CAIE Further Paper 3 2024 November — Question 5 8 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTwo projectiles meeting - 2D flight
DifficultyChallenging +1.2 This is a multi-part projectile question requiring trajectory equation manipulation, energy considerations, and simultaneous projectile analysis. While it involves several steps and the coordination of two projectiles, the techniques are standard for Further Maths mechanics: using tan θ = 1/3 to find trajectory coefficients, applying energy conservation or kinematic equations, and setting up simultaneous conditions. The algebraic manipulation is moderate but systematic, making this harder than average A-level but not requiring exceptional insight.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model
A particle \(P\) is projected from a point \(O\) on horizontal ground with speed \(u\) at an angle \(\theta\) above the horizontal, where \(\tan \theta = \frac{1}{3}\). The particle \(P\) moves freely under gravity and passes through the point with coordinates \((3a, \frac{4}{5}a)\) relative to horizontal and vertical axes through \(O\) in the plane of the motion.
  1. Use the equation of the trajectory to show that \(u^2 = 25ag\). [2]
  2. Express \(V^2\) in the form \(kag\), where \(k\) is a rational number. [6]
At the instant when \(P\) is moving horizontally, a particle \(Q\) is projected from \(O\) with speed \(V\) at an angle \(\alpha\) above the horizontal. The particles \(P\) and \(Q\) reach the ground at the same point and at the same time.
Question 5:
AnswerMarks
5(a)4 1 g  1
Use correct equation of trajectory: a=3a − (3a)21+ 
AnswerMarks Guidance
5 3 2u2  9M1 No (implied) sight of trajectory equation M0.
4 5ga2 5ga 1
a=a− , = , u2 =25ga
AnswerMarks Guidance
5 u2 u2 5A1 At least one step of intermediate working must be
seen.
AG
2
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
5(b)For P, time of flight T and range R
For Q, time of flight ½ T and range R
 2usin  10a
T = =
 
AnswerMarks Guidance
 g  gB1 Time of flight for P or Q.
2 3
[From motion of P, R= 25ag =] 1 5a
AnswerMarks Guidance
g 10B1 Range for P.
1 2R
For Q: →R=vcos T, vcos=
AnswerMarks Guidance
2 TM1 Obtain an expression for vcos. May involve u
and .
2
1 1 1  1
 0=vsin T − g  T  , vsin= gT
AnswerMarks Guidance
2 2 2  4M1 Obtain an expression for vsin. May involve u
and .
2R 2 1  2  5 
Square and add: v2 =   +  gT   =90ag+ ag 
AnswerMarks
 T  4   8 M1
725
v2 = ag
AnswerMarks Guidance
8A1 1
gT
4 1
tan= =
2R 12
T
AnswerMarks Guidance
QuestionAnswer Marks
5(b)Alternative method for question 5(b)
For P, time of flight T and range R
For Q, time of flight ½ T and range R
Horizontal motion for P and Q
T
R=ucosT and R=(vcos)
AnswerMarks Guidance
2M1 Both.
Vertical motion for P and Q
gT gT
usin= and vsin=
AnswerMarks Guidance
2 4M1 1
Both, may come from using s=ut+ at2.
2
Equate two expressions for R:
AnswerMarks Guidance
vcos=2ucosA1 6
vcos= u
10
Equate two expressions for vertical motion:
1
vsin= usin
AnswerMarks Guidance
2A1 1
vsin= u
2 10
 1   29 
Square and add: v2 =u2 4cos2+ sin2 = u2
 
AnswerMarks
 4   8 M1
29 725
25ag = ag
AnswerMarks
8 8A1
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | 4 1 g  1
Use correct equation of trajectory: a=3a − (3a)21+ 
5 3 2u2  9 | M1 | No (implied) sight of trajectory equation M0.
4 5ga2 5ga 1
a=a− , = , u2 =25ga
5 u2 u2 5 | A1 | At least one step of intermediate working must be
seen.
AG
2
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | For P, time of flight T and range R
For Q, time of flight ½ T and range R
 2usin  10a
T = =
 
 g  g | B1 | Time of flight for P or Q.
2 3
[From motion of P, R= 25ag =] 1 5a
g 10 | B1 | Range for P.
1 2R
For Q: →R=vcos T, vcos=
2 T | M1 | Obtain an expression for vcos. May involve u
and .
2
1 1 1  1
 0=vsin T − g  T  , vsin= gT
2 2 2  4 | M1 | Obtain an expression for vsin. May involve u
and .
2R 2 1  2  5 
Square and add: v2 =   +  gT   =90ag+ ag 
 T  4   8  | M1
725
v2 = ag
8 | A1 | 1
gT
4 1
tan= =
2R 12
T
Question | Answer | Marks | Guidance
5(b) | Alternative method for question 5(b)
For P, time of flight T and range R
For Q, time of flight ½ T and range R
Horizontal motion for P and Q
T
R=ucosT and R=(vcos)
2 | M1 | Both.
Vertical motion for P and Q
gT gT
usin= and vsin=
2 4 | M1 | 1
Both, may come from using s=ut+ at2.
2
Equate two expressions for R:
vcos=2ucos | A1 | 6
vcos= u
10
Equate two expressions for vertical motion:
1
vsin= usin
2 | A1 | 1
vsin= u
2 10
 1   29 
Square and add: v2 =u2 4cos2+ sin2 = u2
 
 4   8  | M1
29 725
25ag = ag
8 8 | A1
6
Question | Answer | Marks | Guidance
A particle $P$ is projected from a point $O$ on horizontal ground with speed $u$ at an angle $\theta$ above the horizontal, where $\tan \theta = \frac{1}{3}$. The particle $P$ moves freely under gravity and passes through the point with coordinates $(3a, \frac{4}{5}a)$ relative to horizontal and vertical axes through $O$ in the plane of the motion.

\begin{enumerate}[label=(\alph*)]
\item Use the equation of the trajectory to show that $u^2 = 25ag$. [2]
\item Express $V^2$ in the form $kag$, where $k$ is a rational number. [6]
\end{enumerate}

At the instant when $P$ is moving horizontally, a particle $Q$ is projected from $O$ with speed $V$ at an angle $\alpha$ above the horizontal. The particles $P$ and $Q$ reach the ground at the same point and at the same time.

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q5 [8]}}
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