M2 June 2013 Q2
2. A particle \(P\) of mass 3 kg moves from point \(A\) to point \(B\) up a line of greatest slope of a fixed rough plane. The plane is inclined at 20\(^\circ\) to the horizontal. The coefficient of friction between \(P\) and the plane is 0.4
Given that \(AB = 15\) m and that the speed of \(P\) at \(A\) is 20 m s\(^{-1}\), find
| Scheme | Marks |
|---|---|
| Work done \(= 15\mu R = 15 \times 0.4 \times 3g\cos 20^\circ\) | M1 M1 |
| \(= 18g\cos 20 = 166\) (J) | A1 |
| (3) |
Notes
M1 \(F_{\max} = \mu \times 3g\cos 20\ (11.05)\). \(R\) must be resolved but condone trig confusion.
M1 \(15 \times\) their \(F_{\max}\). Independent M. \(15 \times F_{\max} + \ldots\) is M0
A1 or 170 (J)
| Scheme | Marks |
|---|---|
| Energy: WD against \(F\) + GPE + final KE = initial KE | |
| their WD \(+ 3g\sin 20^\circ \times 15 + \dfrac{1}{2}3v^2 = \dfrac{1}{2}3 \times 20^2\) | M1A2ft |
| \(v = 13.7\) (m s\(^{-1}\)) | A1 |
| (4) | |
| (7 marks) |
Notes
M1A2ft Must include all four correct terms (including resolving). Condone sign errors and trig confusion. Any sign errors in the KE terms count as a single error. Follow their WD. A2ft: -1ee Follow their WD
A1 or 14
Or 2b
| \(3a = -\ 0.4 \times 3g\cos 20 + 3g\sin 20\) and use of \(v^2 = u^2 + 2as\) | M1 A1ft |
| \(v^2 = 20^2 + 2 \times a \times 15\ (= 188.93\ldots)\) | A1ft |
| \(v = 13.7\) (m s\(^{-1}\)) | A1 |
M1 Complete method. Their \(F_{\max}\) +component of weight
A1ft A correct equation with their \(F_{\max}\). Allow for \(a = +7.03\ldots\) acting down the slope \(a = -7.035\ldots\)
A1ft Correct equation for their \(a\)
A1 or 14 (m s\(^{-1}\))