| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Moderate -0.3 This is a straightforward circular motion problem requiring application of Newton's second law in the horizontal plane (F = mv²/r) and vertical equilibrium. The setup is clearly described, the radius and speed are given directly, and students need only resolve forces and apply standard formulas. It's slightly easier than average because it requires no energy considerations, no variable forces, and the geometry is simple with all values provided. |
| Spec | 3.03d Newton's second law: 2D vectors6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5a = 1 + 0.3x^2 - 8e^{-x}\) | M1 | For using Newton's second law |
| \(v(dv/dx) = 2 + 0.6x^2 - 16e^{-x}\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For separating variables and integrating | |
| \(v^2/2 = 2x + 0.2x^3 + 16e^{-x}\) (+c) | A1 | |
| M1 | For using \(v(0) = 6\) | |
| \(v^2/2 = 2x + 0.2x^3 + 16e^{-x} + 2\) | A1 | |
| Velocity is \(5.33 \text{ ms}^{-1}\) | A1 | [5] |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5a = 1 + 0.3x^2 - 8e^{-x}$ | M1 | For using Newton's second law |
| $v(dv/dx) = 2 + 0.6x^2 - 16e^{-x}$ | A1 | **[2]** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For separating variables and integrating |
| $v^2/2 = 2x + 0.2x^3 + 16e^{-x}$ (+c) | A1 | |
| | M1 | For using $v(0) = 6$ |
| $v^2/2 = 2x + 0.2x^3 + 16e^{-x} + 2$ | A1 | |
| Velocity is $5.33 \text{ ms}^{-1}$ | A1 | **[5]** |
---3 A particle $P$ of mass 0.5 kg moves along the $x$-axis on a horizontal surface. When the displacement of $P$ from the origin $O$ is $x \mathrm {~m}$ the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in the positive $x$-direction. Two horizontal forces act on $P$; one force has magnitude $\left( 1 + 0.3 x ^ { 2 } \right) \mathrm { N }$ and acts in the positive $x$-direction, and the other force has magnitude $8 \mathrm { e } ^ { - x } \mathrm {~N}$ and acts in the negative $x$-direction.\\
(i) Show that $v \frac { \mathrm {~d} v } { \mathrm {~d} x } = 2 + 0.6 x ^ { 2 } - 16 \mathrm { e } ^ { - x }$.\\
(ii) The velocity of $P$ as it passes through $O$ is $6 \mathrm {~ms} ^ { - 1 }$. Find the velocity of $P$ when $x = 3$.\\
(i)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5109244c-3062-4f5f-9277-fc6b5b28f2d4-3_259_745_278_740}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A small sphere $A$ of mass 0.15 kg is moving inside a fixed smooth hollow cylinder whose axis is vertical. $A$ moves with constant speed $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a horizontal circle of radius 0.35 m , and is continuously in contact with both the plane base and the curved surface of the cylinder. Fig. 1 shows a vertical cross-section of the cylinder through its axis. Find the magnitude of the force exerted on $A$ by\\
\hfill \mbox{\textit{CAIE M2 2008 Q3 [7]}}