| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Acceleration from power and speed |
| Difficulty | Moderate -0.3 This is a straightforward multi-part mechanics question testing standard power-force relationships and energy methods. Part (i) is direct recall (P=Fv), part (ii) requires resolving forces on an incline with Newton's second law, and part (iii) uses work-energy principle. All are routine M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | Power = 350 × 15 = 5250 W | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(ii) | B1 | Using Driving force DF = P/15 |
| DF + 1200g sin 1 – 350 = 1200 × 0.12 | M1 | For using Newton’s 2nd law down the slope |
| P = 4270 W (4268.56...) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(iii) | [1200g sin 1 – 350 = 1200a] | M1 |
| A1 | Correct equation | |
| [182 = 202 + 2as] | M1 | Using constant acceleration formulae with a complete method to find |
| Answer | Marks | Guidance |
|---|---|---|
| Distance travelled s = 324 m (324.39) | A1 | |
| Question | Answer | Marks |
| 6(iii) | Alternative method for Q6(iii) |
| Answer | Marks | Guidance |
|---|---|---|
| KE loss = ½ × 1200 × (202 – 182) | M1 | Attempt either PE loss or KE loss |
| A1 | Both PE loss and KE loss correct | |
| [1200g × s sin 1 + ½ × 1200 × (202 – 182) = 350s] | M1 | Apply work-energy equation to the car |
| Distance travelled s = 324 m (324.39) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
--- 6(i) ---
6(i) | Power = 350 × 15 = 5250 W | B1 | Allow 5.25 kW
1
--- 6(ii) ---
6(ii) | B1 | Using Driving force DF = P/15
DF + 1200g sin 1 – 350 = 1200 × 0.12 | M1 | For using Newton’s 2nd law down the slope
P = 4270 W (4268.56...) | A1
3
--- 6(iii) ---
6(iii) | [1200g sin 1 – 350 = 1200a] | M1 | Using Newton’s 2nd law down the slope
A1 | Correct equation
[182 = 202 + 2as] | M1 | Using constant acceleration formulae with a complete method to find
distance, s, travelled.
Distance travelled s = 324 m (324.39) | A1
Question | Answer | Marks | Guidance
6(iii) | Alternative method for Q6(iii)
PE loss = 1200g × s sin 1
KE loss = ½ × 1200 × (202 – 182) | M1 | Attempt either PE loss or KE loss
A1 | Both PE loss and KE loss correct
[1200g × s sin 1 + ½ × 1200 × (202 – 182) = 350s] | M1 | Apply work-energy equation to the car
Distance travelled s = 324 m (324.39) | A1
4
Question | Answer | Marks | GuidanceA car of mass $1200$ kg is driving along a straight horizontal road at a constant speed of $15$ m s$^{-1}$. There is a constant resistance to motion of $350$ N.
\begin{enumerate}[label=(\roman*)]
\item Find the power of the car's engine. [1]
\end{enumerate}
The car comes to a hill inclined at $1°$ to the horizontal, still travelling at $15$ m s$^{-1}$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item The car starts to descend the hill with reduced power and with an acceleration of $0.12$ m s$^{-2}$. Given that there is no change in the resistance force, find the new power of the car's engine at the instant when it starts to descend the hill. [3]
\item When the car is travelling at $20$ m s$^{-1}$ down the hill, the power is cut off and the car gradually slows down. Assuming that the resistance force remains $350$ N, find the distance travelled from the moment when the power is cut off until the speed of the car is reduced to $18$ m s$^{-1}$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2018 Q6 [8]}}