| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Motion with exponential force |
| Difficulty | Challenging +1.8 This is a Further Maths mechanics question involving exponential force and variable acceleration requiring v dv/dx = a, separation of variables, and integration of exponential functions. Part (a) requires setting up F=ma with exponential resistance, integrating to show a given result, and part (b) requires separating variables again to find x(t). The mathematical techniques are advanced (exponential integration, implicit differentiation) but the setup is relatively standard for FM mechanics. The multi-part structure and need to manipulate exponential expressions elevate this above typical A-level questions. |
| Spec | 6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(4v\frac{dv}{dx} = -(4e^{-x} + 12)e^{-x}\) | B1 | |
| \(\frac{1}{2}v^2 = \frac{1}{2}e^{-2x} + 3e^{-x}(+A)\) | M1 | Expression of the correct form |
| \(v=4\), \(x=0\), \(A = \frac{9}{2}\) | A1 | |
| \(v^2 = e^{-2x} + 6e^{-x} + 9 = (3 + e^{-x})^2\); \(v = 3 + e^{-x} = \frac{1+3e^x}{e^x}\) | A1 | AG. Must see the factorisation. Condone lack of justification for taking positive square root |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dx}{dt} = \frac{1+3e^x}{e^x}\) so \(\int \frac{e^x}{3e^x+1}dx = \int 1\, dt\); \(\frac{1}{3}\ln(3e^x+1) = t(+B)\) | M1* A1 | Integration to obtain ln term. Correct answer with constant of integration |
| \(t=0\), \(x=0\), \(B = \frac{1}{3}\ln 4\); \(3t = \ln\frac{3e^x+1}{4}\) | DM1 | Find the constant and substitute into their general solution |
| \(x = \ln\left(\frac{4}{3}e^{3t} - \frac{1}{3}\right)\) | A1 | OE |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $4v\frac{dv}{dx} = -(4e^{-x} + 12)e^{-x}$ | B1 | |
| $\frac{1}{2}v^2 = \frac{1}{2}e^{-2x} + 3e^{-x}(+A)$ | M1 | Expression of the correct form |
| $v=4$, $x=0$, $A = \frac{9}{2}$ | A1 | |
| $v^2 = e^{-2x} + 6e^{-x} + 9 = (3 + e^{-x})^2$; $v = 3 + e^{-x} = \frac{1+3e^x}{e^x}$ | A1 | AG. Must see the factorisation. Condone lack of justification for taking positive square root |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dx}{dt} = \frac{1+3e^x}{e^x}$ so $\int \frac{e^x}{3e^x+1}dx = \int 1\, dt$; $\frac{1}{3}\ln(3e^x+1) = t(+B)$ | M1* A1 | Integration to obtain ln term. Correct answer with constant of integration |
| $t=0$, $x=0$, $B = \frac{1}{3}\ln 4$; $3t = \ln\frac{3e^x+1}{4}$ | DM1 | Find the constant and substitute into their general solution |
| $x = \ln\left(\frac{4}{3}e^{3t} - \frac{1}{3}\right)$ | A1 | OE |
---5 A particle $P$ of mass 4 kg is moving in a horizontal straight line. At time $t$ s the velocity of $P$ is $v \mathrm {~ms} ^ { - 1 }$ and the displacement of $P$ from a fixed point $O$ on the line is $x \mathrm {~m}$. The only force acting on $P$ is a resistive force of magnitude $\left( 4 \mathrm { e } ^ { - x } + 12 \right) \mathrm { e } ^ { - x } \mathrm {~N}$. When $\mathrm { t } = 0 , \mathrm { x } = 0$ and $v = 4$.
\begin{enumerate}[label=(\alph*)]
\item Show by integration that $\mathrm { v } = \frac { 1 + 3 \mathrm { e } ^ { \mathrm { x } } } { \mathrm { e } ^ { \mathrm { x } } }$.
\item Find an expression for $x$ in terms of $t$.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad8b126c-d739-4e2a-8ce3-7811a61f5876-10_510_889_269_580}\\
$A B$ and $B C$ are two fixed smooth vertical barriers on a smooth horizontal surface, with angle $\mathrm { ABC } = 60 ^ { \circ }$. A particle of mass $m$ is moving with speed $u$ on the surface. The particle strikes $A B$ at an angle $\theta$ with $A B$. It then strikes $B C$ and rebounds at an angle $\beta$ with $B C$ (see diagram). The coefficient of restitution between the particle and each barrier is $e$ and $\tan \theta = 2$.
The kinetic energy of the particle after the first collision is $40 \%$ of its kinetic energy before the first collision.\\
(a) Find the value of $e$.\\
(b) Find the size of angle $\beta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad8b126c-d739-4e2a-8ce3-7811a61f5876-12_965_1059_267_502}
A uniform cylinder with a rough surface and of radius $a$ is fixed with its axis horizontal. Two identical uniform rods $A B$ and $B C$, each of weight $W$ and length $2 a$, are rigidly joined at $B$ with $A B$ perpendicular to $B C$. The rods rest on the cylinder in a vertical plane perpendicular to the axis of the cylinder with $A B$ at an angle $\theta$ to the horizontal. $D$ and $E$ are the midpoints of $A B$ and $B C$ respectively and also the points of contact of the rods with the cylinder (see diagram). The rods are about to slip in a clockwise direction. The coefficient of friction between each rod and the cylinder is $\mu$.
The normal reaction between $A B$ and the cylinder is $R$ and the normal reaction between $B C$ and the cylinder is $N$.\\
(a) Find the ratio $R : N$ in terms of $\mu$.\\
(b) Given that $\mu = \frac { 1 } { 3 }$, find the value of $\tan \theta$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q5 [8]}}