M1 June 2014 (R) Q7
7.

A particle \(P\) of mass 2.7 kg lies on a rough plane inclined at 40\(^\circ\) to the horizontal. The particle is held in equilibrium by a force of magnitude 15 N acting at an angle of 50\(^\circ\) to the plane, as shown in Figure 4. The force acts in a vertical plane containing a line of greatest slope of the plane. The particle is in equilibrium and is on the point of sliding down the plane.
Find
The force of magnitude 15 N is removed.

| Scheme | Marks |
|---|---|
| Perpendicular to the slope: \(R = 2.7g\cos 40 + 15\cos 40\) | M1A2 |
| \(=\) 31.8 (N) or 32 (N) | A1 |
| (4) |
Notes
N.B. Only penalise over- or under-accuracy after using \(g = 9.8\), (or use of \(g = 9.81\)), once in whole question.
First M1 for resolving perpendicular to the slope, with correct no. of terms, and both the \(2.7g\) and 15 terms resolved.
First A2 for a correct equation; -1 each error.
Third A1 for 32 (N) or 31.8 (N)
| Scheme | Marks |
|---|---|
| Parallel to the slope: \(F = 2.7g\sin 40 - 15\cos 50\) \((F = 7.366..)\) | M1A2 |
| Use of \(F = \mu R\) | M1 |
| \(\mu = \dfrac{2.7g\sin 40 - 15\cos 50}{R} = 0.23\) or 0.232 | A1 |
| (5) |
Notes
N.B. Only penalise over- or under-accuracy after using \(g = 9.8\), (or use of \(g = 9.81\)), once in whole question.
First M1 for resolving parallel to the slope, with correct no. of terms, and both the \(2.7g\) and 15 terms resolved.
First A2 for a correct equation; -1 each error.
Second M1 for use of \(F = \mu R\)
Third A1 for 0.23 or 0.232
| Scheme | Marks |
|---|---|
| Component of wt parallel to slope \(= 2.7g\sin 40^\circ\ (= 17.0)\) | B1 |
| \(F_{\max} = 0.232 \times 2.7 \times g \times \cos 40^\circ = 4.7\ldots\) (N) | M1A1 |
| \(17.0 > 4.70\) so the particle moves | A1 |
| (4) | |
| (13 marks) |
Notes
N.B. Only penalise over- or under-accuracy after using \(g = 9.8\), (or use of \(g = 9.81\)), once in whole question.
B1 for component of weight down the plane \(2.7g\sin 40^\circ\) (17 or better)
M1 for using their NEW \(R\) and \(\mu\) to find max friction (M0 if they use \(R\) from (a))
First A1 for 4.7 (or better) (should be 4.701242531)
Second A1 for comparison and correct conclusion.
N.B. If first A mark is 0, the second A mark must also be 0.