CAIE M2 2002 June — Question 1 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2002
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeVertical elastic string: released from rest, string starts taut
DifficultyStandard +0.3 This is a straightforward energy conservation problem with elastic strings. Students must identify the extension at release (0.4m), calculate elastic PE loss using ½λx²/l, find gravitational PE gain (mgh for 0.4m), then apply energy conservation to find speed. All steps are standard M2 techniques with no novel insight required, making it slightly easier than average.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

1 One end of a light elastic string of natural length 1.6 m and modulus of elasticity 25 N is attached to a fixed point \(A\). A particle \(P\) of mass 0.15 kg is attached to the other end of the string. \(P\) is held at rest at a point 2 m vertically below \(A\) and is then released.
  1. For the motion from the instant of release until the string becomes slack, find the loss of elastic potential energy and the gain in gravitational potential energy.
  2. Hence find the speed of \(P\) at the instant the string becomes slack.

(i)
AnswerMarks Guidance
Uses the correct EPE formula \([25 \times 0.4^2/(2 \times 1.6)]\)M1
Obtains 1.25 JA1
Obtains \(\text{GPE} = 0.15g \times 0.4 = 0.6\) JB1 3 marks
(ii)
AnswerMarks Guidance
Attempts to form an energy equation involving EPE, GPE and KE terms \([1.25 = \frac{1}{2}(0.15v^2 + 0.6]\)M1
Obtains speed as 2.94 m s\(^{-1}\) (2.943920)A1 2 marks
**(i)**
| Uses the correct EPE formula $[25 \times 0.4^2/(2 \times 1.6)]$ | M1 |
| Obtains 1.25 J | A1 |
| Obtains $\text{GPE} = 0.15g \times 0.4 = 0.6$ J | B1 | 3 marks |

**(ii)**
| Attempts to form an energy equation involving EPE, GPE and KE terms $[1.25 = \frac{1}{2}(0.15v^2 + 0.6]$ | M1 |
| Obtains speed as 2.94 m s$^{-1}$ (2.943920) | A1 | 2 marks |

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1 One end of a light elastic string of natural length 1.6 m and modulus of elasticity 25 N is attached to a fixed point $A$. A particle $P$ of mass 0.15 kg is attached to the other end of the string. $P$ is held at rest at a point 2 m vertically below $A$ and is then released.\\
(i) For the motion from the instant of release until the string becomes slack, find the loss of elastic potential energy and the gain in gravitational potential energy.\\
(ii) Hence find the speed of $P$ at the instant the string becomes slack.

\hfill \mbox{\textit{CAIE M2 2002 Q1 [5]}}