M2 June 2014 (R) Q1
1. A van of mass 600 kg is moving up a straight road inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{16}\). The resistance to motion of the van from non-gravitational forces has constant magnitude \(R\) newtons. When the van is moving at a constant speed of 20 m s\(^{-1}\), the van’s engine is working at a constant rate of 25 kW.
The power developed by the van’s engine is now increased to 30 kW. The resistance to motion from non-gravitational forces is unchanged. At the instant when the van is moving up the road at 20 m s\(^{-1}\), the acceleration of the van is \(a\) m s\(^{-2}\).
| Scheme | Marks |
|---|---|
| Tractive force \(= \dfrac{25000}{20} = 1250\) N | B1 |
| \(1250 = R + 600g\sin\theta\) | M1 |
| \(R = 1250 - 600g \times \dfrac{1}{16}\ (= 882.5)\) | A1ft |
| \(= 883\) or 880 N | A1 |
| (4) |
Notes
B1 Seen or implied
M1 Equation in \(R\). Condone sign errors and sin/cos confusion
A1ft Correct unsimplified expression for \(R\). Allow with their 1250
A1 2 or 3 s.f. 882.5 is A0
| Scheme | Marks |
|---|---|
| T F \(= \dfrac{30000}{20} = 1500\) N | |
| \(1500 - 600g \times \dfrac{1}{16} - R = 600a\) | M1 A2 |
| \(a = \dfrac{1500 - 600 \times 9.8 \div 16 - 882.5}{600}\ (= 0.4166\ldots)\) | |
| \(= 0.42\) or 0.417 m s\(^{-2}\) | A1 |
| (4) | |
| (8 marks) |
Notes
M1 Equation of motion. Must have all the terms. Condone sign errors and sin/cos confusion
A2 -1 each error
A1 2 or 3 s.f. \(\dfrac{5}{12}\) is A0
882.5 and \(\dfrac{5}{12}\) is A0 at the end of (a) and A1 at the end of (b) – penalise once only.
Use of 9.81 is an accuracy error – penalise at the end of the first part affected.
(Corrected from the printed mark scheme: the units of the final answer are printed as m s\(^{-1}\).)