M3 June 2016 Q3
3. One end of a light elastic string, of natural length 1.5 m and modulus of elasticity 14.7 N, is attached to a fixed point \(O\) on a ceiling. A particle \(P\) of mass 0.6 kg is attached to the free end of the string. The particle is held at \(O\) and released from rest. The particle comes to instantaneous rest for the first time at the point \(A\).
Find
| Scheme | Marks |
|---|---|
| EPE gained \(= \dfrac{14.7x^2}{2 \times 1.5}\) | B1 |
| \(\dfrac{14.7x^2}{3} = 0.6g \times (x + 1.5)\) | M1A1ft |
| \(5x^2 - 6x - 9 = 0\) | |
| \(x = \dfrac{6 \pm \sqrt{36 + 180}}{10} = 2.069\ldots\) (or \(-0.869\)) | DM1A1 |
| \(OA = 3.569\ldots = 3.6\) or 3.57 | A1 |
| (6) |
Notes
B1 Correct EPE when extension is \(x\)
M1 Equating EPE to GPE lost EPE to be of the form \(k\dfrac{\lambda x^2}{l}\), where \(k\) is a rational no.
A1ft Correct equation ft their EPE. Use of unknown must be consistent.
DM1 Solve their equation (3TQ) (quadratic formula must be correct)
A1 \(x = 2.069\ldots\) neg value not needed
A1 Add 1.5 to 2.069... and give final answer to 2 or 3 sf. (No "exact" answers allowed here due to use of \(g\).)
ALT 1: Using extension \((x - 1.5)\):
B1 Correct EPE when extension is \((x - 1.5)\) where \(x\) is total length
M1 Equating EPE to GPE lost EPE of form shown above
A1ft Correct equation ft their EPE. Use of unknown must be consistent. No simplification needed. Equation is \(\dfrac{14.7(x - 1.5)^2}{3} = 0.6gx\)
A1 Simplify to \(x^2 - 4.2x + 2.25 = 0\) or equivalent 3TQ
DM1 Solve their equation (3TQ) (quadratic formula must be correct)
A1 \(x = 3.57\) or 3.6 Must be 2 or 3 sf
ALT 2:
Use \(v^2 = u^2 + 2as\) or energy to obtain speed at natural length, then energy to \(A\).
B1 for EPE at \(A\)
No more marks until an energy equation with EPE, GPE and KE terms seen.
M1A1ft Energy equation ft their EPE and initial speed EPE of form shown above
DM1A1 A1 As main scheme
NB: Solution of quadratic by calculator: Method mark only available if solution is correct ( 2.069 or 3.569)
| Scheme | Marks |
|---|---|
| \(\dfrac{14.7 \times 2.069}{1.5} - 0.6 \times 9.8 = 0.6a \quad\) or \(\quad 0.6 \times 9.8 - \dfrac{14.7 \times 2.069}{1.5} = 0.6a\) | M1A1ft |
| \(a = 23.993\ldots = 24\) or 24.0 | A1 cao |
| (3) | |
| (9 marks) |
Notes
M1 Use NL2 at \(A\) inc use of Hooke's law. Formula for HL to be correct. Ext can be \((3.569 - 1.5)\)
A1ft Correct numbers in the equation, ft their extension
A1 24 or 24.0 only. (No negatives allowed.)