CAIE M1 2021 June — Question 1 3 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2021
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicWork done and energy
TypeEnergy method - smooth inclined plane (no resistance)
DifficultyModerate -0.8 This is a straightforward application of the work-energy theorem with only two energy components (kinetic and gravitational potential). The setup is simple with given values that substitute directly into ΔKE + ΔPE = 0, requiring basic algebraic manipulation to find the final speed. No problem-solving insight needed beyond recognizing which energy method to apply.
Spec6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle
1 A particle of mass 0.6 kg is projected with a speed of \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) down a line of greatest slope of a smooth plane inclined at \(10 ^ { \circ }\) to the horizontal. Use an energy method to find the speed of the particle after it has moved 15 m down the plane.
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
Initial \(KE = \frac{1}{2} \times 0.6 \times 4^2\) \([=4.8]\); Final \(KE = \frac{1}{2} \times 0.6 \times v^2\); \(PE \text{ loss} = 0.6 \times g \times 15\sin 10\) \([=15.628]\)B1 Any one of the three expressions correct
\(0.6 \times g \times 15\sin 10 + \frac{1}{2} \times 0.6 \times 4^2 = \frac{1}{2} \times 0.6 \times v^2\)M1 Apply energy equation, 3 terms, dimensions correct
\(v = 8.25 \text{ ms}^{-1}\)A1
[3] 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial $KE = \frac{1}{2} \times 0.6 \times 4^2$ $[=4.8]$; Final $KE = \frac{1}{2} \times 0.6 \times v^2$; $PE \text{ loss} = 0.6 \times g \times 15\sin 10$ $[=15.628]$ | **B1** | Any one of the three expressions correct |
| $0.6 \times g \times 15\sin 10 + \frac{1}{2} \times 0.6 \times 4^2 = \frac{1}{2} \times 0.6 \times v^2$ | **M1** | Apply energy equation, 3 terms, dimensions correct |
| $v = 8.25 \text{ ms}^{-1}$ | **A1** | |
| | **[3]** | |

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1 A particle of mass 0.6 kg is projected with a speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ down a line of greatest slope of a smooth plane inclined at $10 ^ { \circ }$ to the horizontal.

Use an energy method to find the speed of the particle after it has moved 15 m down the plane.\\

\hfill \mbox{\textit{CAIE M1 2021 Q1 [3]}}
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