CAIE Further Paper 3 2023 June — Question 7 9 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTwo projectiles meeting - 2D flight
DifficultyChallenging +1.2 This is a standard projectiles problem requiring simultaneous equations from horizontal and vertical motion. While it involves two moving objects and multiple steps (finding T, then velocity components), the setup is straightforward with given tan θ = 3/4 making calculations clean. The collision condition provides clear equations to solve, and finding final direction is routine differentiation. More involved than basic projectile questions but uses standard A-level techniques without requiring novel insight.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model
The points \(O\) and \(P\) are on a horizontal plane, a distance \(8\) m apart. A ball is thrown from \(O\) with speed \(u\) m s\(^{-1}\) at an angle \(\theta\) above the horizontal, where \(\tan \theta = \frac{3}{4}\). At the same instant, a model aircraft is launched with speed \(5\) m s\(^{-1}\) parallel to the horizontal plane from a point \(4\) m vertically above \(P\). The model aircraft moves in the same vertical plane as the ball and in the same horizontal direction as the ball. The model aircraft moves horizontally with a constant speed of \(5\) m s\(^{-1}\). After \(T\) s, the ball and the model aircraft collide.
  1. Find the value of \(T\). [6]
  2. Find the direction in which the ball is moving immediately before the collision. [3]
Question 7:
AnswerMarks Guidance
7(a)For aircraft, d 5T B1
1
For ball, 4usin T  10T2
AnswerMarks Guidance
2B1 To point of collision.
For ball, ucos T d 85T 8B1
Eliminate u:
 45T2
u4 1 5 55T 8
4 T  10T2, u and u
5 2 4T 3T
AnswerMarks Guidance
3  45T2 45T 8*M1 Dependent on LHS of second B1 being 4,
expression involving only T
AnswerMarks Guidance
3T2 4T 40DM1 Dependent on previous M1.
Obtain and solve 3-term quadratic.
AnswerMarks Guidance
T 2A1 Single correct answer.
62
Note d 8 used leads to T  B1B1B0M1M1A0
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)usin10T
tan1
AnswerMarks Guidance
ucosM1 OE
Accept ‘tan = ……’
8
tan1
AnswerMarks Guidance
9A1 OE
Direction is 41.6° below the horizontalA1 CAO
Note: d 8 used leads to 20.9 above the
horizontal.
3
Question 7:
--- 7(a) ---
7(a) | For aircraft, d 5T | B1
1
For ball, 4usin T  10T2
2 | B1 | To point of collision.
For ball, ucos T d 85T 8 | B1
Eliminate u:
 45T2
u4 1 5 55T 8
4 T  10T2, u and u
5 2 4T 3T
3  45T2 45T 8 | *M1 | Dependent on LHS of second B1 being 4,
expression involving only T
3T2 4T 40 | DM1 | Dependent on previous M1.
Obtain and solve 3-term quadratic.
T 2 | A1 | Single correct answer.
6 | 2
Note d 8 used leads to T  B1B1B0M1M1A0
3
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | usin10T
tan1
ucos | M1 | OE
Accept ‘tan = ……’
8
tan1
9 | A1 | OE
Direction is 41.6° below the horizontal | A1 | CAO
Note: d 8 used leads to 20.9 above the
horizontal.
3
The points $O$ and $P$ are on a horizontal plane, a distance $8$ m apart. A ball is thrown from $O$ with speed $u$ m s$^{-1}$ at an angle $\theta$ above the horizontal, where $\tan \theta = \frac{3}{4}$. At the same instant, a model aircraft is launched with speed $5$ m s$^{-1}$ parallel to the horizontal plane from a point $4$ m vertically above $P$. The model aircraft moves in the same vertical plane as the ball and in the same horizontal direction as the ball. The model aircraft moves horizontally with a constant speed of $5$ m s$^{-1}$. After $T$ s, the ball and the model aircraft collide.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $T$. [6]
\item Find the direction in which the ball is moving immediately before the collision. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q7 [9]}}
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