| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Instantaneous change in power or force |
| Difficulty | Moderate -0.3 This is a straightforward multi-part mechanics question requiring standard formulas (work = force × distance, power = force × velocity, F = ma) with no geometric insight or complex problem-solving. All parts follow directly from given information using routine M1 techniques, making it slightly easier than average but not trivial due to the multiple steps and unit conversions required. |
| Spec | 3.03d Newton's second law: 2D vectors6.02a Work done: concept and definition6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| \([\text{WD} = 1250 \times 36 \times 8]\) | M1 | For using Work Done = Force × Distance |
| \(\text{WD} = 360000\) J | A1 | or 360 kJ |
| Answer | Marks | Guidance |
|---|---|---|
| Power \(= 1250 \times 36\) or \(P = \frac{360000}{8}\ [= 45000\text{ J}]\) | B1 FT | FT Work Done from \(\frac{a(i)}{8}\) |
| \(= 45\) kW | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{DF} = \frac{57000}{36}\ [= 1583.3...]\) | M1 | Use changed Power in \(P = \text{DF} \times v\) |
| \(\frac{57000}{36} - 1250 = 1400a\) | M1 | For using Newton's 2nd law applied to the car |
| \(a = 0.238\ \text{ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{64000}{32} = 1250 + 1400g\sin\theta\) | M1 | For using DF = resistance + component of weight of the car |
| \(\theta = 3.1\ [3.0708...]\) | A1 |
## Question 4:
### Part (a)(i):
| $[\text{WD} = 1250 \times 36 \times 8]$ | M1 | For using Work Done = Force × Distance |
| $\text{WD} = 360000$ J | A1 | or 360 kJ |
### Part (a)(ii):
| Power $= 1250 \times 36$ or $P = \frac{360000}{8}\ [= 45000\text{ J}]$ | B1 FT | FT Work Done from $\frac{a(i)}{8}$ |
| $= 45$ kW | B1 | |
### Part (a)(iii):
| $\text{DF} = \frac{57000}{36}\ [= 1583.3...]$ | M1 | Use changed Power in $P = \text{DF} \times v$ |
| $\frac{57000}{36} - 1250 = 1400a$ | M1 | For using Newton's 2nd law applied to the car |
| $a = 0.238\ \text{ms}^{-2}$ | A1 | |
### Part (b):
| $\frac{64000}{32} = 1250 + 1400g\sin\theta$ | M1 | For using DF = resistance + component of weight of the car |
| $\theta = 3.1\ [3.0708...]$ | A1 | |
---4 A car of mass 1400 kg is moving on a straight road against a constant force of 1250 N resisting the motion.
\begin{enumerate}[label=(\alph*)]
\item The car moves along a horizontal section of the road at a constant speed of $36 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Calculate the work done against the resisting force during the first 8 seconds.
\item Calculate, in kW , the power developed by the engine of the car.
\item Given that this power is suddenly increased by 12 kW , find the instantaneous acceleration of the car.
\end{enumerate}\item The car now travels at a constant speed of $32 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a section of the road inclined at $\theta ^ { \circ }$ to the horizontal, with the engine working at 64 kW .
Find the value of $\theta$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q4 [9]}}