CAIE Further Paper 3 2023 November — Question 5 9 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTwo projectiles meeting - 2D flight
DifficultyChallenging +1.2 Part (a) is a straightforward substitution into standard projectile equations with given angle and point. Part (b) requires coordinating two projectiles' trajectories and timing, involving multiple steps with range/time equations, but follows standard Further Maths projectile methods without requiring novel geometric insight. The 7-mark allocation and coordination aspect elevate it slightly above average difficulty.
Spec3.02i Projectile motion: constant acceleration model
A particle \(P\) is projected with speed \(u\text{ ms}^{-1}\) at an angle \(\theta\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. During its flight \(P\) passes through the point which is a horizontal distance \(3a\) from \(O\) and a vertical distance \(\frac{3}{8}a\) above the horizontal plane. It is given that \(\tan\theta = \frac{1}{3}\).
  1. Show that \(u^2 = 8ag\). [2]
A particle \(Q\) is projected with speed \(V\text{ ms}^{-1}\) at an angle \(\alpha\) above the horizontal from \(O\) at the instant when \(P\) is at its highest point. Particles \(P\) and \(Q\) both land at the same point on the horizontal plane at the same time.
  1. Find \(V\) in terms of \(a\) and \(g\). [7]
Question 5:
AnswerMarks
5(a)gx2
Use equation of trajectory: y=xtan− sec2:
2u2
(3a)2
3 1  1
a=3a −g 1+ 
AnswerMarks
8 3 2u2  9M1
5 5ga
a= , u2 =8ga
AnswerMarks Guidance
8 u2A1 At least one step of working.
AG
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
5(b)2usin
For P, time of flight T =
AnswerMarks
gB1
For P, range = TucosB1
1 1
For Q, time of flight = T, so range = TVcos
2 2
AnswerMarks
Equate: Vcos=2ucos (1)B1
1
gT2
1 2 4Vsin
For Q vertically: 0=Vsin T − , so T =
AnswerMarks
2 4 gM1
Equate with result for P:
4Vsin 2usin 1
= so Vsin= usin (2)
AnswerMarks
g g 2A1
 1 
From (1) and (2): V2 =u2 4(cos)2 + (sin)2 
AnswerMarks
 4 M1
 9 1 1  145
V2 =u2 4 +  = u2
 10 4 10 40
AnswerMarks
V = 29agA1
7
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | gx2
Use equation of trajectory: y=xtan− sec2:
2u2
(3a)2
3 1  1
a=3a −g 1+ 
8 3 2u2  9 | M1
5 5ga
a= , u2 =8ga
8 u2 | A1 | At least one step of working.
AG
2
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | 2usin
For P, time of flight T =
g | B1
For P, range = Tucos | B1
1 1
For Q, time of flight = T, so range = TVcos
2 2
Equate: Vcos=2ucos (1) | B1
1
gT2
1 2 4Vsin
For Q vertically: 0=Vsin T − , so T =
2 4 g | M1
Equate with result for P:
4Vsin 2usin 1
= so Vsin= usin (2)
g g 2 | A1
 1 
From (1) and (2): V2 =u2 4(cos)2 + (sin)2 
 4  | M1
 9 1 1  145
V2 =u2 4 +  = u2
 10 4 10 40
V = 29ag | A1
7
Question | Answer | Marks | Guidance
A particle $P$ is projected with speed $u\text{ ms}^{-1}$ at an angle $\theta$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. During its flight $P$ passes through the point which is a horizontal distance $3a$ from $O$ and a vertical distance $\frac{3}{8}a$ above the horizontal plane. It is given that $\tan\theta = \frac{1}{3}$.

\begin{enumerate}[label=(\alph*)]
\item Show that $u^2 = 8ag$. [2]
\end{enumerate}

A particle $Q$ is projected with speed $V\text{ ms}^{-1}$ at an angle $\alpha$ above the horizontal from $O$ at the instant when $P$ is at its highest point. Particles $P$ and $Q$ both land at the same point on the horizontal plane at the same time.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $V$ in terms of $a$ and $g$. [7]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q5 [9]}}
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