CAIE M2 2014 November — Question 2 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2014
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeHorizontal elastic string on smooth surface
DifficultyStandard +0.3 This is a standard energy conservation problem with elastic strings requiring application of EPE = λx²/2l and KE formulas. While it involves two parts and careful bookkeeping of energy terms, the method is straightforward and commonly practiced in M2 courses—slightly above average difficulty due to algebraic manipulation but no novel insight required.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

2 A particle \(P\) of mass 0.2 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 64 N . The other end of the string is attached to a fixed point \(A\) on a smooth horizontal surface. \(P\) is placed on the surface at a point 0.8 m from \(A\). The particle \(P\) is then projected with speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) directly away from \(A\).
  1. Calculate the distance \(A P\) when \(P\) is at instantaneous rest.
  2. Calculate the speed of \(P\) when it is 1.0 m from \(A\).

Question 2:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{0.2\times10^2}{2} = \frac{64e^2}{2\times0.8}\)M1, A1 KE loss = EE gain, \(e = 0.5\)
\(AP = 1.3\)A1 [3]
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{0.2\times10^2}{2} = \frac{64\times0.2^2}{2\times0.8} + \frac{0.2v^2}{2}\)M1, A1 KE/EE balance
\(v = 9.17\text{ ms}^{-1}\)A1 [3]
## Question 2:

### Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{0.2\times10^2}{2} = \frac{64e^2}{2\times0.8}$ | M1, A1 | KE loss = EE gain, $e = 0.5$ |
| $AP = 1.3$ | A1 | [3] |

### Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{0.2\times10^2}{2} = \frac{64\times0.2^2}{2\times0.8} + \frac{0.2v^2}{2}$ | M1, A1 | KE/EE balance |
| $v = 9.17\text{ ms}^{-1}$ | A1 | [3] |

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2 A particle $P$ of mass 0.2 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 64 N . The other end of the string is attached to a fixed point $A$ on a smooth horizontal surface. $P$ is placed on the surface at a point 0.8 m from $A$. The particle $P$ is then projected with speed $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ directly away from $A$.\\
(i) Calculate the distance $A P$ when $P$ is at instantaneous rest.\\
(ii) Calculate the speed of $P$ when it is 1.0 m from $A$.

\hfill \mbox{\textit{CAIE M2 2014 Q2 [6]}}