June 2025 Paper 3 Q12
12
A rectangular block \(B\) of mass 10 kg lies at rest in limiting equilibrium on a rough plane \(\Pi\) inclined at 30° to the horizontal. A horizontal force of magnitude \(T\) N, acting above a line of greatest slope, is applied to \(B\) (see Fig. 1).
The coefficient of friction between \(B\) and the plane is 0.8.
For the remainder of the question, you should you use this value of \(T\).
Block \(B\) is now cut at an angle of 30° to the horizontal into two smaller blocks. The upper block has a mass of 4 kg, and the lower block has a mass of 6 kg. The two blocks are held at rest with the lower block on \(\Pi\). The horizontal force of magnitude \(T\) N is now applied to the lower block (see Fig. 2).
The two blocks are released from rest and in the subsequent motion the upper block starts to move with acceleration 3.5 m s−2.
| Scheme | Marks | AO |
|---|---|---|
| Resolving perpendicular to the plane to form an equation | M1* | 3.3 |
| \(R + T\sin 30 = 10g\cos 30\) | A1 | 1.1 |
| Resolving parallel to the plane to form an equation | M1* | 3.3 |
| \(T\cos 30 + 10g\sin 30 = F\) or \(T\cos 30 + 10g\sin 30 = 0.8 \times \text{‘}R\text{’}\) | A1 | 1.1 |
| \(T\cos 30 + 10g\sin 30 = 0.8(10g\cos 30 - T\sin 30)\) | M1dep* | 3.4 |
| \(T = \dfrac{40\sqrt{3} - 50}{5\sqrt{3} + 4}g = 14.9\) (N) | A1 | 2.1 |
| [6] |
Notes
M1*: Correct number of dimensionally correct terms – must be using \(m = 10\) for M1
Allow sign errors and sin/cos mix only
A1: oe e.g. \(R = 49\sqrt{3} - 0.5T\) or \(R = 84.8(7048\ldots) - 0.5T\) etc.
\(R\) is the normal contact force
M1*: Correct number of dimensionally correct terms – must be using \(m = 10\) for M1
Allow sign errors and sin/cos mix only
A1: oe \(F = T \times \frac{\sqrt{3}}{2} + 49\) or \(F = 0.866(0254\ldots)T + 49\) etc.
May have replaced \(F\) with \(0.8R\) and be using their (possibly incorrect) \(R\)
\(F\) is the frictional contact force
For reference only: \(R = 77.4076\ldots\) and \(F = 61.92608\ldots\)
M1dep*: Apply \(F = 0.8R\) to obtain an equation in \(T\) only. A correct value of \(T\) to at least 4 significant figures rot can imply this and the next mark www
A correct equation in \(T\) implies the first 5 marks
A1: AG – awrt 14.9 www - sufficient working must be shown as answer given – for this mark we must see either the correct answer to at least 4 significant figures rot or a correct explicit expression for \(T\) e.g. \(T = \dfrac{8g\cos 30 - 10g\sin 30}{\cos 30 + 0.8\sin 30}\) before the given answer of 14.9
14.925760…
Alternative for part (a) – resolving vertically and horizontally
| Scheme | Marks | AO |
|---|---|---|
| Resolving vertically to form an equation | M1* | |
| \(F\sin 30 + R\cos 30 = 10g\) | A1 | |
| Resolving horizontally to form an equation | M1* | |
| \(T + R\sin 30 = F\cos 30\) | A1 | |
| \(R(0.8\sin 30 + \cos 30) = 10g\) \(\Rightarrow T + \dfrac{10g \times \sin 30}{0.8\sin 30 + \cos 30} = \dfrac{10g \times 0.8\cos 30}{0.8\sin 30 + \cos 30}\) | M1dep* | |
| \(T = \dfrac{40\sqrt{3} - 50}{5\sqrt{3} + 4}g = 14.9\) (N) | A1 |
M1*: Correct number of dimensionally correct terms – must be using \(m = 10\) for M1
Allow sign errors and sin/cos mix only
A1: oe e.g. \(F + R\sqrt{3} = 20g\) or \(0.8R + R\sqrt{3} = 20g\) etc.
M1*: Correct number of dimensionally correct terms
Allow sign errors and sin/cos mix only
A1: oe e.g. \(2T + R = F\sqrt{3}\) or \(2T + R = 0.8R\sqrt{3}\) etc.
M1dep*: Apply \(F = 0.8R\) to obtain an equation in \(T\) only. A correct value of \(T\) to at least 4 significant figures rot can imply this and the next mark www
For reference: \(R = 77.4076\ldots\) and \(F = 61.92608\ldots\)
A1: AG – awrt 14.9 www - sufficient working must be shown as answer given – for this mark we must see either the correct answer to at least 4 significant figures or a correct explicit expression for \(T\) e.g. \(T = \dfrac{8g\cos 30 - 10g\sin 30}{\cos 30 + 0.8\sin 30}\) before the given answer of 14.9
14.925760…
| Scheme | Marks | AO |
|---|---|---|
| \(4g\sin 30 - F_1 = 4 \times 3.5\) | M1* | 3.1b |
| \(R_1 = 4g\cos 30\) | B1 | 2.1 |
| \(4g\sin 30 - 4 \times 3.5 = \mu \times 4g\cos 30\) \((\Rightarrow 5.6 = \mu \times 4g\cos 30)\) | M1dep* | 3.4 |
| \((\mu =)\ 0.165\) | A1 | 1.1 |
| [4] |
Notes
M1*: Applying N2L for the upper block – allow sign errors and sin/cos mix only - must be using \(m = 4\) for M1 (if correct then \(F_1 = 5.6\))
where \(F_1\) is the frictional contact force between the two blocks
B1: Correctly resolve forces perpendicular to the line of contact between the two blocks
where \(R_1\) is the normal contact force between the two blocks
M1dep*: Apply \(F = \mu R\) with their \(F_1\) to obtain an un-simplified equation/expression in \(\mu\) only where \(R\) must be replaced by either \(4g\cos 30\) or \(4g\cos 60\) or \(4g\sin 30\) or \(4g\sin 60\)
A1: awrt 0.165 or allow exact \(\frac{2\sqrt{3}}{21}\)
0.16495721…
| Scheme | Marks | AO |
|---|---|---|
| \((F_{\max} =)\ 0.8(10g\cos 30 - 14.9\sin 30)\ (= 61.9\ldots)\) or \((F_{\max} =)\ 14.9\cos 30 + 10g\sin 30\ (= 61.9\ldots)\) | B1* | 3.1b |
| (Resultant force down the plane is) \(4g\sin 30 - 4 \times 3.5 + 14.9\cos 30 + 6g\sin 30\ (= 47.9\ldots)\) or \(5.6 + 14.9\cos 30 + 6g\sin 30\ (= 47.9\ldots)\) | B1* | 3.1b |
| \(F_{\max} \gt\) Resultant force down the plane and so therefore the lower block does not move or \(47.9\ldots - 61.9\ldots = 6a \Rightarrow a \lt 0 \therefore\) doesn’t move | B1dep* | 3.2a |
| [3] |
Notes
B1*: Correct expression for the maximum frictional force between the lower block and the plane – note that \(77.4204\ldots \times 0.8\) is B1
If expressions not seen then values must be correct to at least 3 sf
\(77.4076\ldots \times 0.8\) is B1
B1*: Correct expression for the resultant force acting down the plane for the lower block
Do not penalise if a more accurate correct value of \(T\) is used (14.925760…)
B1dep*: Correctly show that the lower block does not move or correctly showing that \(a\), or even \(6a\) is negative
Correct values to at least 3 sf rot must be seen for this mark