| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2023 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - falling from rest or projected downward |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question on variable force with air resistance proportional to v². Part (a) requires setting up and solving a separable differential equation (2dv/dt = 20 - 0.2v²), which is a routine technique at this level. Part (b) asks for terminal velocity behavior, which follows directly from the solution. The question involves multiple steps but uses well-practiced methods without requiring novel insight—slightly above average difficulty due to the algebraic manipulation and integration involved. |
| Spec | 3.03u Static equilibrium: on rough surfaces6.02h Elastic PE: 1/2 k x^26.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\frac{dv}{dt} = 2g - 0.2v^2\) | B1 | |
| Separate variables and attempt to integrate \(\frac{dv}{0.1(100-v^2)} = dt\) | M1 | Integrate to a ln term of the correct form |
| \(\frac{1}{20}\ln\left(\frac{10+v}{10-v}\right) = 0.1t + c\) | A1 | |
| \(t=0, v=5, \quad c = \frac{1}{20}\ln 3\) | M1 | Use initial condition |
| \(2t = \ln\frac{10+v}{3(10-v)},\quad e^{2t} = \frac{10+v}{3(10-v)}\) | M1 | Rearrange, removing ln |
| \(v = \frac{30-10e^{-2t}}{3+e^{-2t}}\) | A1 | AEF |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v \to 10\) | B1FT | FT from expression of correct form |
| Total: 1 |
## Question 2(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\frac{dv}{dt} = 2g - 0.2v^2$ | B1 | |
| Separate variables and attempt to integrate $\frac{dv}{0.1(100-v^2)} = dt$ | M1 | Integrate to a ln term of the correct form |
| $\frac{1}{20}\ln\left(\frac{10+v}{10-v}\right) = 0.1t + c$ | A1 | |
| $t=0, v=5, \quad c = \frac{1}{20}\ln 3$ | M1 | Use initial condition |
| $2t = \ln\frac{10+v}{3(10-v)},\quad e^{2t} = \frac{10+v}{3(10-v)}$ | M1 | Rearrange, removing ln |
| $v = \frac{30-10e^{-2t}}{3+e^{-2t}}$ | A1 | AEF |
| **Total: 6** | | |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v \to 10$ | B1FT | FT from expression of correct form |
| **Total: 1** | | |2 A ball of mass 2 kg is projected vertically downwards with speed $5 \mathrm {~ms} ^ { - 1 }$ through a liquid. At time $t \mathrm {~s}$ after projection, the velocity of the ball is $v \mathrm {~ms} ^ { - 1 }$ and its displacement from its starting point is $x \mathrm {~m}$. The forces acting on the ball are its weight and a resistive force of magnitude $0.2 v ^ { 2 } \mathrm {~N}$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $t$.
\item Deduce what happens to $v$ for large values of $t$.\\
\includegraphics[max width=\textwidth, alt={}, center]{e7091f6c-af72-49f3-b825-cdce9fb2c06f-06_803_652_251_703}
A uniform square lamina of side $2 a$ and weight $W$ is suspended from a light inextensible string attached to the midpoint $E$ of the side $A B$. The other end of the string is attached to a fixed point $P$ on a rough vertical wall. The vertex $B$ of the lamina is in contact with the wall. The string $E P$ is perpendicular to the side $A B$ and makes an angle $\theta$ with the wall (see diagram). The string and the lamina are in a vertical plane perpendicular to the wall. The coefficient of friction between the wall and the lamina is $\frac { 1 } { 2 }$.
Given that the vertex $B$ is about to slip up the wall, find the value of $\tan \theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{e7091f6c-af72-49f3-b825-cdce9fb2c06f-08_581_576_269_731}
A light elastic string has natural length $8 a$ and modulus of elasticity $5 m g$. A particle $P$ of mass $m$ is attached to the midpoint of the string. The ends of the string are attached to points $A$ and $B$ which are a distance $12 a$ apart on a smooth horizontal table. The particle $P$ is held on the table so that $A P = B P = L$ (see diagram). The particle $P$ is released from rest. When $P$ is at the midpoint of $A B$ it has speed $\sqrt { 80 a g }$.\\
(a) Find $L$ in terms of $a$.\\
(b) Find the initial acceleration of $P$ in terms of $g$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q2 [7]}}