| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2017 |
| Session | December |
| Marks | 9 |
| Topic | Power and driving force |
| Type | Towing system: horizontal road |
| Difficulty | Standard +0.8 This is a multi-part Further Mechanics question requiring understanding of oblique tension geometry (using Pythagoras to find the angle), work-energy principles, and power-force relationships. While the individual techniques are standard, the combination of non-parallel tracks creating an angled chain, applying work done correctly to the carriage's motion, and linking power to acceleration makes this more challenging than typical A-level mechanics problems. |
| Spec | 6.02a Work done: concept and definition6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| KE gain of carriage is \(\frac{1}{2} \times 6000 \times 5.6^2\) (= 94 080) Work done by tension = KE gain of carriage | B1, M1 | Need not be evaluated at this stage; May just be stated, or can be implied by answer for work done |
| 94 100 (J) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(5.6^2 = 2 \times a \times 250 \Rightarrow a = 0.0627\) \(T \cos\theta = 6000a\) and Work done \(= T\cos\theta \times 250\) \(94 100\) (J) | B1, M1, A1 [3] | Correct acceleration from suvat; Both steps soi |
| Answer | Marks | Guidance |
|---|---|---|
| \(T\cos\theta \times 250 = 94 080\) \(\cos\theta = \frac{12}{13}\) Tension is 408 (N) | B1, A1, A1 [3] | M0 if \(\cos\theta\) taken as 1; oe, eg 0.923...; may occur in (i); Marks may be implied by earlier work if alternative solution used in (i) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = \frac{68000}{8.4}\) (= 8095.23...) \(a = \frac{8095}{16000}\) \(= 0.506\) | B1, M1, A1 [3] | Need not be evaluated; Use of Newton II to find \(a\); Accept 0.51 |
## (i)
KE gain of carriage is $\frac{1}{2} \times 6000 \times 5.6^2$ (= 94 080) Work done by tension = KE gain of carriage | B1, M1 | Need not be evaluated at this stage; May just be stated, or can be implied by answer for work done
94 100 (J) | A1 |
**Alternative solution**
$5.6^2 = 2 \times a \times 250 \Rightarrow a = 0.0627$ $T \cos\theta = 6000a$ and Work done $= T\cos\theta \times 250$ $94 100$ (J) | B1, M1, A1 [3] | Correct acceleration from suvat; Both steps soi
## (ii)
$T\cos\theta \times 250 = 94 080$ $\cos\theta = \frac{12}{13}$ Tension is 408 (N) | B1, A1, A1 [3] | M0 if $\cos\theta$ taken as 1; oe, eg 0.923...; may occur in (i); Marks may be implied by earlier work if alternative solution used in (i)
## (iii)
$F = \frac{68000}{8.4}$ (= 8095.23...) $a = \frac{8095}{16000}$ $= 0.506$ | B1, M1, A1 [3] | Need not be evaluated; Use of Newton II to find $a$; Accept 0.51
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$A$ and $B$ are two long straight parallel horizontal sections of railway track. An engine on track $A$ is attached to a carriage of mass 6000 kg on track $B$ by a light inextensible chain which remains horizontal and taut in the ensuing motion. The chain is 13 m in length and the points of attachment on the engine and carriage are a perpendicular distance of 5 m apart. The engine and carriage start at rest and then the engine accelerates uniformly to a speed of $5.6 \mathrm {~ms} ^ { - 1 }$ while travelling 250 m . It is assumed that any resistance to motion can be ignored.\\
(i) Find the work done on the carriage by the tension in the chain.\\
(ii) Find the magnitude of the tension in the chain.
The mass of the engine is 10000 kg .\\
(iii) At a point further along the track the engine and the carriage are moving at a speed of $8.4 \mathrm {~ms} ^ { - 1 }$ and the power of the engine is 68 kW . Find the acceleration of the engine at this instant.
\hfill \mbox{\textit{OCR FM1 AS 2017 Q4 [9]}}