CAIE Further Paper 3 2021 November — Question 7 8 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile passing through given point
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring multiple sophisticated techniques: energy conservation to find speed at detachment, projectile trajectory equation in a non-standard coordinate system, and circular motion tension analysis. The trajectory part (finding V) requires careful geometric setup and algebraic manipulation, while part (b) needs applying circular motion principles at the initial position. The multi-stage nature and need to connect circular motion with projectile motion elevates this above standard A-level questions.
Spec3.02i Projectile motion: constant acceleration model6.05d Variable speed circles: energy methods

One end of a light inextensible string of length \(a\) is attached to a fixed point \(O\). The other end of the string is attached to a particle \(P\) of mass \(m\). The particle \(P\) is held vertically below \(O\) with the string taut and then projected horizontally. When the string makes an angle of \(60°\) with the upward vertical, \(P\) becomes detached from the string. In its subsequent motion, \(P\) passes through the point \(A\) which is a distance \(a\) vertically above \(O\).
  1. The speed of \(P\) when it becomes detached from the string is \(V\). Use the equation of the trajectory of a projectile to find \(V\) in terms of \(a\) and \(g\). [4]
  2. Find, in terms of \(m\) and \(g\), the tension in the string immediately after \(P\) is initially projected horizontally. [4]

Question 7:

AnswerMarks Guidance
7(a)Coordinates of A: x=asin60, y=a−acos60 B1
( )
2
a 3
g
a a 3 22
= 3−
2 2 1
2V2.
AnswerMarks Guidance
4M1 Substitute their (x, y) into correct trajectory equation.
Rearrange to find V2.M1
3 3
V2 = ag, V = ag
AnswerMarks
2 2A1
4

AnswerMarks Guidance
7(b)1 mu2 − 1 mV2 =mga ( 1+cos60 )
2 2M1 Energy equation.
9
u2 = ag
AnswerMarks Guidance
2A1 u is the speed at P.
m
T −mg = u2
AnswerMarks Guidance
aM1 N2L
11
T = mg
AnswerMarks
2A1
4
Question 7:
--- 7(a) ---
7(a) | Coordinates of A: x=asin60, y=a−acos60 | B1
( )
2
a 3
g
a a 3 22
= 3−
2 2 1
2V2.
4 | M1 | Substitute their (x, y) into correct trajectory equation.
Rearrange to find V2. | M1
3 3
V2 = ag, V = ag
2 2 | A1
4
--- 7(b) ---
7(b) | 1 mu2 − 1 mV2 =mga ( 1+cos60 )
2 2 | M1 | Energy equation.
9
u2 = ag
2 | A1 | u is the speed at P.
m
T −mg = u2
a | M1 | N2L
11
T = mg
2 | A1
4
One end of a light inextensible string of length $a$ is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass $m$. The particle $P$ is held vertically below $O$ with the string taut and then projected horizontally. When the string makes an angle of $60°$ with the upward vertical, $P$ becomes detached from the string. In its subsequent motion, $P$ passes through the point $A$ which is a distance $a$ vertically above $O$.

\begin{enumerate}[label=(\alph*)]
\item The speed of $P$ when it becomes detached from the string is $V$. Use the equation of the trajectory of a projectile to find $V$ in terms of $a$ and $g$. [4]

\item Find, in terms of $m$ and $g$, the tension in the string immediately after $P$ is initially projected horizontally. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q7 [8]}}