| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Constant speed up/down incline |
| Difficulty | Standard +0.3 This is a standard multi-part work-energy question requiring application of P=Fv and energy conservation principles. Part (i) is routine recall, part (ii) involves resolving forces on an incline (standard M1 content), and part (iii) requires an energy method but follows a predictable structure. All techniques are textbook exercises with no novel insight required, making it slightly easier than average overall. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03f Weight: W=mg6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | [P = DF × v = 850 × 36] | M1 |
| Answer | Marks |
|---|---|
| Power = rate of working = 30.6 kW | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(ii) | [DF = 1250 g × 0.1 + 850] | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| v | M1 | P |
| Answer | Marks |
|---|---|
| v = 30 so speed of car is 30 ms–1 | A1 |
| Answer | Marks |
|---|---|
| 6(iii) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | [= 110 000] |
| Loss in PE = 1250 g × 176 × 0.1 | B1 | [= 220 000] |
| WD by car’s engine = 20 000 × 8 | B1 | [= 160 000] |
| Answer | Marks | Guidance |
|---|---|---|
| WD against resistance + 110 000] | M1 | 4 term work energy equation |
| WD = 270 000 J = 270 kJ | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
--- 6(i) ---
6(i) | [P = DF × v = 850 × 36] | M1 | Apply P = DF × v with
DF = Resistance force
Power = rate of working = 30.6 kW | A1
2
--- 6(ii) ---
6(ii) | [DF = 1250 g × 0.1 + 850] | M1 | Driving force comprising of resistance
plus a weight component
63000
DF =
v | M1 | P
DF =
v
v = 30 so speed of car is 30 ms–1 | A1
3
--- 6(iii) ---
6(iii) | 1
Gain in KE = × 1250 × (242 – 202)
2 | B1 | [= 110 000]
Loss in PE = 1250 g × 176 × 0.1 | B1 | [= 220 000]
WD by car’s engine = 20 000 × 8 | B1 | [= 160 000]
[160 000 + 220 000 =
WD against resistance + 110 000] | M1 | 4 term work energy equation
WD = 270 000 J = 270 kJ | A1
5
Question | Answer | Marks | GuidanceA car has mass $1250 \text{ kg}$.
\begin{enumerate}[label=(\roman*)]
\item The car is moving along a straight level road at a constant speed of $36 \text{ m s}^{-1}$ and is subject to a constant resistance of magnitude $850 \text{ N}$. Find, in kW, the rate at which the engine of the car is working. [2]
\item The car travels at a constant speed up a hill and is subject to the same resistance as in part (i). The hill is inclined at an angle of $\theta°$ to the horizontal, where $\sin \theta° = 0.1$, and the engine is working at $63 \text{ kW}$. Find the speed of the car. [3]
\item The car descends the same hill with the engine of the car working at a constant rate of $20 \text{ kW}$. The resistance is not constant. The initial speed of the car is $20 \text{ m s}^{-1}$. Eight seconds later the car has speed $24 \text{ m s}^{-1}$ and has moved $176 \text{ m}$ down the hill. Use an energy method to find the total work done against the resistance during the eight seconds. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2018 Q6 [10]}}