M1 June 2013 (R) Q5
5.
A particle \(P\) of mass 0.6 kg slides with constant acceleration down a line of greatest slope of a rough plane, which is inclined at 25\(^\circ\) to the horizontal. The particle passes through two points \(A\) and \(B\), where \(AB = 10\) m, as shown in Figure 3. The speed of \(P\) at \(A\) is 2 m s\(^{-1}\). The particle \(P\) takes 3.5 s to move from \(A\) to \(B\). Find
| Scheme | Marks |
|---|---|
| \(s = \dfrac{u + v}{2}t\) \(10 = \dfrac{2 + v}{2} \times 3.5\) | M1A1 |
| \(v = \dfrac{20}{3.5} - 2 = \dfrac{26}{7} = 3.71\) (m s\(^{-1}\)) | A1 |
| (3) |
Notes
First M1 for producing an equation in \(v\) only.
First A1 for a correct equation
Second A1 for 26/7 oe, 3.7 or better (ms\(^{-1}\))
| Scheme | Marks |
|---|---|
| \(a = \dfrac{v - u}{t} = \dfrac{\dfrac{26}{7} - 2}{3.5} = \dfrac{24}{49} = 0.490\) (m s\(^{-2}\)) | M1A1 |
| (2) |
Notes
M1 for producing an equation in \(a\) only.
A1 for 24/49, 0.49 or better (ms\(^{-2}\))
| Scheme | Marks |
|---|---|
| Normal reaction : \(R = 0.6g\cos 25^\circ\) | B1 |
| Resolve parallel to the slope : \(0.6g\sin 25^\circ - \mu \times R = 0.6 \times a\) | M1A2 |
| \(\mu = 0.41\) or 0.411 | A1 |
| (5) | |
| (10 marks) |
Notes
B1 for \(R = 0.6g\cos 25^\circ\)
M1 for resolving along the plane, correct no. of terms etc.
A2 (-1 each error) \(R\) and \(a\) do not need to be substituted
Third A1 for 0.41 or 0.411