M2 June 2014 Q8
8. The points \(A\) and \(B\) are 10 m apart on a line of greatest slope of a fixed rough inclined plane, with \(A\) above \(B\). The plane is inclined at 25\(^\circ\) to the horizontal. A particle \(P\) of mass 5 kg is released from rest at \(A\) and slides down the slope. As \(P\) passes \(B\), it is moving with speed 7 m s\(^{-1}\).
| Scheme | Marks |
|---|---|
| Work done against friction = Loss in GPE– Gain in KE \(= 5 \times 9.8 \times 10\sin 25 - \dfrac{1}{2} \times 5 \times 7^2 = 84.58\ldots\) | M1 A2 |
| \(= 85\) (J) (84.6) | A1 |
| (4) |
Notes
M1 Must be using Work-energy principle Needs to consider (work done) KE & GPE and no other terms. Condone sign errors. Watch out for incorrect solutions including both change in GPE and the work done against the weight – this is a method error.
A2 -1 each error
A1 Max 3 s.f. Must be +ve. Accept as \(10F = 84.6\) or equiv.
| Scheme | Marks |
|---|---|
| \(F = \mu R = \mu \times 5g\cos 25\) | M1 A1 |
| Work done \(= 10F = 10\mu \times 5g\cos 25 =\) their85 | M1 A1ft |
| \(\mu = 0.19\) | A1 |
| (5) | |
| (9 marks) |
Notes
M1 Resolve to find \(F_{\max.}\) \(g\) missing is an accuracy error
A1 Correct unsimplified
M1 Use of work done = force x distance to form an equation for \(\mu\)
A1ft Correct unsimplified equation for their \(10F\)
A1 Accept 0.190
altb
| \(F = \mu R = \mu \times 5g\cos 25\) | M1 A1 |
| \(v^2 = u^2 + 2as \rightarrow 49 = 20a \rightarrow a = \dfrac{49}{20}\) | M1 |
| N2L \(\rightarrow 5 \times \dfrac{49}{20} = 5g\sin 25 - \mu \times 5g\cos 25\) | A1 |
| \(\mu = 0.19\) | A1 |
M1 Resolve to find \(F_{\max.}\) \(g\) missing is an accuracy error
A1 Correct unsimplified
M1 Complete method to an equation in \(\mu\)
A1 Correct unsimplified equation
A1 Accept 0.190