Standard +0.8 This is a multi-step energy conservation problem requiring careful geometric analysis to find string extensions at two positions, then applying conservation of energy with both gravitational and elastic potential energy terms. It demands spatial reasoning, Pythagoras theorem application, and algebraic manipulation beyond routine exercises, but follows a standard energy method framework for A-level mechanics.
5
\includegraphics[max width=\textwidth, alt={}, center]{a20a6641-d771-4c89-b40f-168a0c61f99d-3_577_693_1740_724}
A particle \(P\) of mass 0.2 kg is attached to the mid-point of a light elastic string of natural length 5.5 m and modulus of elasticity \(\lambda \mathrm { N }\). The ends of the string are attached to fixed points \(A\) and \(B\) which are at the same horizontal level and 6 m apart. \(P\) is held at rest at a point 1.25 m vertically above the mid-point of \(A B\) and then released. \(P\) travels a distance 5.25 m downwards before coming to instantaneous rest (see diagram). By considering the changes in gravitational potential energy and elastic potential energy as \(P\) travels downwards, find the value of \(\lambda\).
For using \(\text{EE} = \lambda x^2 \div (2L)\); \(L\) must be correct (2.75 or 5.5)
For any correct expression for Initial EPE or Final EPE: \([2\times0.5^2\lambda \div (2\times2.75)]\) for initial or \(2\times2.25^2\lambda \div (2\times2.75)\) for final
A1 ft
ft incorrect \(AP\)
Gain in EPE \(= (81-4)\lambda/44 = 1.75\lambda\)
A1
Any correct expression
M1
For applying conservation of energy
\(1.75\lambda = 10.5\)
A1 ft
For any correct equation in \(\lambda\); ft only if initial and final EPE are used
\(\lambda = 6\)
A1
[8]
## Question 5:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Loss in GPE $= 0.2g \times 5.25$ (10.5 J) | B1 | |
| $AP$ is 3.25 initially and 5 finally | B1 | |
| | M1 | For using $\text{EE} = \lambda x^2 \div (2L)$; $L$ must be correct (2.75 or 5.5) |
| For any correct expression for Initial EPE or Final EPE: $[2\times0.5^2\lambda \div (2\times2.75)]$ for initial or $2\times2.25^2\lambda \div (2\times2.75)$ for final | A1 ft | ft incorrect $AP$ |
| Gain in EPE $= (81-4)\lambda/44 = 1.75\lambda$ | A1 | Any correct expression |
| | M1 | For applying conservation of energy |
| $1.75\lambda = 10.5$ | A1 ft | For any correct equation in $\lambda$; ft only if initial and final EPE are used |
| $\lambda = 6$ | A1 | **[8]** |
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\includegraphics[max width=\textwidth, alt={}, center]{a20a6641-d771-4c89-b40f-168a0c61f99d-3_577_693_1740_724}
A particle $P$ of mass 0.2 kg is attached to the mid-point of a light elastic string of natural length 5.5 m and modulus of elasticity $\lambda \mathrm { N }$. The ends of the string are attached to fixed points $A$ and $B$ which are at the same horizontal level and 6 m apart. $P$ is held at rest at a point 1.25 m vertically above the mid-point of $A B$ and then released. $P$ travels a distance 5.25 m downwards before coming to instantaneous rest (see diagram). By considering the changes in gravitational potential energy and elastic potential energy as $P$ travels downwards, find the value of $\lambda$.
\hfill \mbox{\textit{CAIE M2 2005 Q5 [8]}}