CAIE M2 2007 June — Question 2 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2007
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeAcceleration as function of displacement (v dv/dx method)
DifficultyStandard +0.8 This is a non-constant acceleration problem requiring the chain rule technique (a = v dv/dx) and integration, which is beyond basic mechanics. Students must recognize when velocity is maximum (a=0), then integrate to find that maximum velocity. While systematic, this requires understanding of calculus-based mechanics not typically seen in standard A-level, making it moderately challenging.
Spec3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods

2 A particle starts from rest at \(O\) and travels in a straight line. Its acceleration is \(( 3 - 2 x ) \mathrm { ms } ^ { - 2 }\), where \(x \mathrm {~m}\) is the displacement of the particle from \(O\).
  1. Find the value of \(x\) for which the velocity of the particle reaches its maximum value.
  2. Find this maximum velocity.

Question 2:
Part (i)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(3 - 2x = 0 \rightarrow x = 1.5\)B1
Part (ii)
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(v(dv/dx) = 3 - 2x\)M1 For using \(a = v(dv/dx)\), and attempting to solve using separation of variables
\(v^2/2 = 3x - x^2 \; (+C)\)A1
\(v^2/2 = 3 \times 1.5 - 1.5^2\)M1 For \(C = 0\) (may be implied) and substituting ans (i)
Maximum value is \(2.12 \text{ ms}^{-1}\)A1 4
Total: 5 marks
## Question 2:

### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3 - 2x = 0 \rightarrow x = 1.5$ | B1 | |

### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v(dv/dx) = 3 - 2x$ | M1 | For using $a = v(dv/dx)$, and attempting to solve using separation of variables |
| $v^2/2 = 3x - x^2 \; (+C)$ | A1 | |
| $v^2/2 = 3 \times 1.5 - 1.5^2$ | M1 | For $C = 0$ (may be implied) and substituting ans (i) |
| Maximum value is $2.12 \text{ ms}^{-1}$ | A1 | 4 | |

**Total: 5 marks**

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2 A particle starts from rest at $O$ and travels in a straight line. Its acceleration is $( 3 - 2 x ) \mathrm { ms } ^ { - 2 }$, where $x \mathrm {~m}$ is the displacement of the particle from $O$.\\
(i) Find the value of $x$ for which the velocity of the particle reaches its maximum value.\\
(ii) Find this maximum velocity.

\hfill \mbox{\textit{CAIE M2 2007 Q2 [5]}}