| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Maximum speed on horizontal road |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring standard application of Newton's second law, power equations (P=Fv), and energy conservation. Part (a) uses F=ma and P=Fv, part (b) recognizes steady speed means zero acceleration, and part (c) applies work-energy theorem with given values. All steps are routine M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(DF = \frac{P}{7}\) | \(B1\) | For \(P = DF \times 7\) OE seen at any point in working. Allow any force term or simply \(DF\), e.g. \(32,\ 80\times0.1,\ 32-80\times0.1,\ 80\times10\) etc. |
| \(D - 32 = 80 \times 0.1\) | \(M1\) | For use of Newton's second law. Must have correct number of terms. Allow sign errors |
| \([\text{Power} = ]\ 280\ \text{W}\) | \(A1\) | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\text{At steady speed driving force} =]\ 32 = \frac{280}{v}\) | \(M1\) | Attempt at equilibrium equation (\(a = 0\)) with their power |
| Steady speed \(= 8.75\ \text{ms}^{-1}\) or \(\frac{35}{4}\ \text{ms}^{-1}\) | \(A1FT\) | OE. FT their power from part (a) \(\frac{their\ 280}{32}\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(120 \times 4 \quad [480]\) | \(B1\) | Work done by cyclist |
| \(\frac{1}{2} \times 80v^2 \quad [40v^2]\) or \(\frac{1}{2} \times 80 \times 7.5^2 \quad [2250]\) | \(B1\) | For at least one KE term |
| \(80g \times \frac{1}{20} \times 32.2 \quad [1288]\) or \([80 \times 10 \times 1.61]\) | \(B1\) | Change in PE |
| Attempt at work-energy equation | \(M1\) | Attempt at work-energy equation with five relevant terms (four relevant terms plus work done against resistance); dimensionally correct. Allow sign errors. Allow sin/cos mix |
| \(120\times4 + 80g\times\frac{1}{20}\times32.2 - 1128 = \frac{1}{2}\times80(v^2 - 7.5^2)\), \([480 + 1288 - 1128 = 40v^2 - 2250]\) | \(A1\) | For correct equation |
| Speed \(= 8.5[0]\ \text{ms}^{-1}\) or \(\frac{17}{2}\) | \(A1\) | OE. Use of constant acceleration scores M0 and cannot score B marks if the method leading to their answer only uses constant acceleration |
| 6 |
## Question 6(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $DF = \frac{P}{7}$ | $B1$ | For $P = DF \times 7$ OE seen at any point in working. Allow any force term or simply $DF$, e.g. $32,\ 80\times0.1,\ 32-80\times0.1,\ 80\times10$ etc. |
| $D - 32 = 80 \times 0.1$ | $M1$ | For use of Newton's second law. Must have correct number of terms. Allow sign errors |
| $[\text{Power} = ]\ 280\ \text{W}$ | $A1$ | |
| | **3** | |
---
## Question 6(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\text{At steady speed driving force} =]\ 32 = \frac{280}{v}$ | $M1$ | Attempt at equilibrium equation ($a = 0$) with their power |
| Steady speed $= 8.75\ \text{ms}^{-1}$ or $\frac{35}{4}\ \text{ms}^{-1}$ | $A1FT$ | OE. FT their power from part (a) $\frac{their\ 280}{32}$ |
| | **2** | |
---
## Question 6(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $120 \times 4 \quad [480]$ | $B1$ | Work done by cyclist |
| $\frac{1}{2} \times 80v^2 \quad [40v^2]$ or $\frac{1}{2} \times 80 \times 7.5^2 \quad [2250]$ | $B1$ | For at least one KE term |
| $80g \times \frac{1}{20} \times 32.2 \quad [1288]$ or $[80 \times 10 \times 1.61]$ | $B1$ | Change in PE |
| Attempt at work-energy equation | $M1$ | Attempt at work-energy equation with five relevant terms (four relevant terms plus work done against resistance); dimensionally correct. Allow sign errors. Allow sin/cos mix |
| $120\times4 + 80g\times\frac{1}{20}\times32.2 - 1128 = \frac{1}{2}\times80(v^2 - 7.5^2)$, $[480 + 1288 - 1128 = 40v^2 - 2250]$ | $A1$ | For correct equation |
| Speed $= 8.5[0]\ \text{ms}^{-1}$ or $\frac{17}{2}$ | $A1$ | OE. Use of constant acceleration scores M0 and cannot score B marks if the method leading to their answer only uses constant acceleration |
| | **6** | |6 A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and her bicycle is 80 kg . There is a constant resistance force of magnitude 32 N to the cyclist's motion. At an instant when she is travelling at $7 \mathrm {~ms} ^ { - 1 }$, her acceleration is $0.1 \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the power output of the cyclist.
\item Find the steady speed that the cyclist can maintain if her power output and the resistance force are both unchanged.\\
\includegraphics[max width=\textwidth, alt={}, center]{c3246fbe-6f77-48f7-98eb-19e9166008bc-08_2718_35_141_2012}\\
\includegraphics[max width=\textwidth, alt={}, center]{c3246fbe-6f77-48f7-98eb-19e9166008bc-09_2724_35_136_20}
The cyclist later descends a straight hill of length 32.2 m , inclined at an angle of $\sin ^ { - 1 } \left( \frac { 1 } { 20 } \right)$ to the horizontal. Her power output is now 120 W , and the resistance force now has variable magnitude such that the work done against this force in descending the hill is 1128 J . The time taken to descend the hill is 4 s .
\item Given that the speed of the cyclist at the top of the hill is $7.5 \mathrm {~ms} ^ { - 1 }$, find her speed at the bottom of the hill.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q6 [11]}}