WJEC Unit 4 2018 June — Question 7 11 marks

Exam BoardWJEC
ModuleUnit 4 (Unit 4)
Year2018
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferential equations
TypeParticle motion - velocity/time (dv/dt = f(v,t))
DifficultyStandard +0.3 This is a standard A-level mechanics differential equation problem involving resistance proportional to velocity. Part (a) requires setting up Newton's second law with given information (routine). Part (b) involves solving a first-order linear DE using separation of variables or integrating factor (standard technique). Part (c) is a simple substitution (v=0). While it requires multiple steps and integration skills, it follows a well-established template for resistance problems with no novel insight needed, making it slightly easier than average.
Spec1.08k Separable differential equations: dy/dx = f(x)g(y)3.03d Newton's second law: 2D vectors

An object of mass \(0 \cdot 5\) kg is thrown vertically upwards with initial speed \(24\) ms\(^{-1}\). The velocity of the object at time \(t\) seconds is \(v\) ms\(^{-1}\). During the upward motion, the object experiences a resistance to motion \(RN\), where \(R\) is proportional to \(v\). When the velocity of the object is \(0 \cdot 2\) ms\(^{-1}\) the resistance to motion is \(0 \cdot 08\) N.
  1. Show that the upward motion of the object satisfies the differential equation $$\frac{\mathrm{d}v}{\mathrm{d}t} = -9 \cdot 8 - 0 \cdot 8\,v.$$ [3]
  2. Find an expression for \(v\) at time \(t\). [6]
  3. Determine the value of \(t\) when the object is at the highest point of the motion. [2]

An object of mass $0 \cdot 5$ kg is thrown vertically upwards with initial speed $24$ ms$^{-1}$. The velocity of the object at time $t$ seconds is $v$ ms$^{-1}$. During the upward motion, the object experiences a resistance to motion $RN$, where $R$ is proportional to $v$. When the velocity of the object is $0 \cdot 2$ ms$^{-1}$ the resistance to motion is $0 \cdot 08$ N.

\begin{enumerate}[label=(\alph*)]
\item Show that the upward motion of the object satisfies the differential equation
$$\frac{\mathrm{d}v}{\mathrm{d}t} = -9 \cdot 8 - 0 \cdot 8\,v.$$ [3]

\item Find an expression for $v$ at time $t$. [6]

\item Determine the value of $t$ when the object is at the highest point of the motion. [2]
\end{enumerate}

\hfill \mbox{\textit{WJEC Unit 4 2018 Q7 [11]}}