A2 June 2023 Q4
4. A light elastic string has natural length \(2a\) and modulus of elasticity \(4mg\).
One end of the elastic string is attached to a fixed point \(O\). A particle \(P\) of mass \(m\) is attached to the other end of the elastic string.
The particle \(P\) hangs freely in equilibrium at the point \(E\), which is vertically below \(O\)
Particle \(P\) is now pulled vertically downwards to the point \(A\), where \(OA = 4a\), and released from rest. The resistance to the motion of \(P\) is a constant force of magnitude \(\dfrac{1}{4}mg\).
Particle \(P\) is now held at \(O\)
Particle \(P\) is released from rest and reaches its maximum speed at the point \(B\).
The resistance to the motion of \(P\) is again a constant force of magnitude \(\dfrac{1}{4}mg\).
| Scheme | Marks | AO |
|---|---|---|
| \(T = \dfrac{4mge}{2a}\) | B1 | 3.3 |
| \(T = mg\) | M1 | 3.1a |
| \(e = \dfrac{1}{2}a\) | A1 | 1.1b |
| \(OE = \dfrac{5a}{2}\) | A1 | 1.1b |
| (4) |
Notes
B1: Hooke’s Law seen with \(4mg\) and \(2a\) substituted.
M1: Resolving vertically. Correct number of terms.
A1: cao for extension.
A1: cao for \(OE\). Note that if the extension is \((OE - 2a)\) in their equation, \(OE\) can be found directly and both A’s can be earned together.
| Scheme | Marks | AO |
|---|---|---|
| GPE term, \(\pm\, mga\) | B1 | 3.4 |
| Work done against resistance, \(\pm\, \dfrac{1}{4}mga\) | B1 | 3.4 |
| Use of EPE formula once. | M1 | 3.4 |
| \(\pm\, \dfrac{4mg}{2 \times 2a}\left\{(2a)^2 - a^2\right\}\) | A1 | 1.1b |
| Work energy equation: | M1 | 3.1a |
| \(\dfrac{1}{4}mga = \dfrac{4mg}{2 \times 2a}\left\{(2a)^2 - a^2\right\} - mga - \dfrac{1}{2}mv^2\) | A1 | 1.1b |
| \(v = \sqrt{\dfrac{7ag}{2}}\) oe | A1 | 1.1b |
| (7) |
Notes
B1: GPE term seen, ignore sign.
B1: Work term seen \(\dfrac{mga}{4}\), ignore sign.
Allow B1 for the case where WD = \(\dfrac{5mga}{4}\). This is a special case where the work done against resistance is included within the term. \(\dfrac{5mga}{4}\) = WD against resistance + WD against weight.
M1: Use of EPE formula. Accept EPE in the form \(\dfrac{\lambda x^2}{ka}\)
A1: Difference between two correct EPE terms seen, unsimplified.
M1: Work-energy equation is formed with all relevant terms and no extras.: KE, GPE, 2EPE, WD. Condone sign errors.
M0: For work-energy equation with WD = \(\dfrac{5mga}{4}\) and a GPE term. This is because weight is considered twice and so the equation contains an extra term.
A1: Correct unsimplified equation
A1: Correct answer in terms of \(a\) and \(g\), do not allow 9.8 for \(g\) \(\quad v = \sqrt{\dfrac{7ag}{2}}\), \(\ v = \dfrac{1}{2}\sqrt{14ag}\)
| Scheme | Marks | AO |
|---|---|---|
| \(mg - T - \dfrac{1}{4}mg = 0\) | M1 | 3.1a |
| \(mg - \dfrac{4mgx}{2a} - \dfrac{1}{4}mg = 0\) | A1 | 1.1b |
| \(x = \dfrac{3a}{8}\) | A1 | 1.1b |
| \(OB = \dfrac{19a}{8}\) oe | A1 | 1.1b |
| (4) | ||
| (15 marks) |
Notes
M1: Vertical equilibrium equation or equation of motion with \(a = 0\). Condone sign errors. Correct no. of terms - all 3 forces must be included although \(\left(mg \pm \dfrac{mg}{4}\right)\) may already be simplified.
Hooke’s Law does not need to be substituted but M0 if the equilibrium position from (a) is used.
A1: Correct equation in one unknown.
A1: cao
A1: cao
Note that if the extension is \((OB - 2a)\) in their equation, \(OB\) can be found directly and both A’s can be earned together.
4(c) Alt 1: Using differentiation with a Work - energy equation from the point of release
M1: Forming work-energy equation with the usual rules: all relevant terms to be included and of the correct form and no extra terms.
\(\dfrac{1}{2}mv^2 = mgh - \dfrac{4mg(h - 2a)^2}{2(2a)} - \dfrac{mgh}{4}\)
A1: Correct equation for \(v^2\) and \(h\) (may use a different letter)
A1: Correct equation after differentiating \(v^2\) or \(v\) with respect to \(h\) and setting it equal to zero.
\(\dfrac{\mathrm{d}}{\mathrm{dh}}\left(v^2\right) = 0 \quad \rightarrow \quad \dfrac{3g}{2} = \dfrac{4g(h - 2a)}{a}\) oe
A1: Correct answer \(OB = \dfrac{19a}{8}\)