| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| 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 Further Mechanics question on power, resistance forces, and Newton's second law. Parts (i)-(iii) involve routine application of P=Fv and F=ma with multiple steps, while part (iv) requires conceptual understanding of how resistance affects motion. Slightly above average difficulty due to the multi-part nature and the need to track different scenarios, but all techniques are standard for Further Mechanics. |
| Spec | 6.02k Power: rate of doing work6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R_C = 40000/42\) | M1 | |
| \(952\text{ N}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R_T = 40000/30 - R_C\) | M1ft | |
| \(381\text{ N}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(D - R_C - R_T = 1400 \times 0.57\) | M1* | Attempt at "\(F = ma\)" for whole system (4 term equation). Allow 1333.3... instead of \(R_C + R_T\). Or \(D - R_C - T = 1200 \times 0.57\) ("\(F=ma\)" for car) |
| Correct equation (unsimplified) | A1 | |
| \(P = D \times 15\) | M1dep | |
| \(32000\) or \(32\text{kW}\) | A1 | NB 31970W |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T - R_T = 200 \times 0.57\) | M1FT | "\(F = ma\)" for trailer. Solution could use "\(F=ma\)" for car. Could be seen in (iii)(a). |
| \(495\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New model will predict a lower time to achieve a speed of \(20\text{ms}^{-1}\) | B1 | |
| Because at low speeds new model has no resistance and so acceleration will be greater | E1 | Resistance and acceleration must be mentioned or implied. Allow e.g. "no resistance means reaching 10m/s would occur faster" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| New model predicts the same | E1 | |
| Greatest speed depends only on (final) resistance (and power) | B1 |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R_C = 40000/42$ | M1 | |
| $952\text{ N}$ | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R_T = 40000/30 - R_C$ | M1ft | |
| $381\text{ N}$ | A1 | |
## Part (iii)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $D - R_C - R_T = 1400 \times 0.57$ | M1* | Attempt at "$F = ma$" for whole system (4 term equation). Allow 1333.3... instead of $R_C + R_T$. Or $D - R_C - T = 1200 \times 0.57$ ("$F=ma$" for car) |
| Correct equation (unsimplified) | A1 | |
| $P = D \times 15$ | M1dep | |
| $32000$ or $32\text{kW}$ | A1 | NB 31970W |
## Part (iii)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T - R_T = 200 \times 0.57$ | M1FT | "$F = ma$" for trailer. Solution could use "$F=ma$" for car. Could be seen in (iii)(a). |
| $495$ | A1 | |
## Part (iv)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New model will predict a lower time to achieve a speed of $20\text{ms}^{-1}$ | B1 | |
| Because at low speeds new model has no resistance and so acceleration will be greater | E1 | Resistance **and** acceleration must be mentioned or implied. Allow e.g. "no resistance means reaching 10m/s would occur faster" |
## Part (iv)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| New model predicts the same | E1 | |
| Greatest speed depends only on (final) resistance (and power) | B1 | |5 The engine of a car of mass 1200 kg produces a maximum power of 40 kW .\\
In an initial model of the motion of the car the total resistance to motion is assumed to be constant.\\
(i) Given that the greatest steady speed of the car on a straight horizontal road is $42 \mathrm {~ms} ^ { - 1 }$, find the magnitude of the resistance force.
The car is attached to a trailer of mass 200 kg by a light rigid horizontal tow bar. The greatest steady speed of the car and trailer on the road is now $30 \mathrm {~ms} ^ { - 1 }$. The resistance to motion of the trailer may also be assumed constant.\\
(ii) Find the magnitude of the resistance force on the trailer.
The car and trailer again travel along the road. At one instant their speed is $15 \mathrm {~ms} ^ { - 1 }$ and their acceleration is $0.57 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(iii) (a) Find the power of the engine of the car at this instant.\\
(b) Find the magnitude of the tension in the tow bar at this instant.
In a refined model of the motion of the car and trailer the resistance to the motion of each is assumed to be zero until they reach a speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. When the speed is $10 \mathrm {~ms} ^ { - 1 }$ or above the same constant resistance forces as in the first model are assumed to apply to each.
The car and trailer start at rest on the road and accelerate, using maximum power.\\
(iv) Without carrying out any further calculations,\\
(a) explain whether the time taken to attain a speed of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ would be predicted to be lower, the same or higher using the refined model compared with the original model,\\
(b) explain whether the greatest steady speed of the system would be predicted to be lower, the same or higher using the refined model compared with the original model.
\hfill \mbox{\textit{OCR Further Mechanics AS 2018 Q5 [14]}}