| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2003 |
| 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 connected particles problem with straightforward application of Newton's second law. Parts (a) and (b) require routine F=ma calculations treating the system as connected, then separately. Part (c) involves basic kinematics (v=u+at) comparing two scenarios. All steps are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors3.03k Connected particles: pulleys and equilibrium6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Car + truck: \(2000a = 2400 - 600 - 400\) | M1 A1 | |
| \(a = 0.7\) m s\(^{-2}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| Car only: \(T - 400 = 800\times 0.7\) | M1 A1 ft | |
| [or truck only: \(2400 - T - 600 = 1200\times 0.7\)] | ||
| \(T = 960\) N | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| New acceleration of truck: \(1200\,a' = 2400 - 600\) | M1 | |
| \(a' = 1.5\) m s\(^{-2}\) | A1 | |
| Time to reach 28 m s\(^{-1}\) \(= \frac{28-20}{1.5} = 5.33\) s | M1 A1 | |
| Time to reach 28 m s\(^{-1}\) if rope had not broken \(= \frac{28-20}{0.7} = 11.43\) s | M1 A1 | |
| Difference \(= 6.1\) s \(\approx 6\) s | A1 | (*) given on paper, (7) |
# Question 8:
## Part (a):
| Working | Marks | Notes |
|---------|-------|-------|
| Car + truck: $2000a = 2400 - 600 - 400$ | M1 A1 | |
| $a = 0.7$ m s$^{-2}$ | A1 | (3) |
## Part (b):
| Working | Marks | Notes |
|---------|-------|-------|
| Car only: $T - 400 = 800\times 0.7$ | M1 A1 ft | |
| [or truck only: $2400 - T - 600 = 1200\times 0.7$] | | |
| $T = 960$ N | A1 | (3) |
## Part (c):
| Working | Marks | Notes |
|---------|-------|-------|
| New acceleration of truck: $1200\,a' = 2400 - 600$ | M1 | |
| $a' = 1.5$ m s$^{-2}$ | A1 | |
| Time to reach 28 m s$^{-1}$ $= \frac{28-20}{1.5} = 5.33$ s | M1 A1 | |
| Time to reach 28 m s$^{-1}$ if rope had not broken $= \frac{28-20}{0.7} = 11.43$ s | M1 A1 | |
| Difference $= 6.1$ s $\approx 6$ s | A1 | (*) given on paper, (7) |8. A car which has run out of petrol is being towed by a breakdown truck along a straight horizontal road. The truck has mass 1200 kg and the car has mass 800 kg . The truck is connected to the car by a horizontal rope which is modelled as light and inextensible. The truck's engine provides a constant driving force of 2400 N . The resistances to motion of the truck and the car are modelled as constant and of magnitude 600 N and 400 N respectively. Find
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the truck and the ear,
\item the tension in the rope.
When the truck and car are moving at $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the rope breaks. The engine of the truck provides the same driving force as before. The magnitude of the resistance to the motion of the truck remains 600 N .
\item Show that the truck reaches a speed of $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ approximately 6 s earlier than it would have done if the rope had not broken.
\section*{END}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2003 Q8 [13]}}