| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Variable force with integration (force as function of position) |
| Difficulty | Challenging +1.2 This is a multi-step mechanics problem requiring application of Newton's second law with variable forces, energy methods, and integration. Part (i) is a standard 'show that' derivation using F=ma in the form v(dv/dx), while part (ii) requires integrating the differential equation and applying boundary conditions. The concepts are standard M2 material (elastic strings, variable resistance), but the algebraic manipulation and integration of the resulting expression requires careful work across multiple steps, placing it moderately above average difficulty. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | T = 16(1.6 – 0.8 – x)/0.8 ( = 16 – 20x) | B1 |
| 0.5vdv/dx = 16(1.6 – 0.8 – x)/0.8 – 48x2 | M1 | Use Newton's Second Law horizontally |
| vdv/dx = 32 – 40x – 48x2 AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | 48x2 + 40x – 32 = 0 | M1 |
| x = 0.5 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (v2/2 = 32x – 40x2/2 – 48x3/3 + c) | M1 | Attempt to integrate the equation from part (i) |
| 4.52/2 = 32×0.5 – 20 ×0.52 – 16×0.53 + c, c = 1.125 | M1 | Substitute x = 0.5 , v = 4.5 to find c |
| v = 1.5 | A1 | Use x = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | T = 16(1.6 – 0.8 – x)/0.8 ( = 16 – 20x) | B1 | Use T = λx/L
0.5vdv/dx = 16(1.6 – 0.8 – x)/0.8 – 48x2 | M1 | Use Newton's Second Law horizontally
vdv/dx = 32 – 40x – 48x2 AG | A1
3
Question | Answer | Marks | Guidance
--- 4(ii) ---
4(ii) | 48x2 + 40x – 32 = 0 | M1 | Put acceleration = 0 for maximum velocity
x = 0.5 | A1
∫vdv = ∫(32 – 40x – 48x2)dx
(v2/2 = 32x – 40x2/2 – 48x3/3 + c) | M1 | Attempt to integrate the equation from part (i)
4.52/2 = 32×0.5 – 20 ×0.52 – 16×0.53 + c, c = 1.125 | M1 | Substitute x = 0.5 , v = 4.5 to find c
v = 1.5 | A1 | Use x = 0
5
Question | Answer | Marks | Guidance\includegraphics{figure_4}
A particle $P$ of mass $0.5\text{ kg}$ is projected along a smooth horizontal surface towards a fixed point $A$. Initially $P$ is at a point $O$ on the surface, and after projection, $P$ has a displacement from $O$ of $x\text{ m}$ and velocity $v\text{ m s}^{-1}$. The particle $P$ is connected to $A$ by a light elastic string of natural length $0.8\text{ m}$ and modulus of elasticity $16\text{ N}$. The distance $OA$ is $1.6\text{ m}$ (see diagram). The motion of $P$ is resisted by a force of magnitude $24x^2\text{ N}$.
\begin{enumerate}[label=(\roman*)]
\item Show that $v\frac{\text{d}v}{\text{d}x} = 32 - 40x - 48x^2$ while $P$ is in motion and the string is stretched. [3]
\item The maximum value of $v$ is $4.5$.
Find the initial value of $v$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2018 Q4 [8]}}