| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find k or constants |
| Difficulty | Moderate -0.3 This is a straightforward application of the power equation P = Fv with resistance force kvΒ². Part (a) requires simple algebraic manipulation (5000 = kΓ20Β³Γ20), part (b) is identical calculation with different values, and part (c) adds a component for gravitational force on an incline but remains routine. All steps are standard textbook exercises requiring recall of formulas rather than problem-solving insight. |
| Spec | 6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | Let the driving force on the car be D N |
| Answer | Marks | Guidance |
|---|---|---|
| 2 0 | B1 | 3.4 |
| π· = π(20)2 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 203 8 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Power is 5 (28)3 = 13720 (W) | |
| 8 | B1 | 1.1 |
| Answer | Marks |
|---|---|
| (c) | Let the mass of the car be m kg |
| Answer | Marks |
|---|---|
| 650 = 250+ππsin2Β° | B1 |
| M1 | 1.1 |
| 1.1 | Correct weight component ππsin2Β° |
| Answer | Marks | Guidance |
|---|---|---|
| Mass is 1170 (kg) | A1 | 1.1 |
Question 1:
1 | (a) | Let the driving force on the car be D N
5 0 0 0
D = ( = 250)
2 0 | B1 | 3.4 | soi
π· = π(20)2 | M1 | 3.3 | Using driving force = resistance
e.g. 5000 = π(20)2Γ20 implies B1 and M1
5000 5
οk= =
203 8 | A1 | 1.1 | AG Must be correctly obtained
[3]
(b) | Power is 5 (28)3 = 13720 (W)
8 | B1 | 1.1 | Accept 13 700 www
Accept 14 000 from correct method
[1]
(c) | Let the mass of the car be m kg
13000 = 5 (20)2+ππsin2Β°
20 8
650 = 250+ππsin2Β° | B1
M1 | 1.1
1.1 | Correct weight component ππsin2Β°
(B0 for πsin2Β° unless π = ππ used later)
Resolving parallel to the plane β attempt at
driving force (using 13000), resistance, and
attempt to resolve weight (condone missing g)
Driving force 13000 is M0
Mass is 1170 (kg) | A1 | 1.1 | 1169.53911β¦
[3]1 Throughout all parts of this question, the resistance to the motion of a car has magnitude $\mathrm { kv } ^ { 2 } \mathrm {~N}$, where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the car and $k$ is a constant.
At first, the car travels along a straight horizontal road with constant speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The power developed by the car at this speed is 5000 W .
\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac { 5 } { 8 }$.
\item Find the power the car must develop in order to maintain a constant speed of $28 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when travelling along the same horizontal road.
The car climbs a hill which is inclined at an angle of $2 ^ { \circ }$ to the horizontal. The power developed by the car is 13000 W , and the car has a constant speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Determine the mass of the car.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2023 Q1 [7]}}