| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Instantaneous change in power or force |
| Difficulty | Moderate -0.8 This is a straightforward application of the power formula P=Fv with constant speed conditions. Part (i)(a) requires recognizing that driving force equals resistance at constant speed, then calculating P=1550×40. Part (i)(b) uses P=Fv to find new driving force, then F=ma for deceleration. Part (ii) involves resolving forces on an incline but again uses constant speed (equilibrium) and P=Fv. All steps are standard textbook procedures with no problem-solving insight required, making it easier than average A-level questions. |
| Spec | 6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Power \(= 1550 \times 40\) W | M1 | Using Power \(= Fv\) where \(F\) = Resistance force |
| Power \(= 62000\) W \(= 62\) kW | A1 | Answer must be in kW. Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((62000 - 22000) = \text{DF} \times 40\), \([\text{DF} = 1000]\) | B1ft | For stating \(P - 22000 = \text{DF} \times 40\) to find new driving force; ft on Power found in (i)(a) |
| \(\text{DF} - 1550 = 1100a\) | M1 | For applying Newton's second law to the car (3 terms) |
| \(a = -0.5\ \text{ms}^{-2}\) or \(d = 0.5\ \text{ms}^{-2}\) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{DF} = 1100g\sin 8 + 1550\ [= 3081]\) | M1 | For stating the equilibrium of the three forces |
| \(80000 = 3081v\) | M1 | For using \(P = Fv\) with \(F\) involving a weight and a resistance term |
| \(v = 26(.0)\ \text{ms}^{-1}\) | A1 | Total: 3 |
## Question 6(i)(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Power $= 1550 \times 40$ W | M1 | Using Power $= Fv$ where $F$ = Resistance force |
| Power $= 62000$ W $= 62$ kW | A1 | Answer must be in kW. **Total: 2** |
## Question 6(i)(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(62000 - 22000) = \text{DF} \times 40$, $[\text{DF} = 1000]$ | B1ft | For stating $P - 22000 = \text{DF} \times 40$ to find new driving force; ft on Power found in (i)(a) |
| $\text{DF} - 1550 = 1100a$ | M1 | For applying Newton's second law to the car (3 terms) |
| $a = -0.5\ \text{ms}^{-2}$ or $d = 0.5\ \text{ms}^{-2}$ | A1 | **Total: 3** |
## Question 6(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{DF} = 1100g\sin 8 + 1550\ [= 3081]$ | M1 | For stating the equilibrium of the three forces |
| $80000 = 3081v$ | M1 | For using $P = Fv$ with $F$ involving a weight and a resistance term |
| $v = 26(.0)\ \text{ms}^{-1}$ | A1 | **Total: 3** |6 A car of mass 1100 kg is moving on a road against a constant force of 1550 N resisting the motion.\\
(i) The car moves along a straight horizontal road at a constant speed of $40 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate, in kW , the power developed by the engine of the car.
\item Given that this power is suddenly decreased by 22 kW , find the instantaneous deceleration of the car.\\
(ii) The car now travels at constant speed up a straight road inclined at $8 ^ { \circ }$ to the horizontal, with the engine working at 80 kW . Assuming the resistance force remains the same, find this constant speed.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2016 Q6 [8]}}