| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 9 |
| Topic | Power and driving force |
| Type | Towing system: horizontal road |
| Difficulty | Standard +0.3 This is a straightforward multi-part mechanics question testing standard power-force-velocity relationships and equilibrium conditions. Part (a) requires integrating P=Fv with F=ma, part (b) is direct substitution into P=Fv at equilibrium, and part (c) involves setting up two simultaneous equations for resistances. All techniques are routine for Further Mechanics students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03v Motion on rough surface: including inclined planes6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| \(15000 \times 6 = \frac{1}{2} \times 800v^2\) | M1 [1.1a] | Finding energy input using \(W = Pt\) and equating to KE gained |
| \(v = \sqrt{2225} = 15\) m s\(^{-1}\) | A1 [1.1] | [2] |
| Or \(\frac{15000}{v} = 800\frac{dv}{dt}\) \(\Rightarrow 75\int_{0}^{6} dt = \int_{0}^{v} 2v \, dv\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Considering forces along the road: \(800g \sin\theta + 150 = D\) | M1 [1.1a] | Where \(D\) is the driving force |
| \(15000 = Dv\) | M1 [1.1] | Use of \(P = Fv\) |
| \(v = \frac{15000}{800g \times \frac{1}{20} + 150}\) awrt 27.7 m s\(^{-1}\) | A1 [1.1] | 27.675276... |
| [3] | Or \(W = 800gd \sin\theta + 150d\), where \(d\) is distance travelled ... in time \(t\) and \(W = 15000t\) and \(v = \frac{d}{t}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(D = \frac{15000}{30} \quad (= 500)\) | M1 [1.1a] | Driving force ... |
| \(\frac{15000}{30} = K_{total}\) | M1 [2.2a] | ... equals total resistance |
| \(R_c = \frac{3}{4} \times 500 = 375\) N | A1 [1.1] | |
| Car: \(500 = T + 375 \Rightarrow T = 125\) N | A1 [1.1] | Or Trailer: \(T = R_T = \frac{1}{4} \times 500 = 125\) N |
| [4] |
### (a)
$15000 \times 6 = \frac{1}{2} \times 800v^2$ | M1 [1.1a] | Finding energy input using $W = Pt$ and equating to KE gained
$v = \sqrt{2225} = 15$ m s$^{-1}$ | A1 [1.1] | [2]
| Or $\frac{15000}{v} = 800\frac{dv}{dt}$ $\Rightarrow 75\int_{0}^{6} dt = \int_{0}^{v} 2v \, dv$ oe
### (b)
Considering forces along the road: $800g \sin\theta + 150 = D$ | M1 [1.1a] | Where $D$ is the driving force
$15000 = Dv$ | M1 [1.1] | Use of $P = Fv$
$v = \frac{15000}{800g \times \frac{1}{20} + 150}$ awrt 27.7 m s$^{-1}$ | A1 [1.1] | 27.675276...
| [3] | Or $W = 800gd \sin\theta + 150d$, where $d$ is distance travelled ... in time $t$ and $W = 15000t$ and $v = \frac{d}{t}$
### (c)
$D = \frac{15000}{30} \quad (= 500)$ | M1 [1.1a] | Driving force ...
$\frac{15000}{30} = K_{total}$ | M1 [2.2a] | ... equals total resistance
$R_c = \frac{3}{4} \times 500 = 375$ N | A1 [1.1]
Car: $500 = T + 375 \Rightarrow T = 125$ N | A1 [1.1] | Or Trailer: $T = R_T = \frac{1}{4} \times 500 = 125$ N
| [4]
---2 A car of mass 800 kg is driven with its engine generating a power of 15 kW .
\begin{enumerate}[label=(\alph*)]
\item The car is first driven along a straight horizontal road and accelerates from rest.
Assuming that there is no resistance to motion, find the speed of the car after 6 seconds.
\item The car is next driven at constant speed up a straight road inclined at an angle $\theta$ to the horizontal. The resistance to motion is now modelled as being constant with magnitude 150 N .
Given that $\sin \theta = \frac { 1 } { 20 }$, find the speed of the car.
\item The car is now driven at a constant speed of $30 \mathrm {~ms} ^ { - 1 }$ along the horizontal road pulling a trailer of mass 150 kg which is attached by means of a light rigid horizontal towbar.
Assuming that the resistance to motion of the car is three times the resistance to motion of the trailer, find
\begin{itemize}
\item the resistance to motion of the car,
\item the magnitude of the tension in the towbar.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q2 [9]}}