| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Maximum speed on incline vs horizontal |
| Difficulty | Standard +0.3 This is a straightforward application of the power-force-velocity relationship (P=Fv) with constant resistance. Part (a) requires F=P/v then F=ma, part (b) uses P=Rv at maximum speed, and part (c) adds a component of weight. All steps are standard textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | 10000 = D 5 => D = 2000 |
| Answer | Marks |
|---|---|
| a = 1.458... so acceleration is awrt 1.46 ms–2 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | Using “P = Fv” to find the |
| Answer | Marks |
|---|---|
| (b) | At max speed, 250 = 20000/v |
| => v = 80 so max speed is 80 ms–1 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | Using NII with a = 0 oe and “P = |
| Answer | Marks |
|---|---|
| (c) | 20000/v = 250 + 1200gsin |
| v = 23.86... so max speed is awrt 23.9 ms–1 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.4 |
| 1.1 | Using NII with a = 0 oe and three |
| Answer | Marks |
|---|---|
| maximum power | 20000 = 838v |
Question 1:
1 | (a) | 10000 = D 5 => D = 2000
2000 – 250 = 1200a
a = 1.458... so acceleration is awrt 1.46 ms–2 | M1
M1
A1
[3] | 3.4
1.1
1.1 | Using “P = Fv” to find the
‘driving force’
Using NII with two force terms in
opposite directions to derive an
equation in a
(b) | At max speed, 250 = 20000/v
=> v = 80 so max speed is 80 ms–1 | M1
A1
[2] | 3.4
1.1 | Using NII with a = 0 oe and “P =
Fv” with maximum power
(c) | 20000/v = 250 + 1200gsin
v = 23.86... so max speed is awrt 23.9 ms–1 | M1
A1
[2] | 3.4
1.1 | Using NII with a = 0 oe and three
force terms and “P = Fv” with
maximum power | 20000 = 838v
Allow sign slip1 A car has mass 1200 kg . The total resistance to the car's motion is constant and equal to 250 N .
\begin{enumerate}[label=(\alph*)]
\item The car is driven along a straight horizontal road with its engine working at 10 kW .
Find the acceleration of the car at the instant that its speed is $5 \mathrm {~ms} ^ { - 1 }$.
The maximum power that the car's engine can generate is 20 kW .
\item Find the greatest constant speed at which the car can be driven along a straight horizontal road.
The car is driven up a straight road which is inclined at an angle $\theta$ above the horizontal where $\sin \theta = 0.05$.
\item Find the greatest constant speed at which the car can be driven up this road.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2022 Q1 [7]}}