One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(\lambda mg\), is attached to a fixed point \(O\). The string lies on a smooth horizontal surface. A particle \(P\) of mass \(m\) is attached to the other end of the string. The particle \(P\) is projected in the direction \(OP\). When the length of the string is \(\frac{4}{3}a\), the speed of \(P\) is \(\sqrt{2ag}\). When the length of the string is \(\frac{5}{3}a\), the speed of \(P\) is \(\frac{1}{2}\sqrt{2ag}\). Find the value of \(\lambda\). [4]

Question 2:
2 | 1
1 m  v2  v2    2 mg    2 a   2   1 a   2  
2  4  a   3  3    | M1 | Kinetic energy = elastic potential energy, 4 terms,
dimensionally correct, allow sign errors.
1
1  1  2 mg  2  2 1  2
m  2ag ag    a    a  
2  2  a  3  3  
  | A1 | With v substituted.
3 3 
Solve [ v2 g  a  ]
4 9  | M1 | Solve to find value for  dependent on energy
equation with 3 or 4 terms
9

2 | A1 | 9
SCB2 for  mg if given  not used
2
4
Question | Answer | Marks | Guidance

One end of a light elastic string, of natural length $a$ and modulus of elasticity $\lambda mg$, is attached to a fixed point $O$. The string lies on a smooth horizontal surface. A particle $P$ of mass $m$ is attached to the other end of the string. The particle $P$ is projected in the direction $OP$. When the length of the string is $\frac{4}{3}a$, the speed of $P$ is $\sqrt{2ag}$. When the length of the string is $\frac{5}{3}a$, the speed of $P$ is $\frac{1}{2}\sqrt{2ag}$.

Find the value of $\lambda$. [4]

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q2 [4]}}