June 2025 Paper 2 Q16
16 In this question use \(g = 9.8\ \text{m s}^{-2}\)
A sledge is pulled in a straight line up a rough path, as shown in the diagram.
The path is inclined at an angle of 20° to the horizontal.
The sledge is pulled by a light, inextensible rope inclined at an angle of 30° to the path.
The mass of the sledge is 10 kilograms.
(a) In one model, the sledge moves at a constant speed and experiences a combined resistance force of 15 newtons.
Find the tension in the rope for this model.
[4 marks](b) In a different model, the sledge experiences no air resistance.
The tension in the rope for this model is 54 N
The coefficient of friction between the sledge and the path is 0.2
Find the acceleration of the sledge for this model.
[7 marks](c) State an assumption you have used to answer both parts (a) and (b). [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(10g\sin 20\) OE | B1 | 3.1b |
| Resolves parallel to the path to form a three-term equation with at least two correct terms | M1 | 3.3 |
| Obtains \(10g\sin 20 + 15 = \mathbf{T}\cos 30\) OE | A1 | 1.1b |
| Obtains AWRT 56 Condone missing units. | A1 | 1.1b |
| (4) |
Typical solution
\[10g\sin 20 + 15 = \mathbf{T}\cos 30\]\[\mathbf{T} = 56.02\]\[\mathbf{T} = 56\ \text{N}\]| Scheme | Marks | AO |
|---|---|---|
| Uses \(\mathbf{F} = \mu\mathbf{R}\) | B1 | 1.1b |
| Obtains \(10g\cos 20\) OE | B1 | 1.1b |
| Resolves perpendicular to the plane to form a three-term dimensionally correct equation with at least two terms correct. | M1 | 3.3 |
| Obtains a correct expression or value for R AWRT 65 PI by friction = AWRT \(\pm 13.02\) | A1 | 1.1b |
| Uses F = \(m\)a to form a four-term dimensionally correct equation, excluding a 15N force, with at least three terms correct. FT their R value Condone inclusion of a 15N force from 16(a) as a 5th term | M1 | 3.4 |
| Forms fully correct equation in a for the particle with all values substituted. FT their R value Note this alone obtains B1B1M1A1M1A1 | A1F | 1.1b |
| Obtains AWRT 0.023 \(\text{m s}^{-2}\) | A1 | 1.1b |
| (7) |
Typical solution
\[\mathbf{R} + 54\sin 30 = 10g\cos 20\]\[\mathbf{R} = 65.0899\ \text{N}\]\[54\cos 30 - \mu\mathbf{R} - 10g\sin 20 = 10a\]\[a = \frac{54\cos 30 - 0.2 \times 65.0899 - 98\sin 20}{10}\]\[a = 0.023\ \text{m s}^{-2}\]| Scheme | Marks | AO |
|---|---|---|
| States the sledge is modelled as a particle If more than one assumption is stated E0 | E1 | 3.5a |
| (1) | ||
| (12 marks) |
Typical solution
The sledge is modelled as a particle.