One end of a light elastic string of natural length \(0.8\) m and modulus of elasticity \(36\) N is attached to a fixed point \(O\) on a smooth plane. The plane is inclined at an angle \(\alpha\) to the horizontal, where \(\sin \alpha = \frac{3}{5}\). A particle \(P\) of mass \(2\) kg is attached to the other end of the string. The string lies along a line of greatest slope of the plane with the particle below the level of \(O\). The particle is projected with speed \(\sqrt{2}\) m s\(^{-1}\) directly down the plane from the position where \(OP\) is equal to the natural length of the string. Find the maximum extension of the string during the subsequent motion. [5]

Question 2:
2 | 1 36x2 22.5x2)
× (=
2 0.8 | B1 | EPE correct.
Loss in GPE + loss in KE = gain in EPE
1 1 36x2
x mgsinα+ m×2= ×
2 2 0.8 | *M1 | Energy equation with only GPE, EPE and KE terms, allow sign
errors, allow missing g for M1 only, weight must be resolved
(allow sin or cos).
All terms correct | A1
 6 x+ 1 = 9 x2  [leading to45x2 −24x−4 = 0]
 
5 5 4  | DM1 | Simplify to 3-term quadratic and attempt to solve.
( )( )
3x−2 15x+2 =0
2
x= only
3 | A1
5
Question | Answer | Marks | Guidance

One end of a light elastic string of natural length $0.8$ m and modulus of elasticity $36$ N is attached to a fixed point $O$ on a smooth plane. The plane is inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = \frac{3}{5}$.
A particle $P$ of mass $2$ kg is attached to the other end of the string. The string lies along a line of greatest slope of the plane with the particle below the level of $O$. The particle is projected with speed $\sqrt{2}$ m s$^{-1}$ directly down the plane from the position where $OP$ is equal to the natural length of the string.

Find the maximum extension of the string during the subsequent motion. [5]

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q2 [5]}}