| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics A AS (Further Mechanics A AS) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Towing system: horizontal road |
| Difficulty | Standard +0.3 This is a standard multi-part mechanics question testing power-force-velocity relationships and equations of motion. Parts (i)-(iii) involve straightforward formula application (P=Fv, F=ma), part (iv) requires integration or numerical methods for variable acceleration, and part (v) is a standard modelling critique. While part (iv) adds some complexity, the overall question follows predictable patterns for AS Further Mechanics with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (i) | π |
| Answer | Marks |
|---|---|
| π = 23 kW | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| [2] | 1.1 | |
| 1.1 | Soi (by 23 000) | P is power of car |
| (ii) | 45000 |
| Answer | Marks |
|---|---|
| R = 1330 (N) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| 1.1 | Must be 3 terms; allow sign errors |
| If 0 scored, sc B1 for sight of 2250 | R is resistance due to βvan |
| (iii) | 68 000 |
| Answer | Marks |
|---|---|
| 48 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Allow M1 if mass or sign wrong or |
| Answer | Marks |
|---|---|
| 0.47916666 | Resistances can be combined |
| Answer | Marks |
|---|---|
| (iv) | 1 |
| Answer | Marks |
|---|---|
| T = 40.7 (seconds) | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | B1 for 68 000T or 2250Γ1024 |
| Answer | Marks |
|---|---|
| 40.658 | Allow sign errors; equation |
| Answer | Marks |
|---|---|
| (v) | Eg: Resistance could have been modelled as a |
| Answer | Marks | Guidance |
|---|---|---|
| resistance is greater at a geater speed). | B1 | |
| [1] | 3.5c | Or other correct reason |
Question 2:
2 | (i) | π
β920 = 0
25
π = 23 kW | M1
A1
[2] | 1.1
1.1 | Soi (by 23 000) | P is power of car
(ii) | 45000
β920βπ
= 0
20
R = 1330 (N) | M1
A1
[2] | 3.3
1.1 | Must be 3 terms; allow sign errors
If 0 scored, sc B1 for sight of 2250 | R is resistance due to βvan
(iii) | 68 000
β920β1330 = 2400π
20
23
a = 0.479 (m sβ2) or (m sβ2)
48 | M1
A1
A1
[3] | 3.3
1.1
1.1 | Allow M1 if mass or sign wrong or
one resistance wrong or omitted
0.47916666 | Resistances can be combined
for M1A1
(iv) | 1
Γ2400Γ202 +68 000πβ2250Γ1024
2
1
= Γ2400Γ282
2
T = 40.7 (seconds) | B1
M1
A1
A1
[4] | 3.4
1.1
1.1
1.1 | B1 for 68 000T or 2250Γ1024
M1 for use of KE (at least once),
work done & work against resistance
40.658 | Allow sign errors; equation
not needed for M1
Use of const accel: (B1)M0
(v) | Eg: Resistance could have been modelled as a
variable force (since it is more realistic to assume
resistance is greater at a geater speed). | B1
[1] | 3.5c | Or other correct reason | Ignore irrelevant reasons2 A car of mass 1350 kg travels along a straight horizontal road. Throughout this question the resistance force to the motion of the car is modelled as constant and equal to 920 N .\\
(i) Calculate the power, in kW , developed by the car when the car is travelling at a constant speed of $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
The car is now used to tow a caravan of mass 1050 kg along the same road. When the car tows the caravan at a constant speed of $20 \mathrm {~ms} ^ { - 1 }$ the power developed by the car is 45 kW .\\
(ii) Find the additional resistance force due to the caravan.
In the remaining parts of this question the power developed by the car is constant and equal to 68 kW and the resistance force due to the caravan is modelled as constant and equal to the value found in part (ii).
When the car and caravan pass a point A on the same straight horizontal road the speed of the car and caravan is $20 \mathrm {~ms} ^ { - 1 }$.\\
(iii) Find the acceleration of the car and caravan at point A .
The car and caravan later pass a point B on the same straight horizontal road with speed $28 \mathrm {~ms} ^ { - 1 }$. The distance $A B$ is $1024 m$.\\
(iv) Find the time taken for the car and caravan to travel from point A to point B .\\
(v) Suggest one way in which any of the modelling assumptions used in this question could have been improved.
\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2018 Q2 [12]}}