ldots)$ | A1 | |
| $[0^2 = 30^2 + 2as]$ | M1 | Use of $v^2 = u^2 + 2as$ to find $s$ |
| Distance $s = 721$ m (720.84...) | A1 | |
| | 4 | |
**OR:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[3200g\sin1.5s]$ or $[\frac{1}{2} \times 3200 \times 900]$ | M1 | For PE gain or KE loss |
| $3200g\sin1.5s$ and $\frac{1}{2} \times 3200 \times 900$ | A1 | For PE gain and KE loss |
| $[\frac{1}{2} \times 3200 \times 900 = 1160s + 3200g\sin1.5s]$ | M1 | For work/energy equation |
| Distance $s = 721$ m (720.84...) | A1 | |
| | 4 | |6 A van of mass 3200 kg travels along a horizontal road. The power of the van's engine is constant and equal to 36 kW , and there is a constant resistance to motion acting on the van.\\
(i) When the speed of the van is $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, its acceleration is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the resistance force.\\
When the van is travelling at $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it begins to ascend a hill inclined at $1.5 ^ { \circ }$ to the horizontal. The power is increased and the resistance force is still equal to the value found in part (i).\\
(ii) Find the power required to maintain this speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iii) The engine is now stopped, with the van still travelling at $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and the van decelerates to rest. Find the distance the van moves up the hill from the point at which the engine is stopped until it comes to rest.\\
\hfill \mbox{\textit{CAIE M1 2018 Q6 [10]}}