CAIE Further Paper 3 2023 November — Question 6 9 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile clearing obstacle
DifficultyStandard +0.8 Part (a) is a standard derivation worth 3 marks. Part (b) requires setting up two simultaneous conditions involving the trajectory equation with different speeds, then solving a system that leads to a quadratic in tan α. This involves algebraic manipulation across multiple steps and careful handling of the 80% condition, making it moderately challenging but still within standard Further Maths scope.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model
A particle \(P\) is projected with speed \(u\) at an angle \(\alpha\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of \(P\) from \(O\) at a subsequent time \(t\) are denoted by \(x\) and \(y\) respectively.
  1. Derive the equation of the trajectory of \(P\) in the form $$y = x \tan \alpha - \frac{gx^2}{2u^2} \sec^2 \alpha.$$ [3]
During its flight, \(P\) must clear an obstacle of height \(h\) m that is at a horizontal distance of \(32\) m from the point of projection. When \(u = 40\sqrt{2}\) m s\(^{-1}\), \(P\) just clears the obstacle. When \(u = 40\) m s\(^{-1}\), \(P\) only achieves \(80\%\) of the height required to clear the obstacle.
  1. Find the two possible values of \(h\). [6]
Question 6:
AnswerMarks
6(a)→ x=ucos t
1
 y=usin t− gt2
AnswerMarks Guidance
2B1 Both correct.
x 1  x  2
Eliminate t: y=usin − g 
AnswerMarks Guidance
ucos 2 ucosαM1 Eliminate.
gx2
y=xtan− sec2 AG
AnswerMarks
2u2A1
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(b)4 g322 16
h=32tan− sec2 [= 32tan− sec2]
AnswerMarks
5 2402 5B1
g322 8
h=32tan− sec2 [= 32tan− sec2]
( )2 5
AnswerMarks
2 40 2B1
8 5 16
32tan− sec2= (32tan− sec2)
5 4 5
8( 1+t2) ( 1+t2)
32t− =40t−4
5
AnswerMarks Guidance
3t2 −10t+3=0M1 Equate expressions for h and obtain a 3-term quadratic in
tan.
1
t=3,
AnswerMarks Guidance
3A1 Both correct
h=80
80
h=
AnswerMarks Guidance
9M1 For using their value of t to work out one value of h.
A1Both correct
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | → x=ucos t
1
 y=usin t− gt2
2 | B1 | Both correct.
x 1  x  2
Eliminate t: y=usin − g 
ucos 2 ucosα | M1 | Eliminate.
gx2
y=xtan− sec2 AG
2u2 | A1
3
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | 4 g322 16
h=32tan− sec2 [= 32tan− sec2]
5 2402 5 | B1
g322 8
h=32tan− sec2 [= 32tan− sec2]
( )2 5
2 40 2 | B1
8 5 16
32tan− sec2= (32tan− sec2)
5 4 5
8( 1+t2) ( 1+t2)
32t− =40t−4
5
3t2 −10t+3=0 | M1 | Equate expressions for h and obtain a 3-term quadratic in
tan.
1
t=3,
3 | A1 | Both correct
h=80
80
h=
9 | M1 | For using their value of t to work out one value of h.
A1 | Both correct
6
Question | Answer | Marks | Guidance
A particle $P$ is projected with speed $u$ at an angle $\alpha$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of $P$ from $O$ at a subsequent time $t$ are denoted by $x$ and $y$ respectively.

\begin{enumerate}[label=(\alph*)]
\item Derive the equation of the trajectory of $P$ in the form
$$y = x \tan \alpha - \frac{gx^2}{2u^2} \sec^2 \alpha.$$ [3]
\end{enumerate}

During its flight, $P$ must clear an obstacle of height $h$ m that is at a horizontal distance of $32$ m from the point of projection. When $u = 40\sqrt{2}$ m s$^{-1}$, $P$ just clears the obstacle. When $u = 40$ m s$^{-1}$, $P$ only achieves $80\%$ of the height required to clear the obstacle.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the two possible values of $h$. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q6 [9]}}
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