| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2002 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv - horizontal motion |
| Difficulty | Standard +0.3 This is a standard M2 variable force question with air resistance proportional to velocity. Part (i) requires setting up F=ma with v dv/dx and integrating, (ii) involves separating variables to find x(t), and (iii) requires taking a limit as tââ. All techniques are routine for M2 students who have practiced resisted motion problems, though the multi-step nature and algebraic manipulation add modest challenge above pure recall. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Uses Newton's 2nd law and \(a = v\frac{dv}{dx}\), and attempts to integrate | M1* | |
| \([(1/10)v\frac{dv}{dx} = -v/200]\) | ||
| \(v = -x/20\) \((+C)\) | A1 | |
| Uses \(v(0) = 5\) to find \(C\) | M1 dep | |
| Obtains \(v = -x/20 + 5\) | A1 | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \(v = dx/dt\), separates the variables and integrates | M1* | |
| \([\int\frac{1}{100-x}dx = \int\frac{-1}{20}dt]\) | ||
| Obtains \(\ln(100-x) = -t/20(+C)\) | A1 | |
| Uses \(x = 0\) when \(t = 0\) to obtain \(t = 20[\ln 100 - \ln(100-x)]\) | A1 ft | |
| ft only if the term in \(x\) is logarithmic | ||
| For taking anti-logarithms throughout the equation | M1 dep | |
| \([100 - x = 100e^{-t/20}]\) | ||
| Obtains \(x = 100(1-e^{-t/20})\) | A1 | 5 marks |
| Answer | Marks |
|---|---|
| Uses Newton's 2nd law with \(a = dv/dt\), separates the variables and integrates | M1* |
| \([\int\frac{1}{v}dv = -\int\frac{1}{20}dt]\) | |
| Obtains \(\ln v = -t/20\) \((+C)\) | A1 |
| Uses \(v = 5\) when \(t = 0\) to obtain \(t = 20[\ln 5 - \ln v]\) | A1 ft |
| ft only if the term in \(v\) is logarithmic | |
| For taking anti-logarithms throughout the equation | M1 dep |
| \([v = 5e^{-t/20}]\) | |
| Uses \(v = dx/dt\) and integrates | M1* |
| \([x = \int 5e^{-t/20}dt]\) | |
| Obtains \(x = -100e^{-t/20}(+C)\) | A1 |
| Uses \(x = 0\) when \(t = 0\) to obtain \(x = 100(1-e^{-t/20})\) | A1 |
| Eliminates the exponential term from \(x = 100(1-e^{-t/20})\) and \(v = 5e^{-t/20}\) to obtain an equation in \(x\) and \(v\) | M1 dep |
| \([x = 100(1-v/5)]\) | |
| Obtains \(v = -x/20 + 5\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 100(1-e^{-t/20})\) and \(e^{-t/20}\) is \(+ve\) for all \(t \Rightarrow x < 100\) | B1 | 1 mark |
**(i)**
Uses Newton's 2nd law and $a = v\frac{dv}{dx}$, and attempts to integrate | M1* |
$[(1/10)v\frac{dv}{dx} = -v/200]$ | |
$v = -x/20$ $(+C)$ | A1 |
Uses $v(0) = 5$ to find $C$ | M1 dep |
Obtains $v = -x/20 + 5$ | A1 | 4 marks |
**(ii)**
Uses $v = dx/dt$, separates the variables and integrates | M1* |
$[\int\frac{1}{100-x}dx = \int\frac{-1}{20}dt]$ | |
Obtains $\ln(100-x) = -t/20(+C)$ | A1 |
Uses $x = 0$ when $t = 0$ to obtain $t = 20[\ln 100 - \ln(100-x)]$ | A1 ft |
ft only if the term in $x$ is logarithmic | |
For taking anti-logarithms throughout the equation | M1 dep |
$[100 - x = 100e^{-t/20}]$ | |
Obtains $x = 100(1-e^{-t/20})$ | A1 | 5 marks |
**Alternatively for the above 9 marks:**
Uses Newton's 2nd law with $a = dv/dt$, separates the variables and integrates | M1* |
$[\int\frac{1}{v}dv = -\int\frac{1}{20}dt]$ | |
Obtains $\ln v = -t/20$ $(+C)$ | A1 |
