| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: horizontal road |
| Difficulty | Moderate -0.3 This is a standard mechanics question testing power-force-velocity relationships and Newton's second law with connected particles. Part (a)(i) is straightforward P=Fv with total resistance. Part (a)(ii) requires F=P/v then F=ma, followed by resolving forces on the caravan to find tension - routine two-body problems. Part (b) adds an incline component but remains a direct application of P=Fv with weight components. All techniques are standard M1 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03l Newton's third law: extend to situations requiring force resolution3.03o Advanced connected particles: and pulleys6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = 650 \times 25\) | M1 | Use \(P = Fv\) with \(F =\) total resistance |
| \(P = 16\,250\text{ W} = 16.25\text{ kW}\) | A1 | Accept \(16\,300\text{ W}\) or \(16.3\text{ kW}\) (3sf) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{DF} = \dfrac{39000}{25}\ (= 1560)\) | B1 | For using \(\text{DF} = P/v\) |
| For applying Newton's 2nd law to the system to form an equation in \(a\), or to the caravan or the car to form an equation in \(T\) and \(a\) | M1 | \([1560 - 650 = 2400 \times a]\) |
| \(1560 - 650 = 2400a\); \(T - 250 = 800a\); \(1560 - 400 - T = 1600a\) | A1 | Two correct equations |
| \(\left[a = \dfrac{(1560 - 650)}{2400}\right]\) | M1 | For solving for \(a\) or for \(T\) |
| \(a = 0.379\text{ ms}^{-2}\) \((0.37916...)\); \(T = 553\text{ N}\) \((553.33...)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([\text{DF} = 650 + 2400 \times 10 \times 0.05]\) | M1 | Newton's 2nd law |
| \(32\,500 = (650 + 24\,000 \times 0.05)v\) | M1 | For using \(P = Fv\) |
| \(v = 17.6\) | A1 | Allow \(v = \dfrac{650}{37}\) |
## Question 6(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = 650 \times 25$ | M1 | Use $P = Fv$ with $F =$ total resistance |
| $P = 16\,250\text{ W} = 16.25\text{ kW}$ | A1 | Accept $16\,300\text{ W}$ or $16.3\text{ kW}$ (3sf) |
## Question 6(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{DF} = \dfrac{39000}{25}\ (= 1560)$ | B1 | For using $\text{DF} = P/v$ |
| For applying Newton's 2nd law to the system to form an equation in $a$, or to the caravan or the car to form an equation in $T$ and $a$ | M1 | $[1560 - 650 = 2400 \times a]$ |
| $1560 - 650 = 2400a$; $T - 250 = 800a$; $1560 - 400 - T = 1600a$ | A1 | Two correct equations |
| $\left[a = \dfrac{(1560 - 650)}{2400}\right]$ | M1 | For solving for $a$ or for $T$ |
| $a = 0.379\text{ ms}^{-2}$ $(0.37916...)$; $T = 553\text{ N}$ $(553.33...)$ | A1 | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[\text{DF} = 650 + 2400 \times 10 \times 0.05]$ | M1 | Newton's 2nd law |
| $32\,500 = (650 + 24\,000 \times 0.05)v$ | M1 | For using $P = Fv$ |
| $v = 17.6$ | A1 | Allow $v = \dfrac{650}{37}$ |6 A car of mass 1600 kg is pulling a caravan of mass 800 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 400 N and 250 N respectively.
\begin{enumerate}[label=(\alph*)]
\item The car and caravan are travelling along a straight horizontal road.
\begin{enumerate}[label=(\roman*)]
\item Given that the car and caravan have a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the power of the car's engine.
\item The engine's power is now suddenly increased to 39 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 of $\sin ^ { - 1 } 0.05$ to the horizontal, at a constant speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car's engine is working at 32.5 kW .
Find $v$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2020 Q6 [10]}}