| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Force from power and speed |
| Difficulty | Moderate -0.3 Part (a) is a straightforward application of P=Fv at constant speed (equilibrium). Part (b) requires work-energy principle with multiple energy terms (KE change, PE gain, work against resistance), but follows a standard template for inclined plane problems with clear given values and direct substitution into ΔKE + ΔPE = Work_driving - Work_resistance. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resistive force \(= DF = \frac{60}{3} = 20\text{ N}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(PE = \pm84g\times150\sin0.8\) | B1 | \(\pm1759.23\ldots\) |
| KE change \(= \pm\left(\frac{1}{2}\times84v^2 - \frac{1}{2}\times84\times3^2\right)\) | B1 | \(\pm\left(\frac{1}{2}\times84v^2 - 378\right)\) |
| Work Done \(= \pm(24\times150 - 13\times150)\) | B1 | \(\pm(3600-1950)=\pm1650\) |
| Attempt at work-energy equation | M1 | 5 terms, dimensionally correct, allow sign errors, sin/cos mix on PE term, PE must include \(\sin0.8\) or \(\cos0.8\) |
| \(84g\times150\sin0.8+\frac{1}{2}\times84v^2-\frac{1}{2}\times84\times3^2=24\times150-13\times150\) \([1759.23\ldots+42v^2-378=3600-1950\rightarrow42v^2=268.765\ldots]\) | A1 | |
| \([v=]\ 2.53\text{ ms}^{-1}\) | A1 | AWRT 2.53; 2.5296… |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Resolve parallel to slope and use Newton's second law | \*M1 | Four terms, allow sign errors, allow sin/cos mix |
| \(24-13-84g\sin0.8=\pm84a\) | A1 | For reference \(a=\pm0.008669\ldots\) |
| Use constant acceleration formula to get an equation in \(v\) or \(v^2\) | DM1 | E.g. \(v^2=3^2+2\times(\text{their }a)\times150\) |
| \([v=]\ 2.53\text{ ms}^{-1}\) | A1 | AWRT 2.53; 2.5296… |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resistive force $= DF = \frac{60}{3} = 20\text{ N}$ | B1 | |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $PE = \pm84g\times150\sin0.8$ | B1 | $\pm1759.23\ldots$ |
| KE change $= \pm\left(\frac{1}{2}\times84v^2 - \frac{1}{2}\times84\times3^2\right)$ | B1 | $\pm\left(\frac{1}{2}\times84v^2 - 378\right)$ |
| Work Done $= \pm(24\times150 - 13\times150)$ | B1 | $\pm(3600-1950)=\pm1650$ |
| Attempt at work-energy equation | M1 | 5 terms, dimensionally correct, allow sign errors, sin/cos mix on PE term, PE must include $\sin0.8$ or $\cos0.8$ |
| $84g\times150\sin0.8+\frac{1}{2}\times84v^2-\frac{1}{2}\times84\times3^2=24\times150-13\times150$ $[1759.23\ldots+42v^2-378=3600-1950\rightarrow42v^2=268.765\ldots]$ | A1 | |
| $[v=]\ 2.53\text{ ms}^{-1}$ | A1 | AWRT 2.53; 2.5296… |
**Special case for use of constant acceleration: Maximum 4 marks**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Resolve parallel to slope and use Newton's second law | \*M1 | Four terms, allow sign errors, allow sin/cos mix |
| $24-13-84g\sin0.8=\pm84a$ | A1 | For reference $a=\pm0.008669\ldots$ |
| Use constant acceleration formula to get an equation in $v$ or $v^2$ | DM1 | E.g. $v^2=3^2+2\times(\text{their }a)\times150$ |
| $[v=]\ 2.53\text{ ms}^{-1}$ | A1 | AWRT 2.53; 2.5296… |
---4 An athlete of mass 84 kg is running along a straight road.
\begin{enumerate}[label=(\alph*)]
\item Initially the road is horizontal and he runs at a constant speed of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The athlete produces a constant power of 60 W .
Find the resistive force which acts on the athlete.
\item The athlete then runs up a 150 m section of the road which is inclined at $0.8 ^ { \circ }$ to the horizontal. The speed of the athlete at the start of this section of road is $3 \mathrm {~ms} ^ { - 1 }$ and he now produces a constant driving force of 24 N . The total resistive force which acts on the athlete along this section of road has constant magnitude 13 N .
Use an energy method to find the speed of the athlete at the end of the 150 m section of road.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2023 Q4 [7]}}