| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Horizontal road towing |
| Difficulty | Standard +0.3 This is a standard M2 mechanics question involving power-force-velocity relationships and Newton's second law applied to connected particles. Part (a) uses the routine formula P=Fv at constant speed (equilibrium). Parts (b)-(d) apply F=ma to the system and individual bodies with straightforward calculations. The multi-part structure and bookwork nature place it slightly easier than average, though the connected particles aspect requires careful force analysis. |
| Spec | 3.03a Force: vector nature and diagrams3.03c Newton's second law: F=ma one dimension6.02a Work done: concept and definition6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(50000 = F \times 25\) \((F = 2000)\) | M1 | or equivalent |
| \(F = R + 750\) | M1 | |
| \(R = 1250\) | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| N2L: \(1500 + 2000 = 2500a\) | M1 A1 | ignore sign of \(a\) |
| \(a = 1.4 \text{ ms}^{-2}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Trailer: \(T + R = 1500 \times 1.4\) or Car: \(T - 1500 - 750 = 1000 \times -1.4\) | M1 | |
| \(T = 850\) (N) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(25^2 = 2 \times 1.4 \times s\) \((s = 223.2...)\) | M1 | |
| \(W = 1500 \times s\) | M1 A1ft | ft their \(s\) |
| \(= 335\) (kJ) | A1 | accept 330 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resistances vary with speeds | B1 |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $50000 = F \times 25$ $(F = 2000)$ | M1 | or equivalent |
| $F = R + 750$ | M1 | |
| $R = 1250$ | A1 | cso |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| N2L: $1500 + 2000 = 2500a$ | M1 A1 | ignore sign of $a$ |
| $a = 1.4 \text{ ms}^{-2}$ | A1 | cao |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Trailer: $T + R = 1500 \times 1.4$ or Car: $T - 1500 - 750 = 1000 \times -1.4$ | M1 | |
| $T = 850$ (N) | A1 | |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $25^2 = 2 \times 1.4 \times s$ $(s = 223.2...)$ | M1 | |
| $W = 1500 \times s$ | M1 A1ft | ft their $s$ |
| $= 335$ (kJ) | A1 | accept 330 |
### Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resistances vary with speeds | B1 | |
---5. A car of mass 1000 kg is towing a trailer of mass 1500 kg along a straight horizontal road. The tow-bar joining the car to the trailer is modelled as a light rod parallel to the road. The total resistance to motion of the car is modelled as having constant magnitude 750 N . The total resistance to motion of the trailer is modelled as of magnitude $R$ newtons, where $R$ is a constant. When the engine of the car is working at a rate of 50 kW , the car and the trailer travel at a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $R = 1250$.
When travelling at $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the driver of the car disengages the engine and applies the brakes. The brakes provide a constant braking force of magnitude 1500 N to the car. The resisting forces of magnitude 750 N and 1250 N are assumed to remain unchanged. Calculate
\item the deceleration of the car while braking,
\item the thrust in the tow-bar while braking,
\item the work done, in kJ , by the braking force in bringing the car and the trailer to rest.
\item Suggest how the modelling assumption that the resistances to motion are constant could be refined to be more realistic.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2005 Q5 [13]}}