| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Find acceleration on incline given power |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem requiring standard application of P=Fv and F=ma with resolution of forces on an incline. Part (a) is a 'show that' requiring simple rearrangement, and part (b) adds one extra step of finding acceleration from net force. Slightly easier than average due to clear structure and routine methods. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Driving force \(= \text{DF} = \dfrac{22500}{v}\) | B1 | |
| \(\text{DF} - 1400g \times 0.1 - 600 = 0\) | M1 | Apply Newton's 2nd law to the car with \(a = 0\), three relevant terms. May see term \(1400g \sin 5.7°\). |
| \(v = 11.25 \text{ ms}^{-1}\) | A1 | AG From exact working only, may be implied if using \(5.7°\). |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(DF - 1400g \sin 2 - 600 = 1400a\) | M1 | Use of Newton's second law for the car, 4 relevant terms |
| \(\frac{22500}{11.25} - 1400g \sin 2 - 600 = 1400a\) | A1 | |
| \(a = 0.651 \text{ ms}^{-2}\) (3sf) | A1 | |
| 3 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Driving force $= \text{DF} = \dfrac{22500}{v}$ | **B1** | |
| $\text{DF} - 1400g \times 0.1 - 600 = 0$ | **M1** | Apply Newton's 2nd law to the car with $a = 0$, three relevant terms. May see term $1400g \sin 5.7°$. |
| $v = 11.25 \text{ ms}^{-1}$ | **A1** | **AG** From exact working only, may be implied if using $5.7°$. |
| | **[3]** | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $DF - 1400g \sin 2 - 600 = 1400a$ | M1 | Use of Newton's second law for the car, 4 relevant terms |
| $\frac{22500}{11.25} - 1400g \sin 2 - 600 = 1400a$ | A1 | |
| $a = 0.651 \text{ ms}^{-2}$ (3sf) | A1 | |
| | **3** | |2 A car of mass 1400 kg is travelling at constant speed up a straight hill inclined at $\alpha$ to the horizontal, where $\sin \alpha = 0.1$. There is a constant resistance force of magnitude 600 N . The power of the car's engine is 22500 W .
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of the car is $11.25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
The car, moving with speed $11.25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, comes to a section of the hill which is inclined at $2 ^ { \circ }$ to the horizontal.
\item Given that the power and resistance force do not change, find the initial acceleration of the car up this section of the hill.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q2 [6]}}