| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Banked track – no friction (find speed or radius) |
| Difficulty | Standard +0.3 This is a standard Further Mechanics banked track problem requiring resolution of forces and application of circular motion equations (tan θ = v²/rg). Part (a) is routine calculation, part (b) tests understanding of modelling assumptions, and part (c) requires recognizing that friction acts up the slope when speed exceeds the no-friction condition. While it's FM2 content, the problem follows a well-established template with straightforward application of formulas, making it slightly easier than average overall. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| VILU SIHI NI IIIUM ION OC | VGHV SIHILNI IMAM ION OO | VJYV SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy to find value of \(\theta\) | M1 | Complete strategy involving resolving in perpendicular directions, change of units, solution of simultaneous equations |
| Resolve vertically | M1 | Complete strategy to form one equation involving \(\theta\); condone sin/cos confusion |
| \(R\cos\theta° = mg\) | A1 | Or equivalent |
| Resolve horizontally | M1 | Complete strategy to form second equation involving \(\theta\); condone sin/cos confusion |
| \(R\sin\theta° = \frac{mv^2}{r}\) | A1 | Correct unsimplified – need not substitute for \(v\) or \(r\) |
| \(v = 80 \text{ km h}^{-1} = \frac{80 \times 1000}{60^2} \text{ m s}^{-1}\) | B1 | Correct conversion km h\(^{-1}\) to m s\(^{-1}\) (22.2) |
| \(\tan\theta° = \frac{v^2}{rg} = \frac{640000}{36^2 \times 500 \times 9.8}\), \(\quad \theta = 5.8\) | A1 | Accept 5.8 or 5.75 (follows use of 9.8) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| All weight acting at a single point | B1 | Any appropriate comment e.g. only one point of contact with road; centre of mass of car is on the road |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Friction acting down the slope | B1 | Need to include the direction |
## Question 2:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy to find value of $\theta$ | M1 | Complete strategy involving resolving in perpendicular directions, change of units, solution of simultaneous equations |
| Resolve vertically | M1 | Complete strategy to form one equation involving $\theta$; condone sin/cos confusion |
| $R\cos\theta° = mg$ | A1 | Or equivalent |
| Resolve horizontally | M1 | Complete strategy to form second equation involving $\theta$; condone sin/cos confusion |
| $R\sin\theta° = \frac{mv^2}{r}$ | A1 | Correct unsimplified – need not substitute for $v$ or $r$ |
| $v = 80 \text{ km h}^{-1} = \frac{80 \times 1000}{60^2} \text{ m s}^{-1}$ | B1 | Correct conversion km h$^{-1}$ to m s$^{-1}$ (22.2) |
| $\tan\theta° = \frac{v^2}{rg} = \frac{640000}{36^2 \times 500 \times 9.8}$, $\quad \theta = 5.8$ | A1 | Accept 5.8 or 5.75 (follows use of 9.8) |
**(7 marks)**
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| All weight acting at a single point | B1 | Any appropriate comment e.g. only one point of contact with road; centre of mass of car is on the road |
**(1 mark)**
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Friction acting down the slope | B1 | Need to include the direction |
**(1 mark)**\begin{enumerate}
\item A car moves round a bend which is banked at a constant angle of $\theta ^ { \circ }$ to the horizontal.
\end{enumerate}
When the car is travelling at a constant speed of $80 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ there is no sideways frictional force on the car. The car is modelled as a particle moving in a horizontal circle of radius 500 m .\\
(a) Find the value of $\theta$.\\
(b) Identify one limitation of this model.
The speed of the car is increased so that it is now travelling at a constant speed of $90 \mathrm { kmh } ^ { - 1 }$ The car is still modelled as a particle moving in a horizontal circle of radius 500 m .\\
(c) Describe the extra force that will now be acting on the car, stating the direction of this force.
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\hfill \mbox{\textit{Edexcel FM2 AS 2018 Q2 [9]}}