| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Particle on smooth curved surface |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem with two standard parts: (i) uses conservation of mechanical energy on a smooth surface between two known heights, and (ii) applies work-energy theorem with constant friction over half the distance. Both require direct application of standard formulae with minimal problem-solving insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}mv_A^2 = \frac{1}{2}mv_C^2 - mg \times 2.7\) and \(\frac{1}{2}mv_C^2 = \frac{1}{2}mv_A^2 - mg \times 3\) | M1 | For using the principle of conservation of energy from A to B or from A to C |
| \([v_A^2 = 8^2 - 20 \times 2.7, v_C^2 = 8^2 - 20 \times 3]\) | M1 | For substituting for \(v_A\) to find \(v_B - v_C\) |
| Loss of speed = 10\(^{-1}\) = 2 = 1.16 ms\(^{-1}\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Work done = \(\frac{1}{2} 0.2 \times 2^2 + 0.2 \times g \times 3\) (= 6.4) | M1 | For using: WD against friction (C to D) = KE at C + loss of PE (C to D) |
| A1 | ||
| M1 | For using WD against friction (M to D) = KE at M + loss of PE (M to D) | |
| \(\frac{1}{2}(0.4 + 6) = \frac{1}{2} 0.2v_M^2 + 0.2g \times 1.5\) | A1 | |
| Speed at midpoint is 1.41 ms\(^{-1}\) | A1 | 5 |
**(i)**
$\frac{1}{2}mv_A^2 = \frac{1}{2}mv_C^2 - mg \times 2.7$ and $\frac{1}{2}mv_C^2 = \frac{1}{2}mv_A^2 - mg \times 3$ | M1 | For using the principle of conservation of energy from A to B or from A to C
$[v_A^2 = 8^2 - 20 \times 2.7, v_C^2 = 8^2 - 20 \times 3]$ | M1 | For substituting for $v_A$ to find $v_B - v_C$
Loss of speed = 10$^{-1}$ = 2 = 1.16 ms$^{-1}$ | A1 | 4
**(ii)**
Work done = $\frac{1}{2} 0.2 \times 2^2 + 0.2 \times g \times 3$ (= 6.4) | M1 | For using: WD against friction (C to D) = KE at C + loss of PE (C to D)
| A1 |
| M1 | For using WD against friction (M to D) = KE at M + loss of PE (M to D)
$\frac{1}{2}(0.4 + 6) = \frac{1}{2} 0.2v_M^2 + 0.2g \times 1.5$ | A1 |
Speed at midpoint is 1.41 ms$^{-1}$ | A1 | 56\\
\includegraphics[max width=\textwidth, alt={}, center]{2bb3c9bb-60f0-440d-a148-b4db3478ca31-3_382_1451_797_347}
The diagram shows the vertical cross-section $A B C D$ of a surface. $B C$ is a circular arc, and $A B$ and $C D$ are tangents to $B C$ at $B$ and $C$ respectively. $A$ and $D$ are at the same horizontal level, and $B$ and $C$ are at heights 2.7 m and 3.0 m respectively above the level of $A$ and $D$. A particle $P$ of mass 0.2 kg is given a velocity of $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at $A$, in the direction of $A B$ (see diagram). The parts of the surface containing $A B$ and $B C$ are smooth.\\
(i) Find the decrease in the speed of $P$ as $P$ moves along the surface from $B$ to $C$.
The part of the surface containing $C D$ exerts a constant frictional force on $P$, as it moves from $C$ to $D$, and $P$ comes to rest as it reaches $D$.\\
(ii) Find the speed of $P$ when it is at the mid-point of $C D$.
\hfill \mbox{\textit{CAIE M1 2012 Q6 [9]}}