| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find constant speed |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring students to find the resistance constant from maximum speed conditions (P=Fv), then apply F=ma with power equation to find speed at given acceleration. It involves routine application of well-practiced formulas with straightforward algebra, making it slightly easier than average. |
| Spec | 6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P = Fv \Rightarrow 12000 = D \times v_{\max}\) | B1 | AO 3.1b; \(P = Fv\) used correctly to link driving force \(D\) and maximum speed; could be embedded in \(F = ma\); or \(\frac{12000}{10}\) seen as driving force |
| At maximum speed \(D - R = 0\) so \(P/v_{\max} - kv_{\max} = 0 \Rightarrow \frac{12000}{10} - 10k = 0 \Rightarrow k = 120\) | M1, A1 | AO 1.1; statement of \(F = ma\) with \(a = 0\), \(R = kv_{\max}\) used; \(v_{\max} = 10\), \(P = 12000\); may be seen embedded |
| \(D - R = ma \Rightarrow \frac{12000}{v} -\) ``\(120\)''\(v = 360 \times 1.5\) | M1* | AO 3.1b; statement of \(F = ma\) with \(P = Dv\), \(R = kv\), \(m = 360\) and \(a = 1.5\) used; may see \(v = \frac{12000}{540+120v}\) using \(v = \frac{P}{D}\) |
| \(\Rightarrow 12000 - 120v^2 = 540v\ [\Rightarrow 2v^2 + 9v - 200 = 0]\) | M1dep | AO 1.1; rearranging to 3-term quadratic |
| \((2v+25)(v-8) = 0 \Rightarrow v = 8\) [or \(v = -12.5\)] | A1 | AO 1.1; both values correct if present; fully correct solution implies previous M marks |
| But speed must be positive, so speed is \(8\ (\text{ms}^{-1})\) | A1FT | AO 3.2a; both values seen and rogue solution explicitly rejected with valid reason; FT solutions to their quadratic if negative solution validly rejected. Insufficient explanations: "going forward"; "cannot be negative as it is accelerating"/"acceleration is in the positive direction"; "a cannot be positive when v is negative, because power is positive" |
# Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P = Fv \Rightarrow 12000 = D \times v_{\max}$ | B1 | AO 3.1b; $P = Fv$ used correctly to link driving force $D$ and maximum speed; could be embedded in $F = ma$; or $\frac{12000}{10}$ seen as driving force |
| At maximum speed $D - R = 0$ so $P/v_{\max} - kv_{\max} = 0 \Rightarrow \frac{12000}{10} - 10k = 0 \Rightarrow k = 120$ | M1, A1 | AO 1.1; statement of $F = ma$ with $a = 0$, $R = kv_{\max}$ used; $v_{\max} = 10$, $P = 12000$; may be seen embedded |
| $D - R = ma \Rightarrow \frac{12000}{v} -$ ``$120$''$v = 360 \times 1.5$ | M1* | AO 3.1b; statement of $F = ma$ with $P = Dv$, $R = kv$, $m = 360$ and $a = 1.5$ used; may see $v = \frac{12000}{540+120v}$ using $v = \frac{P}{D}$ |
| $\Rightarrow 12000 - 120v^2 = 540v\ [\Rightarrow 2v^2 + 9v - 200 = 0]$ | M1dep | AO 1.1; rearranging to 3-term quadratic |
| $(2v+25)(v-8) = 0 \Rightarrow v = 8$ [or $v = -12.5$] | A1 | AO 1.1; both values correct if present; fully correct solution implies previous M marks |
| But speed must be positive, so speed is $8\ (\text{ms}^{-1})$ | A1FT | AO 3.2a; both values seen and rogue solution explicitly rejected with valid reason; FT solutions to their quadratic if negative solution validly rejected. Insufficient explanations: "going forward"; "cannot be negative as it is accelerating"/"acceleration is in the positive direction"; "a cannot be positive when v is negative, because power is positive" |
---6 A motorbike and its rider, together denoted by $M$, have a combined mass of 360 kg . The resistive force experienced by $M$ when it is in motion is modelled as being proportional to the speed it is moving at. All motion of $M$ is on a straight horizontal road.
It is found that with the engine of the motorbike working at a rate of 12 kW , the maximum constant speed that $M$ can move at is $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Determine the speed of $M$ such that with the engine working at a rate of 12 kW the acceleration of $M$ is $1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\hfill \mbox{\textit{OCR Further Mechanics AS 2024 Q6 [7]}}