M3 June 2012 Q7
7. A particle \(B\) of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.75 m and modulus of elasticity 24.5 N. The other end of the string is attached to a fixed point \(A\). The particle is hanging in equilibrium at the point \(E\), vertically below \(A\).
(a) Show that \(AE = 0.9\) m. (3)
The particle is held at \(A\) and released from rest. The particle first comes to instantaneous rest at the point \(C\).
(b) Find the distance \(AC\). (5)
(c) Show that while the string is taut, \(B\) is moving with simple harmonic motion with centre \(E\). (4)
(d) Calculate the maximum speed of \(B\). (2)
| Scheme | Marks |
|---|---|
| Use of \(T = \dfrac{\lambda x}{a} = mg\) | M1 |
| \(T = \dfrac{24.5x}{0.75} = 0.5g\) | A1 |
| \(x = \dfrac{0.75 \times 0.5g}{24.5} = 0.15,\ \ AE = 0.75 + 0.15 = 0.9\) (m) (**) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Using gain in EPE = loss in GPE | M1 |
| \(\dfrac{\lambda x^2}{2a} = \dfrac{24.5x^2}{1.5} = \ldots\ldots\) | A1 |
| \(\ldots\ldots = 0.5g(0.75 + x)\) | A1 |
| Form quadratic in \(x\) and attempt to solve for \(x\) : | DM1 |
| \(24.5x^2 = 5.5125 + 7.35x,\ \ 24.5x^2 - 7.35x - 5.5125 = 0,\) | |
| \(x = \dfrac{7.35 \pm \sqrt{7.35^2 + 4 \times 24.5 \times 5.5125}}{49}\) | |
| \(\left(\text{or } 40x^2 - 12x - 9 = 0,\ \ x = \dfrac{12 \pm \sqrt{144 + 1440}}{80}\right)\) | |
| \(x = 0.647\ldots\) (m) \(AC \approx 1.4\) (m) | A1 |
| (5) |
Notes
(Corrected from the printed mark scheme: the alternative formula is printed as \(\dfrac{12 \pm \sqrt{144 + 3600}}{80}\); \(4 \times 40 \times 9 = 1440\).)
| Scheme | Marks |
|---|---|
| Using \(F = ma\) and displacement \(x\) from \(E\): | M1 |
| \(0.5g - \dfrac{24.5(x + 0.15)}{0.75} = 0.5\ddot{x}\) | A2,1,0 |
| \(\ddot{x} = -\dfrac{196}{3}x\), so SHM | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Max speed = their \(a\) \(\times\) their \(\omega\) | M1 |
| \(= (0.647 - 0.15) \times \sqrt{\dfrac{196}{3}}\) | |
| \(\approx 4.0\) m s\(^{-1}\) (4.02) | A1 |
| (2) | |
| (14 marks) |