| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Moderate -0.3 This is a straightforward application of the power formula P=Fv with standard mechanics setup. Part (a)(i) is direct substitution, (a)(ii) uses P=Fv to find new driving force then F=ma, and (b) requires resolving forces on an incline with variable resistance but at constant speed (so forces balance). All techniques are standard M1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Power \(= 19.5\) kW | B2 | Or B1 for either \(650 \times 30\) or \(19\ 500\). |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(their\ 19500 + 9000 = DF \times 30\) | B1FT | Oe FT *their* 19.5 in watts only. |
| \(DF - 650 = 1300a\) | M1 | Newton's second law horizontally; 3 relevant terms; dimensionally correct but allow sign errors; allow with *their* driving force or just \(DF\). |
| \(a = \dfrac{3}{13} = 0.231\ \text{ms}^{-2}\) | A1 | \(0.23076923\ldots\) |
| Alternative scheme: | ||
| \(9000 = DF \times 30\) | \*B1 | Oe e.g. \(DF = \dfrac{9000}{30}\) |
| \(DF = 1300a\) | DM1 | Resolving horizontally using Newton's second law; 2 relevant terms; dimensionally correct but allow sign errors. |
| \(a = \dfrac{3}{13} = 0.231\ \text{ms}^{-2}\) | A1 | \(0.23076923\ldots\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(DF = \dfrac{11500}{v}\) | B1 | oe e.g. \(DF \times v = 11500\) |
| Attempt at Newton's second law | M1 | 4 relevant terms, *their DF* or just \(DF\); allow sign errors; allow sin/cos mix; allow \(g\) missing |
| \(\dfrac{11500}{v} + 1300 \times g \times 0.08 - (1000 + 20v) = 0\) | A1 | Correct equation |
| Speed \(= 25 \text{ ms}^{-1}\) | A1 | |
| 4 |
## Question 6:
### Part 6(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Power $= 19.5$ kW | **B2** | Or B1 for either $650 \times 30$ or $19\ 500$. |
| | **2** | |
### Part 6(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $their\ 19500 + 9000 = DF \times 30$ | **B1FT** | Oe FT *their* 19.5 in watts only. |
| $DF - 650 = 1300a$ | **M1** | Newton's second law horizontally; 3 relevant terms; dimensionally correct but allow sign errors; allow with *their* driving force or just $DF$. |
| $a = \dfrac{3}{13} = 0.231\ \text{ms}^{-2}$ | **A1** | $0.23076923\ldots$ |
| **Alternative scheme:** | | |
| $9000 = DF \times 30$ | **\*B1** | Oe e.g. $DF = \dfrac{9000}{30}$ |
| $DF = 1300a$ | **DM1** | Resolving horizontally using Newton's second law; 2 relevant terms; dimensionally correct but allow sign errors. |
| $a = \dfrac{3}{13} = 0.231\ \text{ms}^{-2}$ | **A1** | $0.23076923\ldots$ |
| | **3** | |
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $DF = \dfrac{11500}{v}$ | **B1** | oe e.g. $DF \times v = 11500$ |
| Attempt at Newton's second law | **M1** | 4 relevant terms, *their DF* or just $DF$; allow sign errors; allow sin/cos mix; allow $g$ missing |
| $\dfrac{11500}{v} + 1300 \times g \times 0.08 - (1000 + 20v) = 0$ | **A1** | Correct equation |
| Speed $= 25 \text{ ms}^{-1}$ | **A1** | |
| | **4** | |
---6 A car of mass 1300 kg is moving on a straight road.
\begin{enumerate}[label=(\alph*)]
\item On a horizontal section of the road, the car has a constant speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and there is a constant force of 650 N resisting the motion.
\begin{enumerate}[label=(\roman*)]
\item Calculate, in kW , the power developed by the engine of the car.
\item Given that this power is suddenly increased by 9 kW , find the instantaneous acceleration of the car.
\end{enumerate}\item On a section of the road inclined at $\sin ^ { - 1 } 0.08$ to the horizontal, the resistance to the motion of the car is $( 1000 + 20 v ) \mathrm { N }$ when the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car travels downwards along this section of the road at constant speed with the engine working at 11.5 kW .
Find this constant speed.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q6 [9]}}