M2 June 2014 (R) Q5
5.

A particle \(P\) of mass 2 kg is released from rest at a point \(A\) on a rough inclined plane and slides down a line of greatest slope. The plane is inclined at 30\(^\circ\) to the horizontal. The point \(B\) is 5 m from \(A\) on the line of greatest slope through \(A\), as shown in Figure 3.
The speed of \(P\) as it reaches \(B\) is 4 m s\(^{-1}\).
The particle \(P\) is now placed at \(A\) and projected down the plane towards \(B\) with speed 3 m s\(^{-1}\). Given that the frictional force remains constant,
| Scheme | Marks |
|---|---|
| PE lost \(= mgh = 2 \times 9.8 \times 5\sin 30 = 49\) J | M1 A1 |
| (2) |
Notes
M1 Condone sin/cos confusion
A1 Accept \(5g\)
| Scheme | Marks |
|---|---|
| \(49 = \dfrac{1}{2} \times 2 \times 4^2 + F_r \times 5\) | M1 A2 |
| \(F_r = 6.6\) N | A1 |
| \(N = 2g\cos 30\) | B1 |
| \(\mu = \dfrac{F_r}{N} = \dfrac{6.6}{2g\cos 30} = 0.389\) or 0.39 | M1 A1 |
| (7) |
Notes
M1 Must be using work-energy. Must be using 5 for distance. Allow with their 49. Condone sign error(s)
A2 -1 each error. Allow with their 49
A1 Accept \(\dfrac{5g - 16}{5}\)
B1 16.97
M1 Use of \(F = \mu N\) with their \(F\) & \(N\)
A1 2 s.f. or 3 s.f. only
| Scheme | Marks |
|---|---|
| \(49 = \dfrac{1}{2} \times 2 \times v^2 - \dfrac{1}{2} \times 2 \times 3^2 + F_r \times 5\) | M1 A1 |
| \(49 + 9 - 33 = v^2\) | DM1 |
| \(v = 5\) m s\(^{-1}\) | A1 |
| (4) | |
| (13 marks) |
Notes
M1 Work Energy equation. All terms required, must be dimensionally correct but condone sign error(s)
A1 Correct equation for their \(F\) – any equivalent form.
DM1 Substitute and solve for \(v\). Dependent on preceding M1
Alt(c)
| \(2g\sin 30 - F = 2a\) | M1 |
| \((a = 1.6)\) | A1 |
| \(v^2 = 9 + 2 \times a \times 5\ (= 25)\) | DM1 |
| \(v = 5\) m s\(^{-1}\) | A1 |
M1 Using N2L with their \(F\). Condone sign error
A1 Correct equation (with their \(F\))
DM1 Use of \(v^2 = u^2 + 2as\) with their \(a\). Dependent on preceding M1
Watch out – there are a lot of incorrect ways of reaching a final answer of 5