M2 June 2014 Q4
4. A truck of mass 1800 kg is towing a trailer of mass 800 kg up a straight road which is inclined to the horizontal at an angle \(\alpha\), where \(\sin\alpha = \dfrac{1}{20}\). The truck is connected to the trailer by a light inextensible rope which is parallel to the direction of motion of the truck. The resistances to motion of the truck and the trailer from non-gravitational forces are modelled as constant forces of magnitudes 300 N and 200 N respectively. The truck is moving at constant speed \(v\) m s\(^{-1}\) and the engine of the truck is working at a rate of 40 kW.
As the truck is moving up the road the rope breaks.
| Scheme | Marks |
|---|---|
 | |
| Constant speed \(F = 200 + 800g\sin\alpha + 300 + 1800g\sin\alpha\ (= 1774)\) | M1 A2 |
| \(40000 = Fv\ (= 1774v)\) | M1 |
| \(v = 22.5\) | A1 |
| (5) |
Notes
M1 Complete method to an equation in “\(F\)”. Requires all the terms, including resolution of the weights. Condone sign errors and sin/cos confusion. \(g\) missing from both weights is a single error. Penalise trig once only.
A2 -1 each error. i.e. A1A1 if no errors A1A0 one error seen, A0A0 two or more errors
M1 Use of \(P = Fv\). Allow with \(F\) or their \(F\). Independent of the first M1
A1 Accept 23. (maximum 3sf following use of 9.8)
| Scheme | Marks |
|---|---|
| "1774"\( - 300 - 1800g\sin\alpha\ (= 592) = 1800a\) | M1 A2 |
| \(F = ma:\ \ a = \dfrac{F}{1800} = 0.32888\ldots = 0.33\) (m s\(^{-2}\)) | A1 |
| (4) | |
| (9 marks) |
Notes
M1 New equation of motion for the truck. Follow their 1774. Requires all the terms, including resolution of the weights. Condone sign errors and sin/cos confusion.
A2 Allow with their 1774. -1 each error. i.e. A1A1 no errors, A1A0 one error, A0A0 two or more errors
A1 Accept 0.329