| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| 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 mechanics problem involving power, forces, and energy with straightforward application of P=Fv and F=ma. While it has multiple parts requiring careful bookkeeping of forces and resistances, each step uses routine A-level mechanics techniques without requiring novel insight or complex problem-solving strategies. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = (440 + 280) \times 30\) | M1 | Using \(P = Fv\) with \(F\) as total resistance |
| \(P = 720 \times 30 = 21.6\) kW | A1 | Answer must be in kW |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = 21600 - 8000\) W, \(\text{DF} = \frac{21600-8000}{30} \left[= \frac{13600}{30} = 453.333...\right]\) | B1FT | Follow through on *their* power from 5(a)(i). Allow Driving Force \(= \frac{8000}{30} = 266.7\) as the force due to solely to the change in power provided correct equation(s) used. |
| Car: \(\text{DF} - 440 - T = 1250a\) Caravan: \(T - 280 = 800a\) System: \(\text{DF} - (440+280) = 2050a\) | M1 | Apply Newton's 2nd law to either the car or to the caravan or to the system. Must be correct number of relevant terms. If \(\text{DF} = \frac{8000}{30}\) is used then the equations must be either \(-\text{DF} = 2050a\) or \(T - 280 = 800a\) |
| Solve for either \(a\) or \(T\) | M1 | Using equation(s) with no missing/extra terms, \(\text{DF} \neq 720\). Solving for \(a\) either from the system equation or from the car AND caravan equation. OR solving for \(T\) from the car AND caravan equation. |
| \(a = -0.13 \text{ ms}^{-2}\) and \(T = 176\) N | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| System: \(\text{DF} = 720 + 2050g \times 0.06 \quad [=1950]\) Car: \(\text{DF} - 440 - T - 1250g \times 0.06 = 0\) Caravan: \(T - 280 - 800g \times 0.06 = 0\) | M1 | Apply Newton's 2nd law with \(a = 0\), either to the system OR by eliminating \(T\) between the equations for the car and the caravan, no extra or missing relevant terms, dimensionally correct, to find DF |
| \(1950v = 28000\) | B1 | \(P = \text{DF} \times v \cdot \frac{28000}{v}\) SOI. |
| \(v = 14.4 \text{ ms}^{-1}\) | A1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{PE} = 800g \times d \times 0.06 = 800g \times 14.4 \times 60 \times 0.06\) | M1 | Using \(\text{PE} = mgh\) with \(h\) being height gained in 60 s, using *their* \(v\) |
| \(\text{PE} = 414\,000\) J or \(\text{PE} = 414\) kJ | A1 | Using \(v = 560/39 = 14.359\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(28\,000 \times 60 = \text{PE of Caravan} + 1250g \times d \times 0.06 + 720 \times d\) and \(d = 60 \times 14.359 = 861.54\) | M1 | For use of \(\text{WD} = P \times t\) to find an expression for PE of caravan and the distance travelled up the incline in 1 minute. |
| \([\text{PE} = 28\,000 \times 60 - 1250g \times 861.54 \times 0.06 - 720 \times 861.54]\) \(\text{PE} = 414\,000\) J or \(\text{PE} = 414\) kJ | A1 | |
| 2 |
## Question 5(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = (440 + 280) \times 30$ | M1 | Using $P = Fv$ with $F$ as total resistance |
| $P = 720 \times 30 = 21.6$ kW | A1 | Answer must be in kW |
| | **2** | |
---
## Question 5(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = 21600 - 8000$ W, $\text{DF} = \frac{21600-8000}{30} \left[= \frac{13600}{30} = 453.333...\right]$ | B1FT | Follow through on *their* power from 5(a)(i). Allow Driving Force $= \frac{8000}{30} = 266.7$ as the force due to solely to the change in power provided correct equation(s) used. |
| Car: $\text{DF} - 440 - T = 1250a$ Caravan: $T - 280 = 800a$ System: $\text{DF} - (440+280) = 2050a$ | M1 | Apply Newton's 2nd law to either the car or to the caravan or to the system. Must be correct number of relevant terms. If $\text{DF} = \frac{8000}{30}$ is used then the equations must be either $-\text{DF} = 2050a$ or $T - 280 = 800a$ |
| Solve for either $a$ or $T$ | M1 | Using equation(s) with no missing/extra terms, $\text{DF} \neq 720$. Solving for $a$ either from the system equation or from the car AND caravan equation. OR solving for $T$ from the car AND caravan equation. |
| $a = -0.13 \text{ ms}^{-2}$ and $T = 176$ N | A1 | |
| | **4** | |
---
## Question 5(b)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| System: $\text{DF} = 720 + 2050g \times 0.06 \quad [=1950]$ Car: $\text{DF} - 440 - T - 1250g \times 0.06 = 0$ Caravan: $T - 280 - 800g \times 0.06 = 0$ | M1 | Apply Newton's 2nd law with $a = 0$, either to the system OR by eliminating $T$ between the equations for the car and the caravan, no extra or missing relevant terms, dimensionally correct, to find DF |
| $1950v = 28000$ | B1 | $P = \text{DF} \times v \cdot \frac{28000}{v}$ SOI. |
| $v = 14.4 \text{ ms}^{-1}$ | A1 | |
| | **3** | |
---
## Question 5(b)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{PE} = 800g \times d \times 0.06 = 800g \times 14.4 \times 60 \times 0.06$ | M1 | Using $\text{PE} = mgh$ with $h$ being height gained in 60 s, using *their* $v$ |
| $\text{PE} = 414\,000$ J or $\text{PE} = 414$ kJ | A1 | Using $v = 560/39 = 14.359$ |
**Alternative method for Question 5(b)(ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $28\,000 \times 60 = \text{PE of Caravan} + 1250g \times d \times 0.06 + 720 \times d$ and $d = 60 \times 14.359 = 861.54$ | M1 | For use of $\text{WD} = P \times t$ to find an expression for PE of caravan and the distance travelled up the incline in 1 minute. |
| $[\text{PE} = 28\,000 \times 60 - 1250g \times 861.54 \times 0.06 - 720 \times 861.54]$ $\text{PE} = 414\,000$ J or $\text{PE} = 414$ kJ | A1 | |
| | **2** | |
---5 A car of mass 1250 kg is pulling a caravan of mass 800 kg along a straight road. The resistances to the motion of the car and caravan are 440 N and 280 N respectively. The car and caravan are connected by a light rigid tow-bar.
\begin{enumerate}[label=(\alph*)]
\item The car and caravan move along a horizontal part of the road at a constant speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Calculate, in kW , the power developed by the engine of the car.
\item Given that this power is suddenly decreased by 8 kW , find the instantaneous deceleration of the car and caravan and the tension in the tow-bar.
\end{enumerate}\item The car and caravan now travel along a part of the road inclined at $\sin ^ { - 1 } 0.06$ to the horizontal. The car and caravan travel up the incline at constant speed with the engine of the car working at 28 kW .
\begin{enumerate}[label=(\roman*)]
\item Find this constant speed.
\item Find the increase in the potential energy of the caravan in one minute.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q5 [11]}}