M1 June 2006 Q6
6. A car is towing a trailer along a straight horizontal road by means of a horizontal tow-rope. The mass of the car is 1400 kg. The mass of the trailer is 700 kg. The car and the trailer are modelled as particles and the tow-rope as a light inextensible string. The resistances to motion of the car and the trailer are assumed to be constant and of magnitude 630 N and 280 N respectively. The driving force on the car, due to its engine, is 2380 N. Find
When the car and trailer are moving at 12 m s\(^{-1}\), the tow-rope breaks. Assuming that the driving force on the car and the resistances to motion are unchanged,
| Scheme | Marks |
|---|---|
| Car + trailer: \(2100a = 2380 - 280 - 630\) | M1 A1 |
| \(= 1470 \;\Rightarrow\; a = 0.7\) m s\(^{-2}\) | A1 |
| (3) |
Notes
(a) M1 for a complete (potential) valid method to get \(a\)
| Scheme | Marks |
|---|---|
| e.g. trailer: \(700 \times 0.7 = T - 280\) | M1 A1ft |
| \(\Rightarrow T = 770\) N | A1 |
| (3) |
Notes
(b) If consider car: then get \(1400a = 2380 - 630 - T\).
Allow M1 A1 for equn of motion for car or trailer wherever seen (e.g. in (a)).
So if consider two separately in (a), can get M1 A1 from (b) for one equation; then M1 A1 from (a) for second equation, and then A1 [(a)] for \(a\) and A1 [(b)] for \(T\).
In equations of motion, M1 requires no missing or extra terms and dimensionally correct (e.g. extra force, or missing mass, is M0). If unclear which body is being considered, assume that the body is determined by the mass used. Hence if ‘\(1400a\)’ used, assume it is the car and mark forces etc accordingly. But allow e.g. 630/280 confused as an A error.
| Scheme | Marks |
|---|---|
| Car: \(1400a' = 2380 - 630\) | M1 A1 |
| \(\Rightarrow a' = 1.25\) m s\(^{-2}\) | A1 |
| distance \(= 12 \times 4 + \tfrac{1}{2} \times 1.25 \times 4^2\) | M1 A1ft |
| \(= 58\) m | A1 |
| (6) |
Notes
(c) Must be finding a new acceleration here. (If they get 1.25 erroneously in (a), and then simply assume it is the same acceln here, it is M0).
| Scheme | Marks |
|---|---|
| Same acceleration for car and trailer | B1 |
| (1) | |
| (13 marks) |
Notes
(d) Allow o.e. but you must be convinced they are saying that it is same acceleration for both bodies. E.g. ‘acceleration constant’ on its own is B0
Ignore extras, but ‘acceleration and tension same at \(A\) and \(B\)’ is B0