| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: inclined road |
| Difficulty | Standard +0.3 This is a standard M2 mechanics problem requiring application of Newton's second law to a two-body system and power calculation (P=Fv). The incline adds mild complexity but sin θ = 1/20 simplifies calculations. All values are given; students follow a routine procedure: resolve forces parallel to slope, apply F=ma to find driving force, then calculate power. Part (b) requires isolating the trailer. Slightly above average due to two-body system and incline, but follows textbook methods with no novel insight required. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equation of motion: \(F + 1000g\sin\theta + 250g\sin\theta - 300 - 100 = 1250a\) | M1A2 | M1 for \(F=ma\) along the plane for whole system; A2 correct equation (-1 each error); omission of \(g\) is an A error |
| \(F + 612.5 - 400 = 1250a = 250\) | M1 | Independent M1 for producing a value for \(F\) |
| \(F = 37.5\) (N) | ||
| Power \(= Fv = 37.5 \times 25 = 940\) W to 2 s.f. (938 W) | M1A1 | M1 for their \(F \times 25\); A1 for 940 or 938 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Motion of trailer: \(T - 100 + 250g\sin\theta = 250a\) | M1A2 | M1 for \(F=ma\) along the plane for either car or trailer; A2 for correct equation (ft on their \(F\)) |
| \(T = 27.5\) (N) or 28 (N) | A1 | A1 for 28 N or 27.5 N |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Motion of car: \(F + 1000g\sin\theta - T - 300 = 1000a\) | M1A2 ft | |
| \(T = 27.5\) (N) or 28 (N) | A1 |
# Question 2:
## Part 2a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equation of motion: $F + 1000g\sin\theta + 250g\sin\theta - 300 - 100 = 1250a$ | M1A2 | M1 for $F=ma$ along the plane for whole system; A2 correct equation (-1 each error); omission of $g$ is an A error |
| $F + 612.5 - 400 = 1250a = 250$ | M1 | Independent M1 for producing a value for $F$ |
| $F = 37.5$ (N) | | |
| Power $= Fv = 37.5 \times 25 = 940$ W to 2 s.f. (938 W) | M1A1 | M1 for their $F \times 25$; A1 for 940 or 938 |
## Part 2b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Motion of trailer: $T - 100 + 250g\sin\theta = 250a$ | M1A2 | M1 for $F=ma$ along the plane for either car or trailer; A2 for correct equation (ft on their $F$) |
| $T = 27.5$ (N) or 28 (N) | A1 | A1 for 28 N or 27.5 N |
## Alt Part 2b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Motion of car: $F + 1000g\sin\theta - T - 300 = 1000a$ | M1A2 ft | |
| $T = 27.5$ (N) or 28 (N) | A1 | |
> **Note:** Deduct only 1 mark in whole question for not giving answer to 2 sf or 3 sf following use of $g=9.8$ or $g=9.81$. Deduct the final A mark in whichever part it first occurs.
---2. A trailer of mass 250 kg is towed by a car of mass 1000 kg . The car and the trailer are travelling down a straight road inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 1 } { 20 }$
The resistance to motion of the car is modelled as a single force of magnitude 300 N acting parallel to the road. The resistance to motion of the trailer is modelled as a single force of magnitude 100 N acting parallel to the road. The towbar joining the car to the trailer is modelled as a light rod which is parallel to the direction of motion. At a given instant the car and the trailer are moving down the road with speed $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and acceleration $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the power being developed by the car's engine at this instant.
\item Find the tension in the towbar at this instant.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q2 [10]}}