A2 June 2025 Q3
3.
A particle \(P\) of mass 0.2 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to the fixed point \(A\).
A second light inextensible string of length 0.6 m has one end attached to \(P\) and the other end attached to the fixed point \(B\).
The point \(B\) is vertically below \(A\) such that \(AB = 0.8\) m, as shown in Figure 4.
The particle moves in a horizontal circle with constant angular speed \(\omega\) radians per second, with both strings taut.
Given that the tension in the string from \(P\) to \(A\) is twice the tension in the string from \(P\) to \(B\), find the value of \(\omega\). (7)
| Scheme | Marks | AO |
|---|---|---|
 | ||
| Resolve vertically | M1 | 3.4 |
| \(\updownarrow\ 2T\sin\theta = 0.2g + T\sin\theta \qquad (T\sin\theta = 0.2g)\) | A1 | 1.1b |
| Equation of motion | M1 | 3.4 |
| \(0.2r\omega^2 = 2T\cos\theta + T\cos\theta \qquad (0.12\omega^2 = 3T)\) | A1 A1 | 1.1b 1.1b |
| Complete strategy to find \(\omega\) | DM1 | 3.1a |
| \(\omega^2 = 25T = \dfrac{5g}{\sin\theta} = \dfrac{15g}{2} \qquad \omega = 8.6\ \ (8.57)\) | A1 | 1.1b |
| (7 marks) |
Notes
M1: Correct number of terms. Condone sin/cos confusion and sign errors
A1: Correct unsimplified equation. Accept \(T_A\sin\theta = 0.2g + T_B\sin\theta\)
Allow \(m\) in place of 0.2 and their numerical \(\sin\theta\) / \(\cos\theta\)
M1: Circular motion. Correct number of terms. Condone sin/cos confusion.
Acceleration must be \(r\omega^2\) or \(v^2/r\)
A1: Unsimplified equation with at most one error.
Allow \(m\) in place of 0.2 and their numerical \(\sin\theta\) / \(\cos\theta\)
A1: Correct unsimplified equation. Accept \(0.2r\omega^2 = T_A\cos\theta + T_B\cos\theta\)
Allow \(m\) in place of 0.2 and their numerical \(\sin\theta\) / \(\cos\theta\)
DM1: Complete strategy to form sufficient equations to solve for \(\omega\) including upper tension being twice lower tension and a method to find \(r\).
Dependent on both previous M marks.
A1: 2 or 3 sf only