CAIE Further Paper 3 2021 June — Question 3 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionJune
Marks7
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TopicHooke's law and elastic energy
TypeVertical elastic string: general symbolic/proof questions
DifficultyStandard +0.8 Part (a) is a routine 1-mark application of Hooke's law at equilibrium. Part (b) requires setting up and solving an energy equation involving elastic potential energy, gravitational potential energy, and kinetic energy with multiple terms and algebraic manipulation. The multi-step energy conservation approach with elastic strings and the algebraic complexity (involving the modulus k from part (a)) elevates this above average difficulty, though it follows a standard method for Further Mechanics.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
One end of a light elastic string, of natural length \(a\) and modulus of elasticity \(kmg\), is attached to a fixed point A. The other end of the string is attached to a particle \(P\) of mass \(4m\). The particle \(P\) hangs in equilibrium a distance \(x\) vertically below A.
  1. Show that \(k = \frac{4a}{x-a}\). [1]
An additional particle, of mass \(2m\), is now attached to \(P\) and the combined particle is released from rest at the original equilibrium position of \(P\). When the combined particle has descended a distance \(\frac{3}{4}a\), its speed is \(\frac{1}{2}\sqrt{ga}\).
  1. Find \(x\) in terms of \(a\). [6]
Question 3:
AnswerMarks
3(a)kmg ( x−a ) 4a
Use Hooke’s Law: 4mg = leading to k =
x−a
AnswerMarks Guidance
aB1 AG. Shown convincingly.
1
AnswerMarks Guidance
3(b)Gain in KE + gain in EPE = loss in GPE B1
1 ×6m× ga + 1 kmg   x+ a −a   2 −( x−a )2  =6mg× a
2 9 2 a   3   3
AnswerMarks
 M1
A1All 3 types of energy included in energy equation.
All terms correct.
AnswerMarks
Simplify and substitute for k from part (a)M1
Obtain linear equation in x and aM1
5
x = a
AnswerMarks Guidance
3A1 (k = 6)
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | kmg ( x−a ) 4a
Use Hooke’s Law: 4mg = leading to k =
x−a
a | B1 | AG. Shown convincingly.
1
--- 3(b) ---
3(b) | Gain in KE + gain in EPE = loss in GPE | B1 | One correct EPE term seen.
1 ×6m× ga + 1 kmg   x+ a −a   2 −( x−a )2  =6mg× a
2 9 2 a   3   3
  | M1
A1 | All 3 types of energy included in energy equation.
All terms correct.
Simplify and substitute for k from part (a) | M1
Obtain linear equation in x and a | M1
5
x = a
3 | A1 | (k = 6)
6
Question | Answer | Marks | Guidance
One end of a light elastic string, of natural length $a$ and modulus of elasticity $kmg$, is attached to a fixed point A. The other end of the string is attached to a particle $P$ of mass $4m$. The particle $P$ hangs in equilibrium a distance $x$ vertically below A.

\begin{enumerate}[label=(\alph*)]
\item Show that $k = \frac{4a}{x-a}$. [1]
\end{enumerate}

An additional particle, of mass $2m$, is now attached to $P$ and the combined particle is released from rest at the original equilibrium position of $P$. When the combined particle has descended a distance $\frac{3}{4}a$, its speed is $\frac{1}{2}\sqrt{ga}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $x$ in terms of $a$. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q3 [7]}}
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