Challenging +1.2 This is a standard Further Maths mechanics question on variable force with resistance proportional to v². Part (a) requires straightforward application of Newton's second law (showing a given result). Part (b) involves separating variables and integrating a rational function with partial fractions - a routine technique at this level. Part (c) uses the chain rule to form dv/dx and integrate to find distance. While it requires multiple techniques and careful algebra, these are all standard methods for FM mechanics with no novel insight required.
An object, of mass 1.8 kg , is falling vertically downwards under gravity. During the motion, it experiences a variable resistance of \(0 \cdot 2 v ^ { 2 } \mathrm {~N}\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of the object at time \(t\) seconds.
Show that \(v\) satisfies the differential equation
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 9 g - v ^ { 2 } } { 9 }$$
At time \(t = 0\), the object passes a point \(A\) with a speed of \(\sqrt { g } \mathrm {~ms} ^ { - 1 }\). The object then hits the ground with a speed of \(8 \mathrm {~ms} ^ { - 1 }\).
Calculate the time taken for the object to hit the ground.
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Given that the distance of the object from \(A\) at time \(t\) is \(x\) metres, form another differential equation to find an expression for \(x\) in terms of \(v\). Hence, find the height of \(A\) above the ground.
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\begin{enumerate}
\item An object, of mass 1.8 kg , is falling vertically downwards under gravity. During the motion, it experiences a variable resistance of $0 \cdot 2 v ^ { 2 } \mathrm {~N}$, where $v \mathrm {~ms} ^ { - 1 }$ is the speed of the object at time $t$ seconds.\\
(a) Show that $v$ satisfies the differential equation
\end{enumerate}
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 9 g - v ^ { 2 } } { 9 }$$
At time $t = 0$, the object passes a point $A$ with a speed of $\sqrt { g } \mathrm {~ms} ^ { - 1 }$. The object then hits the ground with a speed of $8 \mathrm {~ms} ^ { - 1 }$.\\
(b) Calculate the time taken for the object to hit the ground.\\
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\includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_67_1614_639_264}\\
\includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_76_1614_717_264}\\
\includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_79_1614_801_264}\\
\includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_72_1609_895_267}\\
\includegraphics[max width=\textwidth, alt={}, center]{36112cfa-20c4-4ba8-b972-6b7b44e5182f-07_75_1614_979_264}\\
(c) Given that the distance of the object from $A$ at time $t$ is $x$ metres, form another differential equation to find an expression for $x$ in terms of $v$. Hence, find the height of $A$ above the ground.\\
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\hfill \mbox{\textit{WJEC Further Unit 6 2024 Q2}}