| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Standard +0.3 This is a standard Further Mechanics power-force-acceleration problem with three straightforward parts: (a) uses F=ma with P=Fv to find R (shown answer), (b) finds terminal velocity when acceleration=0, (c) applies the revised resistance model on an incline. All parts follow routine procedures with no novel insight required, though it's slightly above average difficulty due to being Further Maths content and requiring careful equation manipulation. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(D = 15000 / 20 = 750\) | B1 | "\(P = Fv\)" used in the solution |
| \(D - R = 800 \times 0.4\) | M1 | Use of NII with a driving force (might be incorrectly derived from power), \(R\) and correct \(ma\) term |
| \(R = 750 - 320 = 430\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Need \(15000 / v_{\max} =\) "430" | M1 | Driving force = resistive force and "\(P = Fv\)" |
| \(v_{\max} = 34.9\) so max speed is \(34.9\) ms\(^{-1}\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \((= 15000/v - 60v - 1568 = 120)\) | M1 | NII with a driving force, \(R\), a component of weight (condone incorrect component) and correct \(ma\) term |
| \(60v^2 + 1688v - 15000 = 0\) | M1 | Reduction to 3 term quadratic equation (must be equation) |
| \(7.10\) or \(-35.2\) | A1 | BC (condone \(7.09\) from incorrect rounding); both roots must be seen for this mark |
| Since \(v > 0\), speed is \(7.10\) ms\(^{-1}\) (3 sf) | A1FT | FT their quadratic, if one positive and one negative root (ie if \(ac < 0\)) for selecting their positive root with valid reason given. SC1 if A0A0 for \(7.10\) ms\(^{-1}\) with no justification |
# Question 2:
## Part (a)
$D = 15000 / 20 = 750$ | **B1** | "$P = Fv$" used in the solution
$D - R = 800 \times 0.4$ | **M1** | Use of NII with a driving force (might be incorrectly derived from power), $R$ and correct $ma$ term
$R = 750 - 320 = 430$ | **A1** | AG
## Part (b)
Need $15000 / v_{\max} =$ "430" | **M1** | Driving force = resistive force and "$P = Fv$"
$v_{\max} = 34.9$ so max speed is $34.9$ ms$^{-1}$ (3 sf) | **A1** |
## Part (c)
$D - R - 800g \times \sin\alpha = 800 \times 0.15$
$(= 15000/v - 60v - 1568 = 120)$ | **M1** | NII with a driving force, $R$, a component of weight (condone incorrect component) and correct $ma$ term
$60v^2 + 1688v - 15000 = 0$ | **M1** | Reduction to 3 term quadratic equation (must be equation)
$7.10$ or $-35.2$ | **A1** | BC (condone $7.09$ from incorrect rounding); both roots must be seen for this mark
Since $v > 0$, speed is $7.10$ ms$^{-1}$ (3 sf) | **A1FT** | FT their quadratic, if one positive and one negative root (ie if $ac < 0$) for selecting their positive root with valid reason given. **SC1** if **A0A0** for $7.10$ ms$^{-1}$ with no justification
---2 A car has a mass of 800 kg . The engine of the car is working at a constant power of 15 kW .
In an initial model of the motion of the car it is assumed that the car is subject to a constant resistive force of magnitude $R N$.
The car is initially driven on a straight horizontal road. At the instant that its speed is $20 \mathrm {~ms} ^ { - 1 }$ its acceleration is $0.4 \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $R = 430$.
\item Hence find the maximum constant speed at which the car can be driven along this road, according to the initial model.
In a revised model the resistance to the motion of the car at any instant is assumed to be 60 v where $v$ is the speed of the car at that instant.
The car is now driven up a straight road which is inclined at an angle $\alpha$ above the horizontal where $\sin \alpha = 0.2$.
\item Determine the speed of the car at the instant that its acceleration is $0.15 \mathrm {~ms} ^ { - 2 }$ up the slope, according to the revised model.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2021 Q2 [9]}}