M1 June 2005 Q7
7.
Figure 4 shows a lorry of mass 1600 kg towing a car of mass 900 kg along a straight horizontal road. The two vehicles are joined by a light towbar which is at an angle of 15\(^\circ\) to the road. The lorry and the car experience constant resistances to motion of magnitude 600 N and 300 N respectively. The lorry’s engine produces a constant horizontal force on the lorry of magnitude 1500 N. Find
(a) the acceleration of the lorry and the car, (3)
(b) the tension in the towbar. (4)
When the speed of the vehicles is 6 m s\(^{-1}\), the towbar breaks. Assuming that the resistance to the motion of the car remains of constant magnitude 300 N,
(c) find the distance moved by the car from the moment the towbar breaks to the moment when the car comes to rest. (4)
(d) State whether, when the towbar breaks, the normal reaction of the road on the car is increased, decreased or remains constant. Give a reason for your answer. (2)
| Scheme | Marks |
|---|---|
| Lorry + Car: \(2500a = 1500 - 300 - 600\) | M1 A1 |
| \(a = 0.24\) m s\(^{-2}\) | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| Car: \(T\cos 15 - 300 = 900a\) OR Lorry: \(1500 - T\cos 15 - 600 = 1600a\) | M1 A1 |
| Sub and solve: \(T \approx 534\) N | M1 A1 |
| (4) |
| Scheme | Marks |
|---|---|
| Deceleration of car \(= 300/900 = 1/3\) m s\(^{-2}\) | M1 A1 |
| Hence \(6^2 = 2 \times 1/3 \times s \;\Rightarrow\; s = 54\) m | M1 A1 |
| (4) |
Notes
(Corrected from the printed mark scheme: the units of the deceleration are printed as m s\(^{-1}\).)
| Scheme | Marks |
|---|---|
| Vertical component of \(T\) now removed | M1 |
| Hence normal reaction is increased | A1 cso |
| (2) | |
| (13 marks) |