| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2018 |
| Session | March |
| Marks | 16 |
| Topic | Power and driving force |
| Type | Towing system: horizontal road |
| Difficulty | Standard +0.8 This is a multi-part Further Mechanics question requiring power-force-velocity relationships, Newton's second law with variable resistance, and connected particles analysis. While the individual techniques are standard (P=Fv, F=ma), the question requires careful systematic application across four parts with increasing complexity, including the inclined plane scenario. The 'show that' part is straightforward, but parts (iii) and (iv) require setting up and solving equations with multiple forces acting on a two-body system, which is moderately challenging for FM1 level. |
| Spec | 3.03d Newton's second law: 2D vectors6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(k \times 25 = \frac{35\,000}{25}\) and \(k = 56\) | M1, A1 [2] | AG |
| \(D - R_c = 850a\) and \(\frac{35\,000}{15} - 56 \times 15 = 850a\) | M1, A1 | NII with 3 terms, signs correct; Correct equation (unsimplified) |
| awrt \(1.76 \text{ m s}^{-2}\) | A1 [3] | |
| \(D - R_c - 350 = (850 + 240) \times 0.2\) | M1 | NII with 4 terms, signs correct |
| \(\frac{30\,000}{v} - 56v - 350 = 218\) | A1 | Correct equation (unsimplified) |
| \(56v^2 + 568v - 30\,000 = 0\) | M1 | Rearrange as quadratic and solve |
| awrt \(18.6 \text{ m s}^{-1}\) | A1 [4] | BC Ignore 2nd root unless presented as part of final answer |
| \(T - 350 = 240 \times 0.2\) | M1 | oe, considering forces on car |
| \(398 \text{ N}\) | A1 [2] | |
| Up: \(\frac{35\,000}{v_u} - 56v_u + 350 + 1090g \sin 8° = 0\) | M1 | 4-term equation, correct signs (\(56v_u^2 + 1836.647v_u - 35\,000 = 0\)) |
| \(v_u = 13.5\) | A1 | BC |
| Down: \(\frac{35\,000}{v_d} + 1090g \sin 8° = 56v_d + 350\) | M1 | 4-term equation, correct signs (\(56v_d^2 - 1136.647v_d - 35\,000 = 0\)) |
| \(v_d = 37.1\) and Difference in speeds \(= \text{awrt } 23.6 \text{ m s}^{-1}\) | A1, A1 [5] | BC; Ignore second roots of the quadratic equations unless they contribute to something appearing in the final answer |
| Answer | Marks | Guidance |
|--------|-------|----------|
| $k \times 25 = \frac{35\,000}{25}$ and $k = 56$ | M1, A1 [2] | AG |
| $D - R_c = 850a$ and $\frac{35\,000}{15} - 56 \times 15 = 850a$ | M1, A1 | NII with 3 terms, signs correct; Correct equation (unsimplified) |
| awrt $1.76 \text{ m s}^{-2}$ | A1 [3] | |
| $D - R_c - 350 = (850 + 240) \times 0.2$ | M1 | NII with 4 terms, signs correct |
| $\frac{30\,000}{v} - 56v - 350 = 218$ | A1 | Correct equation (unsimplified) |
| $56v^2 + 568v - 30\,000 = 0$ | M1 | Rearrange as quadratic and solve |
| awrt $18.6 \text{ m s}^{-1}$ | A1 [4] | BC Ignore 2nd root unless presented as part of final answer |
| $T - 350 = 240 \times 0.2$ | M1 | oe, considering forces on car |
| $398 \text{ N}$ | A1 [2] | |
| Up: $\frac{35\,000}{v_u} - 56v_u + 350 + 1090g \sin 8° = 0$ | M1 | 4-term equation, correct signs ($56v_u^2 + 1836.647v_u - 35\,000 = 0$) |
| $v_u = 13.5$ | A1 | BC |
| Down: $\frac{35\,000}{v_d} + 1090g \sin 8° = 56v_d + 350$ | M1 | 4-term equation, correct signs ($56v_d^2 - 1136.647v_d - 35\,000 = 0$) |
| $v_d = 37.1$ and Difference in speeds $= \text{awrt } 23.6 \text{ m s}^{-1}$ | A1, A1 [5] | BC; Ignore second roots of the quadratic equations unless they contribute to something appearing in the final answer |4 A car has a mass of 850 kg and its engine can generate a maximum power of 35 kW . The total resistance to motion of the car is modelled as $k v \mathrm {~N}$ where $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is the speed of the car and $k$ is a constant.
When the car is moving in a straight line on a straight horizontal road, the greatest constant speed that it can attain is $25 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\begin{enumerate}[label=(\roman*)]
\item Show that $k = 56$.
\item Find the greatest possible acceleration of the car on the road at an instant when it is moving with a speed of $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
A trailer of mass 240 kg is attached to the car by means of a light inextensible tow bar which is parallel to the surface of the road. The resistance to motion of the trailer is modelled as a constant force of magnitude 350 N .
The car and trailer move on the horizontal road. At a certain instant the car's engine is working at a rate of 30 kW and the acceleration of the car is $0.2 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item (a) Find the speed of the car at this instant.\\
(b) Find the magnitude of the tension in the tow bar at this instant.
The car and trailer now move in a straight line on a straight road inclined at $8 ^ { \circ }$ to the horizontal.
\item Find the difference between their greatest possible constant speed travelling up the slope and their greatest possible constant speed travelling down the slope.
\end{enumerate}
\hfill \mbox{\textit{OCR FM1 AS 2018 Q4 [16]}}