| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| 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 P=Fv at constant speed for part (a), then using work-energy principle for part (b). The calculations are routine with clear setups, though part (b) requires solving a quadratic equation from energy considerations. Slightly above average due to the multi-step nature and need to connect power, work and kinematics, but no novel insight required. |
| 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 |
|---|---|---|
| 5 | (a) | Let the caravan have mass m kg and the driving force on the |
| Answer | Marks | Guidance |
|---|---|---|
| No acceleration, so D=( 1600+m ) gsin4.7°. | M1 | 3.3 |
| 120000=( 1600+m ) gsin4.7°×32 | M1 | 3.4 |
| ⇒m≈3070 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | Let the speed of the car at B be v ms-1. | |
| Total work done by engine =80000×3.54 (=283200 ) J | B1 | 1.1 |
| Gain in GPE =1600×9.8sin4.7°×80 (≈102784 ) J | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| ⇒v≈25.0 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | Resistance forces could be taken into account. | B1 |
Question 5:
5 | (a) | Let the caravan have mass m kg and the driving force on the
car have magnitude D N.
No acceleration, so D=( 1600+m ) gsin4.7°. | M1 | 3.3
120000=( 1600+m ) gsin4.7°×32 | M1 | 3.4 | Power = force×velocity
⇒m≈3070 | A1 | 1.1
[3]
(b) | Let the speed of the car at B be v ms-1.
Total work done by engine =80000×3.54 (=283200 ) J | B1 | 1.1 | Need not be seen separately.
Gain in GPE =1600×9.8sin4.7°×80 (≈102784 ) J | B1 | 1.1 | Need not be seen separately.
1⋅1600⋅202 +283200= 1⋅1600⋅v2 +102784
2 2 | M1 | 3.3 | WEP: all terms present; condone
one sign error / term on wrong side.
⇒v≈25.0 | A1 | 1.1 | Accept 25 | 25.0104…
[4]
(c) | Resistance forces could be taken into account. | B1 | 3.5c
[1]5 A car of mass 1600 kg is travelling uphill along a straight road inclined at $4.7 ^ { \circ }$ to the horizontal. The power developed by the car is constant and equal to 120 kW . The car is towing a caravan and together they have a maximum speed of $32 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ uphill.
In this question you may model any resistances to motion as negligible.
\begin{enumerate}[label=(\alph*)]
\item Determine the mass of the caravan.
The caravan is now detached from the car. Continuing up the same road, the car passes a point A at a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The car later passes through a point $B$ on the same road such that $A B = 80 \mathrm {~m}$ and the car takes 3.54 seconds to travel from A to B . The power developed by the car while travelling from A to B is constant and equal to 80 kW .
\item Determine the speed of the car at B .
\item State one possible refinement to the model used in parts (a) and (b).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2021 Q5 [8]}}