| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force up incline, find KE/PE changes as sub-parts |
| Difficulty | Moderate -0.3 This is a straightforward application of the work-energy principle with standard formulas (KE = ½mv², PE = mgh) and clear given values. Parts (i) and (ii) are direct substitutions, while part (iii) requires setting up work done against resistance but involves only basic algebra. Slightly easier than average due to minimal problem-solving required. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \cdot 70 \cdot 0.4^2\) | M1 | |
| \(560\) J | A1 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(70 \times 9.8 \times 6\) | M1 | |
| \(4120\) | A1 2 | \(4116\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(60d\) | B1 | |
| \(8000 = 560 + 4120 + 60d\) | M1 | 4 terms |
| A1\(\checkmark\) | \(\checkmark\) their KE and PE | |
| \(55.4\) m | A1 4 |
## Question 5(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \cdot 70 \cdot 0.4^2$ | M1 | |
| $560$ J | A1 **2** | |
## Question 5(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $70 \times 9.8 \times 6$ | M1 | |
| $4120$ | A1 **2** | $4116$ |
## Question 5(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $60d$ | B1 | |
| $8000 = 560 + 4120 + 60d$ | M1 | 4 terms |
| | A1$\checkmark$ | $\checkmark$ their KE and PE |
| $55.4$ m | A1 **4** | |
---5 A cyclist and her bicycle have a combined mass of 70 kg . The cyclist ascends a straight hill $A B$ of constant slope, starting from rest at $A$ and reaching a speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $B$. The level of $B$ is 6 m above the level of $A$. For the cyclist's motion from $A$ to $B$, find\\
(i) the increase in kinetic energy,\\
(ii) the increase in gravitational potential energy.
During the ascent the resistance to motion is constant and has magnitude 60 N . The work done by the cyclist in moving from $A$ to $B$ is 8000 J .\\
(iii) Calculate the distance $A B$.
\hfill \mbox{\textit{OCR M2 2007 Q5 [8]}}