A2 October 2020 Q2
2. A truck of mass 1200 kg is moving along a straight horizontal road.
At the instant when the speed of the truck is \(v\ \text{m s}^{-1}\), the resistance to the motion of the truck is modelled as a force of magnitude \((900 + 9v)\) N.
The engine of the truck is working at a constant rate of 25 kW.
Later on, the truck is moving up a straight road that is inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{20}\)
At the instant when the speed of the truck is \(v\ \text{m s}^{-1}\), the resistance to the motion of the truck from non-gravitational forces is modelled as a force of magnitude \((900 + 9v)\) N.
When the engine of the truck is working at a constant rate of 25 kW the truck is moving up the road at a constant speed of \(V\ \text{m s}^{-1}\).
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion: \(F - (900 + 9 \times 25) = 1200a\) | M1 | 3.3 |
| Use of \(25000 = F \times 25\) | M1 | 3.4 |
| \(\dfrac{25000}{25} - (900 + 225) = 1200a\) | A1 | 1.1b |
| \(a = -\dfrac{5}{48}\) deceleration \(= \dfrac{5}{48}\ \ (= 0.10416..)\ \ (\text{m s}^{-2})\) | A1 | 1.1b |
| (4) |
Notes
M1: Dimensionally correct. Condone sign errors
M1: Correct use of \(P = Fv\). Allow in (b) if not seen in (a).
A1: Correct unsimplified equation
A1: 0.10 or better. Final answer must be positive.
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion: | M1 | 3.3 |
| \(\dfrac{25000}{V} - 1200g\sin\theta - (900 + 9V) = 0\) | A1 A1 | 1.1b 1.1b |
| Form quadratic and solve for \(V\): | M1 | 1.1b |
| \((9V^2 + 1488V - 25000 = 0)\) \(V = 15.4\ (15)\) | A1 | 1.1b |
| (5) | ||
| (9 marks) |
Notes
M1: Need all terms. Dimensionally correct. Condone sign errors
A1 A1: Unsimplified equation with at most one error
Correct unsimplified equation
M1: Complete method to solve for \(V\)
A1: Correct to 2 sf or 3 sf