Uses $v = 5$ when $t = 0$ to obtain $t = 20[\ln 5 - \ln v]$ | A1 ft |
ft only if the term in $v$ is logarithmic | |
For taking anti-logarithms throughout the equation | M1 dep |
$[v = 5e^{-t/20}]$ | |
Uses $v = dx/dt$ and integrates | M1* |
$[x = \int 5e^{-t/20}dt]$ | |
Obtains $x = -100e^{-t/20}(+C)$ | A1 |
Uses $x = 0$ when $t = 0$ to obtain $x = 100(1-e^{-t/20})$ | A1 |
Eliminates the exponential term from $x = 100(1-e^{-t/20})$ and $v = 5e^{-t/20}$ to obtain an equation in $x$ and $v$ | M1 dep |
$[x = 100(1-v/5)]$ | |
Obtains $v = -x/20 + 5$ | A1 |
**(iii)**
$x = 100(1-e^{-t/20})$ and $e^{-t/20}$ is $+ve$ for all $t \Rightarrow x < 100$ | B1 | 1 mark |
*NB: In (i) it is solved as in 3c, where the DE is solved using the alternative method; the 5 marks awarded for (ii) from the alternative method are M1* (do), A1 (no dep), M1 (do_dp), M1* (when $v = dx/dt$ and integrate or sub for V (from do); (A1) A1 (no dep) (d_J).*6\\
\includegraphics[max width=\textwidth, alt={}, center]{fcf239a6-6558-43ec-b404-70aa349af6a9-3_177_880_1658_635}
A particle $P$ of mass $\frac { 1 } { 10 } \mathrm {~kg}$ travels in a straight line on a smooth horizontal surface. It passes through the fixed point $O$ with velocity $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t = 0$. After $t$ seconds its displacement from $O$ is $x \mathrm {~m}$ and its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 } . P$ is subject to a single force of magnitude $\frac { v } { 200 } \mathrm {~N}$ which acts in a direction opposite to the motion (see diagram).\\
(i) Find an expression for $v$ in terms of $x$.\\
(ii) Find an expression for $x$ in terms of $t$.\\
(iii) Show that the value of $x$ is less than 100 for all values of $t$.\\
(i)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fcf239a6-6558-43ec-b404-70aa349af6a9-4_477_684_264_774}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
Fig. 1 shows the cross section through the centre of mass $C$ of a uniform L-shaped prism. $C$ is $x \mathrm {~cm}$ from $O Y$ and $y \mathrm {~cm}$ from $O X$. Find the values of $x$ and $y$.\\
(ii)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fcf239a6-6558-43ec-b404-70aa349af6a9-4_257_428_1064_902}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The prism is placed on a rough plane with $O X$ in contact with the plane. The plane is tilted from the horizontal so that $O X$ lies along a line of greatest slope, as shown in Fig. 2. When the angle of inclination of the plane is sufficiently great the prism starts to slide (without toppling). Show that the coefficient of friction between the prism and the plane is less than $\frac { 7 } { 5 }$.\\
(iii)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{fcf239a6-6558-43ec-b404-70aa349af6a9-4_303_414_1710_909}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
The prism is now placed on a rough plane with $O Y$ in contact with the plane. The plane is tilted from the horizontal so that $O Y$ lies along a line of greatest slope, as shown in Fig. 3. When the angle of inclination of the plane is sufficiently great the prism starts to topple (without sliding). Find the least possible value of the coefficient of friction between the prism and the plane. [3]
\hfill \mbox{\textit{CAIE M2 2002 Q6 [10]}}