CAIE Further Paper 3 2022 November — Question 5 9 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2022
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTwo possible trajectories through point
DifficultyStandard +0.3 Part (a) is a standard 'show that' derivation of the trajectory equation using parametric equations and the sec²θ identity—routine bookwork for Further Maths. Part (b) requires substituting the point into the trajectory equation and solving a quadratic in tan θ, which is a straightforward application once set up. This is a typical textbook exercise testing standard projectile techniques with no novel insight required, making it slightly easier than average even for Further Maths.
Spec3.02i Projectile motion: constant acceleration model
A particle \(P\) is projected with speed \(u\) m s\(^{-1}\) at an angle of \(\theta\) 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 \(t\) s are denoted by \(x\) m and \(y\) m respectively.
  1. Show that the equation of the trajectory is given by $$y = x \tan \theta - \frac{gx^2}{2u^2}(1 + \tan^2 \theta).$$ [4]
In the subsequent motion \(P\) passes through the point with coordinates \((30, 20)\).
  1. Given that one possible value of \(\tan \theta\) is \(\frac{4}{3}\), find the other possible value of \(\tan \theta\). [5]
Question 5:
AnswerMarks
5Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
AnswerMarks Guidance
5(a)→ x=ucos t B1
1
 y=usin t− gt2
AnswerMarks
2B1
x 1  x  2
Eliminate t: y=usin  − g  
AnswerMarks
cos 2 ucos M1
gx2
y=xtan− (1+tan2)
AnswerMarks Guidance
2u2A1 Must be an intermediate line of working.
AG
4
AnswerMarks
5(b)4 302  4 2
20=30 −10 1+  
3 2u2  3 
AnswerMarks Guidance
 M1 Substituting values correctly.
u2 =625, u=25 A1
Substitute back into trajectory equation,
g302 36
20=30tan − sec2 =30tan − (1+tan2 )
2.252 5
AnswerMarks Guidance
18tan2−75tan+68=0M1 Obtain a 3-term quadratic.
4
One solution is , (3tan−4)(6tan−17)=0
AnswerMarks
3M1
17
Giving tan=
AnswerMarks
6A1
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | → x=ucos t | B1 | Result quoted from MF19 scores 0/4.
1
 y=usin t− gt2
2 | B1
x 1  x  2
Eliminate t: y=usin  − g  
cos 2 ucos  | M1
gx2
y=xtan− (1+tan2)
2u2 | A1 | Must be an intermediate line of working.
AG
4
--- 5(b) ---
5(b) | 4 302  4 2
20=30 −10 1+  
3 2u2  3 
  | M1 | Substituting values correctly.
u2 =625, u=25  | A1
Substitute back into trajectory equation,
g302 36
20=30tan − sec2 =30tan − (1+tan2 )
2.252 5
18tan2−75tan+68=0 | M1 | Obtain a 3-term quadratic.
4
One solution is , (3tan−4)(6tan−17)=0
3 | M1
17
Giving tan=
6 | A1
5
Question | Answer | Marks | Guidance
A particle $P$ is projected with speed $u$ m s$^{-1}$ at an angle of $\theta$ 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 $t$ s are denoted by $x$ m and $y$ m respectively.

\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the trajectory is given by
$$y = x \tan \theta - \frac{gx^2}{2u^2}(1 + \tan^2 \theta).$$ [4]
\end{enumerate}

In the subsequent motion $P$ passes through the point with coordinates $(30, 20)$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Given that one possible value of $\tan \theta$ is $\frac{4}{3}$, find the other possible value of $\tan \theta$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q5 [9]}}
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