June 2024 Paper 3 Q11

Notes

M1: Attempt at moments about \(A\) – at least two terms (one term for the weight and one term with a component of \(R_D\)) – must be written as an equation to score any marks in this part (see Answer column for the only acceptable forms)
with \(k_1, k_2 \gt 0\) but not \(= \pm 1\)

A1: Correct moment for the contact force at \(D\)

A1: Correct moment for the weight of the lamina (See Appendix) (e.g. \(0.5g \times (3 - \tan 50)\cos 50\) or \(0.5g \times \cos 50 \times 1.808\ldots\) etc. e.g. \(0.5g \times \sqrt{10}\sin\left(40 - \arctan\left(\tfrac{1}{3}\right)\right)\) or \(0.5g \times \sqrt{10}\sin(21.565\ldots)\) etc.) – if value stated with no or unclear working, then must see at least 5.6953… to award this mark - if moment stated as 5.6993… then A0 (and A0 for the next mark)
For reference 5.695360…

A1: AG – as this answer is given then working must be checked carefully – all previous marks must have been awarded
1.23912743…
A fully correct equation followed by stating 1.24 can score full marks

Additional guidance for 11(a) (Appendix)

There are many ways that candidates are correctly approaching this part:
For example,
\(6(R_D\sin 50) = 3(0.5g\cos 50) - (0.5g\sin 50)\)
\(6(R_D\sin 50) = 0.5g \times (3 - \tan 50)\cos 50\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\sin\left(40 - \arctan\left(\tfrac{1}{3}\right)\right)\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos(140 - \arctan(3))\)
\(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos\left(50 + \arctan\left(\tfrac{1}{3}\right)\right)\)

Any of these followed by the correct answer of 1.24 would score full marks (we do not need to see any intermediate working). If these exact expressions are not seen then sufficient working with values that are correct to at least 3 significant figures should be awarded full marks (and those that have 2 significant figures with sufficient working should get partial credit), e.g.

• \(6(R_D\sin 50) = 9.448977\ldots - 3.7536177\ldots\) so seeing \(6(R_D\sin 50) = 9.45 - 3.75\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 9.4 - 3.8\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.22 not 1.24)
• \(6(R_D\sin 50) = 5.7(0)\) is no marks (assume working backwards from the AG and no indication that the RHS contains a weight component)
• \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.808246\ldots\) so seeing \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.81\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times \cos 50 \times 1.80\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.23 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 0.643 \times 1.81\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times 0.64 \times 1.8\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.23 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 1.16\) followed by 1.24 is M1 A1 A0 A0 (assume working backwards from the given answer on the RHS)
• \(6(R_D\sin 50) = 0.5g \times \sqrt{10}\sin(40 - 18.434948\ldots)\) so seeing \(6(R_D\sin 50) = 0.5g \times 3.16 \times \sin(21.6)\) followed by 1.24 is M1 A1 A1 A1
• \(6(R_D\sin 50) = 0.5g \times 3.16 \times \sin(22)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.26 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times 3.2 \times \sin(22)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.28 not 1.24)
• \(6(R_D\sin 50) = 0.5g \times \sqrt{10}\cos(68.4349\ldots)\) so seeing \(6(R_D\sin 50) = 0.5g \times \sqrt{10} \times \cos(68)\) followed by 1.24 is M1 A1 A1 A0 (leads to 1.26 not 1.24)

Notes

B1*: Resolving horizontally and vertically (possibly implied by later working) – must be using given 1.24 (or a more accurate correct value) only (so not their incorrect value from part (a))
If taking moments about another point e.g. \(D\), then the corresponding equation(s) must be correct

M1dep*: Use of \(F = \mu R\) or \(F \leqslant \mu R\) with \(0.5g\) and 1.24 (or better) for \(F\)

A1: awrt 0.253
If final answer is \(\mu \geqslant 0.253\) then A0
Using the exact value or 1.24 for \(R_D\) leads to the same answer to 3 significant figures