CAIE M2 2015 June — Question 2 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2015
SessionJune
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 Part (i) is a straightforward equilibrium problem using Hooke's law (T = λx/l) and balancing forces, requiring only substitution. Part (ii) involves energy conservation with elastic potential energy, requiring identification of when the string becomes slack (at natural length) and careful bookkeeping of energy terms, but follows a standard method for this topic with no novel insight required.
Spec6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings

2 One end of a light elastic string of natural length 0.5 m and modulus of elasticity 30 N is attached to a fixed point \(O\). The other end of the string is attached to a particle \(P\) which hangs in equilibrium vertically below \(O\), with \(O P = 0.8 \mathrm {~m}\).
  1. Show that the mass of \(P\) is 1.8 kg . The particle is pulled vertically downwards and released from rest from the point where \(O P = 1.2 \mathrm {~m}\).
  2. Find the speed of \(P\) at the instant when the string first becomes slack.

Question 2:
Part (i):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(mg = 30(0.8 - 0.5)/0.5\)M1
\(m = 1.8\text{ kg}\) AGA1
Total: 2 marks
Part (ii):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(EE = 30(1.2 - 0.5)^2/(2 \times 0.5)\)B1
\(\frac{1.8v^2}{2} = \frac{30(1.2-0.5)^2}{2 \times 0.5} - 1.8 \times (1.2 - 0.5)g\)M1 KE/EE/PE equation, 3 terms; RHS = 2.1
\(v = 1.53\text{ ms}^{-1}\)A1
Total: 3 marks
## Question 2:

### Part (i):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $mg = 30(0.8 - 0.5)/0.5$ | M1 | |
| $m = 1.8\text{ kg}$ AG | A1 | |

**Total: 2 marks**

### Part (ii):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $EE = 30(1.2 - 0.5)^2/(2 \times 0.5)$ | B1 | |
| $\frac{1.8v^2}{2} = \frac{30(1.2-0.5)^2}{2 \times 0.5} - 1.8 \times (1.2 - 0.5)g$ | M1 | KE/EE/PE equation, 3 terms; RHS = 2.1 |
| $v = 1.53\text{ ms}^{-1}$ | A1 | |

**Total: 3 marks**

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2 One end of a light elastic string of natural length 0.5 m and modulus of elasticity 30 N is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ which hangs in equilibrium vertically below $O$, with $O P = 0.8 \mathrm {~m}$.\\
(i) Show that the mass of $P$ is 1.8 kg .

The particle is pulled vertically downwards and released from rest from the point where $O P = 1.2 \mathrm {~m}$.\\
(ii) Find the speed of $P$ at the instant when the string first becomes slack.

\hfill \mbox{\textit{CAIE M2 2015 Q2 [5]}}