| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: horizontal road |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question involving power, forces, and energy. Part (a)(i) uses P=Fv with constant speed (forces balanced). Part (a)(ii) requires F=ma with the power equation. Part (b) adds an incline component but follows the same methodology. All parts are routine applications of well-practiced formulas with straightforward multi-step calculations, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03o Advanced connected particles: and pulleys6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P = (650 + 150)\times 24\) | M1 | Use of \(P = \text{DF}\times v\) |
| \(19200\) W or \(19.2\) kW | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(40000 = \text{DF}\times 24\) | B1 | Correct use of \(P = \text{DF}\times v\) |
| \(\dfrac{40000}{24} - 800 = 2250a\) | M1 | Use of Newton's Second Law for the system or for the caravan or for the car |
| \(T - 150 = 500a\) | A1 | Two correct equations |
| \(a = \left(\dfrac{40000}{24} - 800\right)\div 2250\) leading to \(a = \ldots\) | M1 | Solves for \(a\) or for \(T\) |
| Acceleration \(= 0.385\) ms\(^{-2}\) and Tension \(= 343\) N | A1 | From \(a = \dfrac{52}{135} = 0.38518\ldots\) and \(T = \dfrac{9250}{27} = 342.59\ldots\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\text{DF} = 800 + 2250\times g\times 0.14\) | M1 | Resolving up hill using DF = Total resistances |
| \(\dfrac{31000}{v} = 800 + 2250\times g\times 0.14\) | M1 | Use of \(P = \text{DF}\times v\) to form equation in \(v\) |
| \(v = 7.85\) | A1 | From \(v = \dfrac{620}{79} = 7.848\ldots\) |
## Question 6(a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P = (650 + 150)\times 24$ | M1 | Use of $P = \text{DF}\times v$ |
| $19200$ W or $19.2$ kW | A1 | |
---
## Question 6(a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $40000 = \text{DF}\times 24$ | B1 | Correct use of $P = \text{DF}\times v$ |
| $\dfrac{40000}{24} - 800 = 2250a$ | M1 | Use of Newton's Second Law for the system or for the caravan or for the car |
| $T - 150 = 500a$ | A1 | Two correct equations |
| $a = \left(\dfrac{40000}{24} - 800\right)\div 2250$ leading to $a = \ldots$ | M1 | Solves for $a$ or for $T$ |
| Acceleration $= 0.385$ ms$^{-2}$ and Tension $= 343$ N | A1 | From $a = \dfrac{52}{135} = 0.38518\ldots$ and $T = \dfrac{9250}{27} = 342.59\ldots$ |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\text{DF} = 800 + 2250\times g\times 0.14$ | M1 | Resolving up hill using DF = Total resistances |
| $\dfrac{31000}{v} = 800 + 2250\times g\times 0.14$ | M1 | Use of $P = \text{DF}\times v$ to form equation in $v$ |
| $v = 7.85$ | A1 | From $v = \dfrac{620}{79} = 7.848\ldots$ |
---6 A car of mass 1750 kg is pulling a caravan of mass 500 kg . The car and the caravan are connected by a light rigid tow-bar. The resistances to the motion of the car and caravan are 650 N and 150 N respectively.
\begin{enumerate}[label=(\alph*)]
\item The car and caravan are moving along a straight horizontal road at a constant speed of $24 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Find the power of the car's engine.
\item The engine's power is now suddenly increased to 40 kW .
Find the instantaneous acceleration of the car and caravan and find the tension in the tow-bar.
\end{enumerate}\item The car and caravan now travel up a straight hill, inclined at an angle $\sin ^ { - 1 } 0.14$ to the horizontal, at a constant speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car's engine is working at 31 kW . The resistances to the motion of the car and caravan are unchanged.
Find $v$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2022 Q6 [10]}}