M3 January 2008 Q7
7. A particle \(P\) of mass 2 kg is attached to one end of a light elastic string, of natural length 1 m and modulus of elasticity 98 N. The other end of the string is attached to a fixed point \(A\). When \(P\) hangs freely below \(A\) in equilibrium, \(P\) is at the point \(E\), 1.2 m below \(A\). The particle is now pulled down to a point \(B\) which is 0.4 m vertically below \(E\) and released from rest.
| Scheme | Marks |
|---|---|
| (Measuring \(x\) from \(E\)) \(2\ddot{x} = 2g - 98(x + 0.2)\), and so \(\ddot{x} = -49x\) | M1 A1, A1 |
| SHM period with \(\omega^2 = 49\) so \(T = \tfrac{2\pi}{7}\) | d M1 A1cso |
| (5) |
Notes
DM1 requires the minus sign.
Special case
\(2\ddot{x} = 2g - 98x\) is M1A1A0M0A0 \(2\ddot{x} = -98x\) is M0A0A0M0A0
No use of \(\ddot{x}\), just \(a\) is M1 A0,A0 then M1 A0 if otherwise correct.
Quoted results are not acceptable.
| Scheme | Marks |
|---|---|
| Max. acceleration \(= 49 \times\) max. \(x = 49 \times 0.4 = 19.6\) m s\(^{-2}\) | B1 |
| (1) |
Notes
Answer must be positive and evaluated for B1
| Scheme | Marks |
|---|---|
| String slack when \(x = -0.2\): \(v^2 = 49\left(0.4^2 - 0.2^2\right)\) | M1 A1 |
| \(\Rightarrow v \approx 2.42\) m s\(^{-1}\) \(= \dfrac{7\sqrt{3}}{5}\) | A1 |
| (3) |
Notes
M1 – Use correct formula with their \(\omega\), \(a\) and \(x\) but not \(x = 0\).
A1 Correct values but allow \(x = +0.2\)
Alternative
It is possible to use energy instead to do this part
| \(\tfrac{1}{2}mv^2 + mg \times 0.6 = \dfrac{\lambda \times 0.6^2}{2l}\) | M1 A1 |
| Scheme | Marks |
|---|---|
| Uses \(x = a\cos\omega t\) or use \(x = a\sin\omega t\) but not with \(x = 0\) or \(\pm a\) | M1 |
| Attempt complete method for finding time when string goes slack \(-0.2 = 0.4\cos 7t \Rightarrow \cos 7t = -\tfrac{1}{2}\) | dM1 A1 |
| \(t = \dfrac{2\pi}{21} \approx 0.299\) s | A1 |
| Time when string is slack \(= \dfrac{(2) \times 2.42}{g} = \dfrac{2\sqrt{3}}{7} \approx 0.495\) s (2 needed for A) | M1 A1ft |
| Total time \(= 2 \times 0.299 + 0.495 \approx 1.09\) s | A1 |
| (7) | |
| (16 marks) |
Notes
If they use \(x = a\sin\omega t\) with \(x = \pm 0.2\) and add \(\tfrac{\pi}{7}\) or \(\tfrac{\pi}{14}\) this is dM1, A1 if done correctly
If they use \(x = a\cos\omega t\) with x = -0.2 this is dM1, then A1 (as in scheme)
If they use \(x = a\cos\omega t\) with x = +0.2 this needs their \(\dfrac{\pi}{7}\) minus answer to reach dM1, then A1