| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2004 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Motion on rough inclined plane |
| Difficulty | Moderate -0.8 This is a straightforward application of energy conservation and work-energy principles with standard values. Part (i) requires only mgh = KE, while part (ii) adds friction work (μmg cos θ × distance), both being routine textbook exercises with no problem-solving insight needed. |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(KE = 0.2g(0.7)\) | M1 | For using \(KE = PE\) lost and \(PE_{lost} = mgh\) |
| Kinetic energy is 1.4 J | A1 (×2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = 0.2\times10\times\cos16.3°\) | B1 | 1.92 |
| \(F = 0.288\) N | B1 ft | From \(0.15R\) (may be implied by subsequent exact value 0.72, 1.36 or 0.68) |
| \(WD = 0.72\) J or \(a = 1.36\) or resultant downward force \(= 0.272\) N | B1 ft | From \(2.5F\) or from \(0.2a = 0.2\times10\times(7/25) - F\) (may be implied by subsequent exact value 0.68) |
| \(KE = 1.4 - 0.72\) or \(KE = \frac{1}{2}(0.2)(2\times1.36\times2.5)\) or \(0.272\times2.5\) | M1 | For using \(KE = PE_{lost} - WD\) or \(KE = \frac{1}{2}mv^2\) and \(v^2 = 2as\) or \(KE =\) resultant downward force \(\times 2.5\) |
| Kinetic energy is 0.68 J | A1 ft (×5) |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $KE = 0.2g(0.7)$ | M1 | For using $KE = PE$ lost and $PE_{lost} = mgh$ |
| Kinetic energy is 1.4 J | A1 (×2) | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = 0.2\times10\times\cos16.3°$ | B1 | 1.92 |
| $F = 0.288$ N | B1 ft | From $0.15R$ (may be implied by subsequent exact value 0.72, 1.36 or 0.68) |
| $WD = 0.72$ J or $a = 1.36$ or resultant downward force $= 0.272$ N | B1 ft | From $2.5F$ or from $0.2a = 0.2\times10\times(7/25) - F$ (may be implied by subsequent exact value 0.68) |
| $KE = 1.4 - 0.72$ or $KE = \frac{1}{2}(0.2)(2\times1.36\times2.5)$ or $0.272\times2.5$ | M1 | For using $KE = PE_{lost} - WD$ or $KE = \frac{1}{2}mv^2$ and $v^2 = 2as$ or $KE =$ resultant downward force $\times 2.5$ |
| Kinetic energy is 0.68 J | A1 ft (×5) | |
---4 The top of an inclined plane is at a height of 0.7 m above the bottom. A block of mass 0.2 kg is released from rest at the top of the plane and slides a distance of 2.5 m to the bottom. Find the kinetic energy of the block when it reaches the bottom of the plane in each of the following cases:\\
(i) the plane is smooth,\\
(ii) the coefficient of friction between the plane and the block is 0.15 .
\hfill \mbox{\textit{CAIE M1 2004 Q4 [7]}}