M4 June 2013 Q4
4.
A small smooth peg \(P\) is fixed at a distance \(d\) from a fixed smooth vertical wire. A particle of mass \(3m\) is attached to one end of a light inextensible string which passes over \(P\). The particle hangs vertically below \(P\). The other end of the string is attached to a small ring \(R\) of mass \(m\), which is threaded on the wire, as shown in Figure 3.
| Scheme | Marks |
|---|---|
| PE of ring \(= -mgx\) | B1 |
| PE of particle \(= -3mg\left(L - \sqrt{x^2 + d^2}\right)\) | M1 A1 |
| \(\Rightarrow V = 3mg\sqrt{x^2 + d^2} - mgx\) + constant. AG | A1 |
| (4) |
Notes
B1 Taking the level of the peg as zero PE
A1 Watch out
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}x} = \dfrac{3mg.2x}{2\sqrt{x^2 + d^2}} - mg\) | M1 |
| \(\dfrac{dV}{dx} = 0 \Rightarrow 3x = \sqrt{x^2 + d^2},\ 9x^2 = x^2 + d^2,\ 8x^2 = d^2\) | M1 |
| \(x = \dfrac{d}{\sqrt{8}} = \left(\dfrac{\sqrt{2}d}{4}\right)\) | A1 |
Notes
M1 Set \(\dfrac{dV}{dx} = 0\) and solve for \(x\)
A1 \(0.354d\) of better
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2} = 3mg\left(\dfrac{\sqrt{x^2 + d^2}.1 - x.\dfrac{2x}{2\sqrt{x^2 + d^2}}}{x^2 + d^2}\right) =\) | M1 |
| \(3mg\left(\dfrac{\sqrt{9x^2}.1 - x.\dfrac{2x}{2\sqrt{9x^2}}}{9x^2}\right) \quad = \dfrac{3mgd^2}{\left(x^2 + d^2\right)^{\frac{3}{2}}}\ (> 0)\) | A1 |
| Stable | A1ft |
| (10 marks) |
Notes
M1 Product or quotient rule \(\dfrac{\mathrm{d}^2V}{\mathrm{d}x^2} = \dfrac{3mg}{\sqrt{x^2 + d^2}} - \dfrac{3mgx}{2}.2x.\left(x^2 + d^2\right)^{-\frac{3}{2}}\)
A1 OR \(= 3mg\left(\dfrac{3x - \dfrac{x}{3}}{9x^2}\right)(> 0)\) Correct unsimplified.
\(\dfrac{16\sqrt{2}mg}{9d},\ \ 2.5\dfrac{mg}{d},\ \ \dfrac{\mathrm{d}^2V}{\mathrm{d}\theta^2} = \dfrac{9mgd}{\sqrt{8}}\)
A1ft Correct conclusion for their expression