| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 This is a straightforward energy method question requiring standard application of KE = ½mv², work-energy principle, and resolving forces on an incline. Part (i) is direct substitution, part (ii) uses work done by gravity equals KE change, and part (iii) requires resolving a horizontal force and applying equilibrium—all standard M1 techniques with no novel insight required. Slightly above average due to the three-part structure and the horizontal force resolution in part (iii). |
| Spec | 3.03v Motion on rough surface: including inclined planes6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([\frac{1}{2} \times 12(7^2 - 3^2)]\) | M1 | For using \(KE = \frac{1}{2}m(v_B^2 - v_A^2)\) |
| Increase is 240 J | A1 | 2 marks total |
| (ii) \(\quad\) | M1 | For using \(mgh =\) KE gain |
| \(12g \times AB\sin 10° = 240\) | A1ft | |
| Distance is 11.5 m | A1 | 3 marks total |
| (iii) \(\quad\) | M1 | For using \(F(AB)\cos 10° =\) PE gain or for using Newton's 2nd law with \(a = 0\) |
| \(F \times 11.5\cos 10° = 240\) or \(F\cos 10° - 12g\sin 10° = 0\) | A1ft | |
| Magnitude is 21.2 N | A1 | 3 marks total |
**(i)** $[\frac{1}{2} \times 12(7^2 - 3^2)]$ | M1 | For using $KE = \frac{1}{2}m(v_B^2 - v_A^2)$
Increase is 240 J | A1 | 2 marks total
**(ii)** $\quad$ | M1 | For using $mgh =$ KE gain
$12g \times AB\sin 10° = 240$ | A1ft |
Distance is 11.5 m | A1 | 3 marks total | SR for candidates who avoid 'hence' (max 2/3): For using Newton's Second Law and $v^2 = u^2 + 2as$: $[12g\sin 10° = 12a$, $7^2 = 3^2 + 2(g\sin 10° \times AB)]$ | M1 |, 11.5 m | A1 |
**(iii)** $\quad$ | M1 | For using $F(AB)\cos 10° =$ PE gain or for using Newton's 2nd law with $a = 0$
$F \times 11.5\cos 10° = 240$ or $F\cos 10° - 12g\sin 10° = 0$ | A1ft |
Magnitude is 21.2 N | A1 | 3 marks total
---5 An object of mass 12 kg slides down a line of greatest slope of a smooth plane inclined at $10 ^ { \circ }$ to the horizontal. The object passes through points $A$ and $B$ with speeds $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively.\\
(i) Find the increase in kinetic energy of the object as it moves from $A$ to $B$.\\
(ii) Hence find the distance $A B$, assuming there is no resisting force acting on the object.
The object is now pushed up the plane from $B$ to $A$, with constant speed, by a horizontal force.\\
(iii) Find the magnitude of this force.
\hfill \mbox{\textit{CAIE M1 2012 Q5 [8]}}