CAIE M1 2024 November — Question 2 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeWork done against friction/resistance - inclined plane or slope
DifficultyModerate -0.3 This is a straightforward application of the work-energy theorem with clearly stated values. Students must account for gravitational PE loss, initial KE, work against resistance, and final KE—all standard M1 techniques requiring no novel insight, just careful bookkeeping of energy terms.
Spec6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle
2 A block of mass 20 kg is held at rest at the top of a plane inclined at \(30 ^ { \circ }\) to the horizontal. The block is projected with speed \(5 \mathrm {~ms} ^ { - 1 }\) down a line of greatest slope of the plane. There is a resistance force acting on the block. As the block moves 2 m down the plane from its point of projection, the work done against this resistance force is 50 J . Find the speed of the block when it has moved 2 m down the plane. \includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-04_2716_38_109_2012}
Question 2:
AnswerMarks Guidance
AnswerMark Guidance
Change in PE \(=\pm 20g\sin 30\times 2\ [=\pm 200]\)B1
\(\pm\frac{1}{2}\times 20v^2\ [=\pm 10v^2]\) and \(\pm\frac{1}{2}\times 20\times 5^2\ [=\pm 250]\)B1 For either expression. Do not allow \(\frac{1}{2}\times 20(v-5)^2\).
\(\frac{1}{2}\times 20v^2-\frac{1}{2}\times 20\times 5^2=20g\sin 30\times 2-50\ \left[10v^2-250=200-50\right]\)M1 Attempt at work-energy equation; 4 terms; dimensionally correct; allow sign errors; PE term must include a component (allow sin/cos mix). Do not allow \(\frac{1}{2}\times 20(v-5)^2\) for KE.
Speed \(=6.32\ \text{ms}^{-1}\) OR \(\sqrt{40}\ \text{ms}^{-1}\)A1 OE, e.g. \(2\sqrt{10}\). 6.324555…. AWRT 6.32.
Special case (constant resistance force):
\(20a=20g\sin 30-\frac{50}{2}\ [\rightarrow a=3.75]\)B1
\(v^2=5^2+2\times 2\times 3.75\Rightarrow v=6.32\ \text{ms}^{-1}\) OR \(\sqrt{40}\ \text{ms}^{-1}\)B1 OE, e.g. \(2\sqrt{10}\). 6.324555…. AWRT 6.32. 
## Question 2:

| Answer | Mark | Guidance |
|--------|------|----------|
| Change in PE $=\pm 20g\sin 30\times 2\ [=\pm 200]$ | B1 | |
| $\pm\frac{1}{2}\times 20v^2\ [=\pm 10v^2]$ and $\pm\frac{1}{2}\times 20\times 5^2\ [=\pm 250]$ | B1 | For either expression. Do not allow $\frac{1}{2}\times 20(v-5)^2$. |
| $\frac{1}{2}\times 20v^2-\frac{1}{2}\times 20\times 5^2=20g\sin 30\times 2-50\ \left[10v^2-250=200-50\right]$ | M1 | Attempt at work-energy equation; 4 terms; dimensionally correct; allow sign errors; PE term must include a component (allow sin/cos mix). Do not allow $\frac{1}{2}\times 20(v-5)^2$ for KE. |
| Speed $=6.32\ \text{ms}^{-1}$ OR $\sqrt{40}\ \text{ms}^{-1}$ | A1 | OE, e.g. $2\sqrt{10}$. 6.324555…. AWRT 6.32. |
| **Special case (constant resistance force):** | | |
| $20a=20g\sin 30-\frac{50}{2}\ [\rightarrow a=3.75]$ | B1 | |
| $v^2=5^2+2\times 2\times 3.75\Rightarrow v=6.32\ \text{ms}^{-1}$ OR $\sqrt{40}\ \text{ms}^{-1}$ | B1 | OE, e.g. $2\sqrt{10}$. 6.324555…. AWRT 6.32. |

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2 A block of mass 20 kg is held at rest at the top of a plane inclined at $30 ^ { \circ }$ to the horizontal. The block is projected with speed $5 \mathrm {~ms} ^ { - 1 }$ down a line of greatest slope of the plane. There is a resistance force acting on the block. As the block moves 2 m down the plane from its point of projection, the work done against this resistance force is 50 J .

Find the speed of the block when it has moved 2 m down the plane.\\

\includegraphics[max width=\textwidth, alt={}, center]{145d93bd-7f56-4e8c-a646-938330511347-04_2716_38_109_2012}

\hfill \mbox{\textit{CAIE M1 2024 Q2 [4]}}
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