AS October 2020 Q2
2.

One end of a string of length \(3a\) is attached to a point \(A\) and the other end is attached to a point \(B\) on a smooth horizontal table. The point \(B\) is vertically below \(A\) with \(AB = a\sqrt{3}\)
A small smooth bead, \(P\), of mass \(m\) is threaded on to the string. The bead \(P\) moves on the table in a horizontal circle, with centre \(B\), with constant speed \(U\). Both portions, \(AP\) and \(BP\), of the string are taut, as shown in Figure 2.
The string is modelled as being light and inextensible and the bead is modelled as a particle.
| Scheme | Marks | AO |
|---|---|---|
| \((a\sqrt{3})^2 + (3a - AP)^2 = AP^2\) | M1 | 1.1b |
| \(AP = 2a\) * | A1* | 1.1b |
| (2) |
Notes
M1: Use of Pythagoras \(3a^2 + 9a^2 - 6a \times AP + AP^2 = AP^2 \Rightarrow 6a \times AP = 12a^2\)
A1: \(AP = 2a\). GIVEN ANSWER
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion horizontally | M1 | 3.1b |
| \(T + T \times \dfrac{1}{2} = \dfrac{mU^2}{a}\) | A1 A1 | 1.1b 1.1b |
| \(T = \dfrac{2mU^2}{3a}\) | A1 | 2.2a |
| (4) |
Notes
M1: Use of horizontal equation to solve the problem, with correct no. of terms etc
A1: Equation with at most one error
A1: Correct equation
A1: Correct answer
| Scheme | Marks | AO |
|---|---|---|
| Resolving vertically | M1 | 3.1b |
| \(R + T \times \dfrac{\sqrt{3}}{2} = mg\) | A1 | 1.1b |
| On the table \(\Rightarrow R \gt 0\) | M1 | 2.1 |
| \(mg - \dfrac{2mU^2\sqrt{3}}{3a \times 2} \gt 0\) | A1 | 1.1b |
| \(U^2 \lt ag\sqrt{3}\) * | A1* | 2.2a |
| (5) |
Notes
M1: Use of vertical resolution to solve the problem, with correct no. of terms etc
A1: Correct equation
M1: Use of \(R \gt 0\)
A1: Correct inequality
A1*: Correctly obtained given answer
| Scheme | Marks | AO |
|---|---|---|
| Bead would lift off the table | B1 | 2.4 |
| (1) |
Notes
B1: Clear comment
| Scheme | Marks | AO |
|---|---|---|
| Tension would vary along the string | B1 | 3.5b |
| (1) | ||
| (13 marks) |
Notes
B1: Clear explanation