M3 June 2017 Q6
6. The ends of a light elastic string, of natural length 0.4 m and modulus of elasticity \(\lambda\) newtons, are attached to two fixed points \(A\) and \(B\) which are 0.6 m apart on a smooth horizontal table. The tension in the string is 8 N.
A particle \(P\) is attached to the midpoint of the string. The particle \(P\) is now pulled horizontally in a direction perpendicular to \(AB\) to a point 0.4 m from the midpoint of \(AB\). The particle is held at rest by a horizontal force of magnitude \(F\) newtons acting in a direction perpendicular to \(AB\), as shown in Figure 5 below.
The particle is released from rest. Given that the mass of \(P\) is 0.3 kg,
| Scheme | Marks |
|---|---|
| \(8 = \dfrac{\lambda \times 0.20}{0.40}\) | M1A1 |
| \(\lambda = 16\) * | A1cso |
| (3) |
Notes
M1 Attempt Hooke's Law using the whole string or a half string.
A1 Correct equation.
A1cso Correct given value of \(\lambda\) obtained with no errors seen.
| Scheme | Marks |
|---|---|
| Length of string = 1 m or 100cm | |
| \(T = \dfrac{\lambda \times 0.6}{0.4},\ \ = 24\) (or use half string) | M1,A1 |
| \(2T\cos\theta = F\) | M1 |
| \(F = 2 \times 24 \times \dfrac{4}{5} = 38.4,\ \ \dfrac{192}{5}\) or \(38\dfrac{2}{5}\) | A1 |
| (4) |
Notes
M1 Use Hooke's Law with the new longer length for the string or half string. \(\lambda\) must be 16, but length need not be correct but use of 0.2 for extension of full string or 0.1 for extension of half string scores M0.
A1 Obtain \(T = 24\)
M1 Resolve parallel to \(F\) or in another direction which gives an equation connecting \(T\) and \(F\).
A1 Obtain the correct value of \(F\)
| Scheme | Marks |
|---|---|
| Initial EPE \(= \dfrac{16 \times 0.6^2}{2 \times 0.4}\left(= \dfrac{36}{5}\right) \qquad\) Final EPE \(= \dfrac{16 \times 0.2^2}{2 \times 0.4}\left(= \dfrac{4}{5}\right)\) | B1 (either) |
| \(\dfrac{16 \times 0.6^2}{2 \times 0.4} - \dfrac{16 \times 0.2^2}{2 \times 0.4} = \dfrac{1}{2}0.3v^2\) | M1A1A1 |
| \(0.3v^2 = 40\left(0.6^2 - 0.2^2\right)\) | |
| \(v = 6.531\ldots\) Accept 6.5 \(\left(\text{m s}^{-1}\right)\) or better or exact value \(8\sqrt{\dfrac{2}{3}}\ \left(\text{m s}^{-1}\right)\) | dM1A1cso |
| (6) | |
| (13 marks) |
Notes
B1 Correct initial or final EPE with one string (\(l = 0.4\)) or two half strings (\(l = 0.2\))
M1 Attempt an energy equation with the difference of 2 EPE terms and a KE term. The EPE terms must be of the form \(k\dfrac{\lambda x^2}{l}\).
A1A1 Deduct one mark per error. (A1A1, A1A0 or A0A0)
dM1 Solve for \(v\). Depends on the previous M mark.
A1cso Correct value of \(v\), min 2 sf or exact value.
Energy terms wrong way round in the equation will lose this mark even if modulus sign inc here.
NB If the energy terms are subtracted the wrong way round, max score is B1M1A1A0M1A0