| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration on incline given power |
| Difficulty | Standard +0.3 This is a standard M1 work-energy-power question requiring application of P=Fv and Newton's second law on an incline. Part (i) involves resolving forces (weight component, resistance, driving force) and using F=ma with the power equation—straightforward but multi-step. Part (ii) is even more routine (constant speed means zero acceleration). The pulley problem is a standard connected particles setup. All techniques are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 3.03o Advanced connected particles: and pulleys6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(DF - 1550 - 1400g\sin4° = 1400 \times 0.4\) | M1 | Use of Newton's Second Law (4 terms) |
| \((DF = 3086.59...)\) | A1 | |
| \([30000 = (1400 \times 0.4 + 1550 + 1400g\sin4°)v]\) | M1 | Use of \(P = Fv\) |
| \(v = 9.72 \text{ ms}^{-1}\) | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([DF - 1550 - 1400g\sin4° = 0]\) | M1 | \((DF = 2526.59...)\) Resolving up the hill |
| \([P_{\max} = (1550 + 1400g\sin4°) \times 40]\) | M1 | Use of \(P = Fv\) |
| \(P = 101000\text{ W}\) or \(101\text{ kW}\) | A1 | \((P = 101063.6...)\) |
| 3 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $DF - 1550 - 1400g\sin4° = 1400 \times 0.4$ | M1 | Use of Newton's Second Law (4 terms) |
| $(DF = 3086.59...)$ | A1 | |
| $[30000 = (1400 \times 0.4 + 1550 + 1400g\sin4°)v]$ | M1 | Use of $P = Fv$ |
| $v = 9.72 \text{ ms}^{-1}$ | A1 | |
| | **4** | |
## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[DF - 1550 - 1400g\sin4° = 0]$ | M1 | $(DF = 2526.59...)$ Resolving up the hill |
| $[P_{\max} = (1550 + 1400g\sin4°) \times 40]$ | M1 | Use of $P = Fv$ |
| $P = 101000\text{ W}$ or $101\text{ kW}$ | A1 | $(P = 101063.6...)$ |
| | **3** | |3 A car of mass 1400 kg is travelling up a hill inclined at an angle of $4 ^ { \circ }$ to the horizontal. There is a constant resistance to motion of magnitude 1550 N acting on the car.\\
(i) Given that the engine of the car is working at 30 kW , find the speed of the car at an instant when its acceleration is $0.4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) The greatest possible constant speed at which the car can travel up the hill is $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the maximum possible power of the engine.\\
\includegraphics[max width=\textwidth, alt={}, center]{539be201-7bfc-4ba0-8378-c7aec4473ac7-06_643_419_255_863}
Two particles $A$ and $B$, of masses 1.3 kg and 0.7 kg respectively, are connected by a light inextensible string which passes over a smooth fixed pulley. Particle $A$ is 1.75 m above the floor and particle $B$ is 1 m above the floor (see diagram). The system is released from rest with the string taut, and the particles move vertically. When the particles are at the same height the string breaks.\\
\hfill \mbox{\textit{CAIE M1 2019 Q3 [7]}}