CAIE M2 2016 November — Question 2 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2016
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeVertical elastic string: released from rest at natural length or above (string initially slack)
DifficultyStandard +0.3 This is a standard two-part elastic string problem requiring Hooke's law for equilibrium (part i) and energy conservation for maximum speed (part ii). While it involves multiple steps and careful consideration of when the string becomes taut, the techniques are routine for M2 students and follow predictable patterns without requiring novel insight.
Spec6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

2 A particle \(P\) of mass 0.5 kg is attached to one end of a light elastic string with modulus of elasticity 24 N and natural length 0.6 m . The other end of the string is attached to a fixed point \(A\). The particle \(P\) hangs in equilibrium vertically below \(A\).
  1. Find the distance \(A P\). The particle \(P\) is raised to \(A\) and released from rest.
  2. Calculate the greatest speed of \(P\) in the subsequent motion.

Question 2:
Part (i):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(5 = 24e/0.6\)M1 Hence \(e = 0.125\)
\(AP = 0.725\) mA1
[2]
Part (ii):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(24 \times 0.125^2/2 \times 0.6\)B1 EE at eqm \((= 0.3125)\)
\(0.5g \times 0.725 =\)M1 KE/EE/PE conservation
\(24 \times 0.125^2/2 \times 0.6 + 0.5v^2/2\)
\(v = 3.64 \text{ ms}^{-1}\)A1
[3]
## Question 2:

### Part (i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $5 = 24e/0.6$ | M1 | Hence $e = 0.125$ |
| $AP = 0.725$ m | A1 | |
| | [2] | |

### Part (ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $24 \times 0.125^2/2 \times 0.6$ | B1 | EE at eqm $(= 0.3125)$ |
| $0.5g \times 0.725 =$ | M1 | KE/EE/PE conservation |
| $24 \times 0.125^2/2 \times 0.6 + 0.5v^2/2$ | | |
| $v = 3.64 \text{ ms}^{-1}$ | A1 | |
| | [3] | |

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2 A particle $P$ of mass 0.5 kg is attached to one end of a light elastic string with modulus of elasticity 24 N and natural length 0.6 m . The other end of the string is attached to a fixed point $A$. The particle $P$ hangs in equilibrium vertically below $A$.\\
(i) Find the distance $A P$.

The particle $P$ is raised to $A$ and released from rest.\\
(ii) Calculate the greatest speed of $P$ in the subsequent motion.

\hfill \mbox{\textit{CAIE M2 2016 Q2 [5]}}