M2 June 2017 Q2
2. A truck of mass 900 kg is towing a trailer of mass 150 kg up an inclined straight road with constant speed 15 m s\(^{-1}\). The trailer is attached to the truck by a light inextensible towbar which is parallel to the road. The road is inclined at an angle \(\theta\) to the horizontal, where \(\sin\theta = \dfrac{1}{9}\). The resistance to motion of the truck from non-gravitational forces has constant magnitude 200 N and the resistance to motion of the trailer from non-gravitational forces has constant magnitude 50 N.
When the truck and trailer are moving up the road at 15 m s\(^{-1}\) the towbar breaks, and the trailer is no longer attached to the truck. The rate at which the engine of the truck is working is unchanged. The resistance to motion of the truck from non-gravitational forces and the resistance to motion of the trailer from non-gravitational forces are still forces of constant magnitudes 200 N and 50 N respectively.
| Scheme | Marks |
|---|---|
| Constant speed \(\Rightarrow\) no acceleration. Driving force \(= 200 + 50 + 900g\sin\theta + 150g\sin\theta\) | M1 |
| Or \(D - T - 200 - 900g\sin\theta = 0\) and \(T - 50 - 150g\sin\theta = 0\) | A1 A1 |
| \(= 250 + 1050g \times \dfrac{1}{9}\ (= 1393.3333...)\) | |
| \(P = \left(250 + 1050g \times \dfrac{1}{9}\right) \times 15\) | M1 |
| \(= 20900\) W (20.9 kW) | A1 |
| (5) |
Notes
M1 Equation of motion of the truck. All terms required & dimensionally correct. Condone sin/cos confusion and sign error(s)
A1 At most one error. Allow for 2 separate equations including \(T\)
A1 Correct unsimplified expression for the driving force (no \(T\))
\(\left(\dfrac{4180}{3}\right)\)
M1 Use of \(P = Fv\) with their \(F\). Independent M1. Could appear in first equation as \(F = \dfrac{P}{v}\).
A1 Accept 21000 W, 21kW. Maximum 3 s.f.
| Scheme | Marks |
|---|---|
| \(\left(\text{their } 1393\dfrac{1}{3}\right) - 200 - 900g \times \dfrac{1}{9} = 900a\) | M1 A1ft |
| \(a = 0.237\) m s\(^{-2}\) | A1 |
| (3) |
Notes
M1 Equation of motion for the truck at instant after the towbar breaks. All terms required & dimensionally correct. Allow for an equation to find acceleration down the slope
A1ft Correct for their driving force \(\left(1393\dfrac{1}{3}\right)\).
A1 Accept 0.24, not \(\dfrac{32}{135}\) must be +ve
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{2} \times 150 \times 15^2 = 50d + 150g\sin\theta d\) | M1 A1 |
| \(\left(16875 = 50d + \dfrac{150}{9}gd\right)\) | A1 |
| \(d = 79\) m (79.1) | A1 |
| (4) | |
| (12 marks) |
Notes
M1 Must be using work-energy (for trailer only) All terms required & dimensionally correct. Condone sin/cos confusion and sign error(s)
A1 Unsimplified equation with at most one error
A1 Correct unsimplified equation for \(d\)
A1 Maximum 3 s.f.