| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Rough inclined plane work-energy |
| Difficulty | Standard +0.3 This is a standard M2 work-energy problem requiring application of the work-energy principle with multiple forces. Part (i) uses energy conservation (KE change + PE change + work against friction = work by pulling force), while part (ii) resolves the pulling force. The calculations are straightforward with clear given values and standard techniques, making it slightly easier than average for M2. |
| Spec | 3.03d Newton's second law: 2D vectors3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component6.02c Work by variable force: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}x80x5^2\) or \(\frac{1}{2}x80x2^2\) | B1 | either KE |
| \(70 \times 25\) | B1 | |
| \(80x9.8x25\sin20°\) | B1 | |
| \(WD=\frac{1}{2}x80x5^2-\frac{1}{2}x80x2^2+70x25+80x9.8x25\sin20°\) | M1 | 4 parts |
| \(9290\) | A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\cos30°x25\) | B1 | or a=0.42 |
| \(P\cos30°.25=9290 / P\cos30°-70-80x9.8\sin20°=80a\) | M1 | |
| \(P = 429 /\text{if } P\text{ound } 1° \text{ then } P\cos30°x25=9290\) ok | A1 | 3 |
**(i)**
$\frac{1}{2}x80x5^2$ or $\frac{1}{2}x80x2^2$ | B1 | either KE | 1000/160
$70 \times 25$ | B1 | | 1750
$80x9.8x25\sin20°$ | B1 | | 6703.6
$WD=\frac{1}{2}x80x5^2-\frac{1}{2}x80x2^2+70x25+80x9.8x25\sin20°$ | M1 | 4 parts
$9290$ | A1 | 5 |
**(ii)**
$P\cos30°x25$ | B1 | or a=0.42
$P\cos30°.25=9290 / P\cos30°-70-80x9.8\sin20°=80a$ | M1 |
$P = 429 /\text{if } P\text{ound } 1° \text{ then } P\cos30°x25=9290$ ok | A1 | 3 | 8 |4 A skier of mass 80 kg is pulled up a slope which makes an angle of $20 ^ { \circ }$ with the horizontal. The skier is subject to a constant frictional force of magnitude 70 N . The speed of the skier increases from $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the point $A$ to $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the point $B$, and the distance $A B$ is 25 m .\\
(i) By modelling the skier as a small object, calculate the work done by the pulling force as the skier moves from $A$ to $B$.\\
(ii)\\
\includegraphics[max width=\textwidth, alt={}, center]{1fbb3693-0beb-47c8-800f-50041f105699-2_451_1019_1425_603}
It is given that the pulling force has constant magnitude $P \mathrm {~N}$, and that it acts at a constant angle of $30 ^ { \circ }$ above the slope (see diagram). Calculate $P$.
\hfill \mbox{\textit{OCR M2 2007 Q4 [8]}}