| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Horizontal road towing |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics problem requiring Newton's second law applied to connected particles, then kinematics after disconnection. The setup is straightforward with clearly stated forces, and all parts follow routine procedures (F=ma for system, then individual particle for tension, then SUVAT). Slightly easier than average due to its predictable structure and lack of conceptual subtlety, though the multi-part nature and numerical work keep it close to typical difficulty. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Car + trailer: \(2100a = 2380 - 280 - 630 = 1470 \Rightarrow a = 0.7 \text{ m s}^{-2}\) | M1 A1 | (3) |
| (b) e.g. trailer: \(700 \times 0.7 = T - 280 \Rightarrow T = 770 \text{ N}\) | M1 A1∇ | A1 (3) |
| (c) Car: \(1400a' = 2380 - 630 \Rightarrow a' = 1.25 \text{ m s}^{-2}\) | M1 A1 | |
| distance \(= 12 \times 4 + \frac{1}{2} \times 1.25 \times 4^2 = 58 \text{ m}\) | M1 A1∇ | A1 (6) |
| (d) Same acceleration for car and trailer | B1 | (1) |
| (a) Car + trailer: $2100a = 2380 - 280 - 630 = 1470 \Rightarrow a = 0.7 \text{ m s}^{-2}$ | M1 A1 | (3) |
| (b) e.g. trailer: $700 \times 0.7 = T - 280 \Rightarrow T = 770 \text{ N}$ | M1 A1∇ | A1 (3) |
| (c) Car: $1400a' = 2380 - 630 \Rightarrow a' = 1.25 \text{ m s}^{-2}$ | M1 A1 | |
| distance $= 12 \times 4 + \frac{1}{2} \times 1.25 \times 4^2 = 58 \text{ m}$ | M1 A1∇ | A1 (6) |
| (d) Same acceleration for car and trailer | B1 | (1) |
**Guidance:**
- (a) M1 for a complete (potential) valid method to get $a$.
- (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.
- (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).
- (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.
---\begin{enumerate}
\item 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\\
(a) the acceleration of the car,\\
(b) the tension in the tow-rope.
\end{enumerate}
When the car and trailer are moving at $12 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the tow-rope breaks. Assuming that the driving force on the car and the resistances to motion are unchanged,\\
(c) find the distance moved by the car in the first 4 s after the tow-rope breaks.\\
(6)\\
(d) State how you have used the modelling assumption that the tow-rope is inextensible.
\hfill \mbox{\textit{Edexcel M1 2006 Q6 [13]}}