M3 June 2005 Q3
3. A light elastic string has natural length \(2l\) and modulus of elasticity \(4mg\). One end of the string is attached to a fixed point \(A\) and the other end to a fixed point \(B\), where \(A\) and \(B\) lie on a smooth horizontal table, with \(AB = 4l\). A particle \(P\) of mass \(m\) is attached to the mid-point of the string.
The particle is released from rest at the point of the line \(AB\) which is \(\dfrac{5l}{3}\) from \(B\). The speed of \(P\) at the mid-point of \(AB\) is \(V\).
| Scheme | Marks |
|---|---|
![]() | M1 A1 |
| \(\dfrac{1}{2}mV^2 + 2 \times \dfrac{4mgl^2}{2l} = \dfrac{4mg\left(\frac{2}{3}l\right)^2}{2l} + \dfrac{4mg\left(\frac{4}{3}l\right)^2}{2l}\) | M1 A1=A1ft |
| \(\dfrac{1}{2}V^2 + 4gl = \dfrac{8}{9}gl + \dfrac{32}{9}gl\) | |
| \(V^2 = \dfrac{8gl}{9}\) solving for \(V^2\) | M1 |
| \(V = \left(\dfrac{8gl}{9}\right)^{\frac{1}{2}}\) or exact equivalents | A1 |
| (7) |
Notes
The scheme brackets the two M1 marks for the energy equation and solving for \(V^2\): the second depends on the first.
Alternative using Newton’s second law
![]() | |
| N2L \(m\ddot{x} = T_2 - T_1 = -\dfrac{8mg}{l}x\) | M1 A1 |
| This is SHM, centre \(M\) | |
| \(a = \dfrac{l}{3}, \quad \omega^2 = \dfrac{8g}{l}\) | A1, A1ft |
| \(v^2 = \omega^2(a^2 - x^2) \Rightarrow v^2 = \dfrac{8g}{l}\left(\dfrac{l^2}{9} - x^2\right)\) Depends on showing SHM | M1 |
| At \(M\), \(x = 0\), \(V^2 = \dfrac{8gl}{9}, \ V = \left(\dfrac{8gl}{9}\right)^{\frac{1}{2}}\) or exact equivalents | M1, A1 |
The scheme brackets the last two M1 marks: the second depends on the first.
| Scheme | Marks |
|---|---|
| The maximum speed occurs when \(a = 0\) | B1 |
| At \(M\) the particle is in equilibrium (the sum of the forces is zero) \(\Rightarrow a = 0\) | B1 |
| (2) | |
| (9 marks) |
Alternative using Newton’s second law
| The particle is performing SHM about the mid-point of \(AB\). | B1 |
| The maximum speed occurs at the centre of the oscillation (when \(x = 0\)) | B1 |
The scheme brackets these two B1 marks.

