M2 June 2015 Q5
5.
A particle \(P\) of mass 10 kg is projected from a point \(A\) up a line of greatest slope \(AB\) of a fixed rough plane. The plane is inclined at angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{5}{12}\) and \(AB = 6.5\) m, as shown in Figure 2. The coefficient of friction between \(P\) and the plane is \(\mu\). The work done against friction as \(P\) moves from \(A\) to \(B\) is 245 J.
The particle is projected from \(A\) with speed 11.5 m s\(^{-1}\). By using the work-energy principle,
| Scheme | Marks |
|---|---|
| Max friction \(= \mu \times 10g\cos\alpha\) | B1 |
| Work done against friction \(= 6.5 \times 10g\mu\cos\alpha\ (= 245)\) | M1 |
| Equation in \(\mu\): \(\ 6.5 \times 10g\mu \times \dfrac{12}{13} = 245,\) | DM1 A1 |
| \(\mu = 0.417\) or 0.42 | A1 |
| (5) |
Notes
B1 Use of \(\mu R\) seen or implied \((\mu \times 90.46)\)
M1 For 6.5 x their \(F\).
DM1 Dependent on preceding M1. A1 Correct substituted equation
A1 Do not accept \(\dfrac{5}{12}\) (cannot have an exact value following the use of 9.8). Use of 9.81 is a rubric infringement, so A0 if seen.
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2} \times 10 \times 11.5^2 - 245 - 10g \times 6.5\sin\alpha = \dfrac{1}{2} \times 10 \times v^2\) or equivalent | M1 A2 |
| \(v = 5.85\ \ (5.9)\) (m s\(^{-1}\)) | A1 |
| (4) | |
| (9 marks) |
Notes
M1 Must be using work-energy equation. All terms required. Condone sign errors and sin/cos confusion.
A2 -1 each error \((661.25 - 245 - 245 = 171.25)\)