| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | November |
| Marks | 7 |
| 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 mechanics formulas. Part (i) requires direct substitution into KE and PE formulas, part (ii) uses energy conservation with given resistance, and part (iii) involves resolving forces. All steps are routine for M1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([\frac{1}{2}20(2.5^2 - 1.5^2), 20x10x10\sin 4.5°]\) | M1 | For using KE loss = \(\frac{1}{2}m(u^2 - v^2)\) or PE gain = \(mg(L\sin\alpha)\) |
| KE loss = 40 J or PE gain = 157J | A1 | |
| PE gain = 157 J or KE loss = 40J | B1 | [3] |
| (ii) \([WD = 157 - 40 + 50]\) | M1 | For using WD by pulling force = PE gain – KE loss + WD against resistance |
| Work done is 167J | A1ft | [2] ft incorrect PE gain + 10, even if –ve |
| (iii) \([167 = Fx10\cos15°]\) | M1 | For using WD = \(F L\cos 15°\) |
| Magnitude is 17.3 N | A1ft | [2] |
| Answer | Marks |
|---|---|
| For magnitude is 17.3 N from \(F\cos 15° - 20g\sin4.5° - 50/10 = 20 \times (-0.2)\) | B1 |
**(i)** $[\frac{1}{2}20(2.5^2 - 1.5^2), 20x10x10\sin 4.5°]$ | M1 | For using KE loss = $\frac{1}{2}m(u^2 - v^2)$ or PE gain = $mg(L\sin\alpha)$
KE loss = 40 J or PE gain = 157J | A1 |
PE gain = 157 J or KE loss = 40J | B1 | [3]
**(ii)** $[WD = 157 - 40 + 50]$ | M1 | For using WD by pulling force = PE gain – KE loss + WD against resistance
Work done is 167J | A1ft | [2] ft incorrect PE gain + 10, even if –ve
**(iii)** $[167 = Fx10\cos15°]$ | M1 | For using WD = $F L\cos 15°$
Magnitude is 17.3 N | A1ft | [2]
**SR(max. 1/2)** for candidates who (implicitly) make the unjustifiable assumption that acceleration is constant and apply Newton's second law
For magnitude is 17.3 N from $F\cos 15° - 20g\sin4.5° - 50/10 = 20 \times (-0.2)$ | B1 |
---4 A block of mass 20 kg is pulled from the bottom to the top of a slope. The slope has length 10 m and is inclined at $4.5 ^ { \circ }$ to the horizontal. The speed of the block is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the bottom of the slope and $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the top of the slope.\\
(i) Find the loss of kinetic energy and the gain in potential energy of the block.\\
(ii) Given that the work done against the resistance to motion is 50 J , find the work done by the pulling force acting on the block.\\
(iii) Given also that the pulling force is constant and acts at an angle of $15 ^ { \circ }$ upwards from the slope, find its magnitude.
\hfill \mbox{\textit{CAIE M1 2010 Q4 [7]}}