One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(\frac{16}{9}Mg\), is attached to a fixed point \(O\). A particle \(P\) of mass \(4M\) is attached to the other end of the string and hangs vertically in equilibrium. Another particle of mass \(2M\) is attached to \(P\) and the combined particle is then released from rest. The speed of the combined particle when it has descended a distance \(\frac{1}{4}a\) is \(v\). Find an expression for \(v\) in terms of \(g\) and \(a\). [6]

Question 3:
3 | 16
Mge
3 3
In equilibrium, =4Mg, e= a
a 4 | B1
In subsequent motion,
Loss in GPE = gain in EPE + gain in KE | M1 | Energy equation with GPE and KE terms correct and at least one
EPE term.
Dimensionally correct.
6Mga 1 16 Mg  3a 2 1
= . . .a2 −  + .6Mv2
4 2 3 a   4   2
  | B1 | EPE correct.
A1 | All correct.
3Mga 8Mg 7
= . a2 +3Mv2 etc
2 3a 16 | M1 | Attempt to find v in terms of a and g.
ga 1
=3v2, v= ga
3 3 | A1
6
Question | Answer | Marks | Guidance

One end of a light elastic string, of natural length $a$ and modulus of elasticity $\frac{16}{9}Mg$, is attached to a fixed point $O$. A particle $P$ of mass $4M$ is attached to the other end of the string and hangs vertically in equilibrium. Another particle of mass $2M$ is attached to $P$ and the combined particle is then released from rest. The speed of the combined particle when it has descended a distance $\frac{1}{4}a$ is $v$.

Find an expression for $v$ in terms of $g$ and $a$.
[6]

\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q3 [6]}}