CAIE Further Paper 3 2024 June — Question 6 8 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypePerpendicular velocity directions
DifficultyStandard +0.8 This is a two-part projectile motion problem requiring vector manipulation and simultaneous equations. Part (a) involves applying speed formulas with velocity components after time t, requiring careful algebraic manipulation to reach the given result. Part (b) adds the perpendicularity condition, requiring dot product of velocity vectors to equal zero, then solving simultaneous equations. While the techniques are standard A-level mechanics (projectile motion, vectors, perpendicularity), the algebraic complexity and the need to coordinate multiple conditions elevates this above routine exercises.
Spec3.02h Motion under gravity: vector form3.02i 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 \(O\) and moves freely under gravity. After 5 seconds the speed of \(P\) is \(\frac{3}{4}u\).
  1. Show that \(\frac{7}{16}u^2 - 100u\sin\theta + 2500 = 0\). [3]
  2. It is given that the velocity of \(P\) after 5 seconds is perpendicular to the initial velocity. Find, in either order, the value of \(u\) and the value of \(\sin\theta\). [5]
Question 6:
AnswerMarks Guidance
6(a)Components of velocity are ucos, usingt B1
2
ucos2 usingt2  3 
 u
 
AnswerMarks Guidance
4 M1 Square and add components of velocity and
2
3 
equate to  u .
4 
7
u2 100usin25000
AnswerMarks Guidance
16A1 AG
At least one correct line of working seen.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
6(b)Let  be angle of direction of motion with horizontal at t5, then
usin5g
tan
AnswerMarks Guidance
ucosB1 Either way up.
 usin5g
tan tan1, so tan   1
AnswerMarks Guidance
 ucos M1 Must be – 1 not +1.
FT their expression for tan.
AnswerMarks
u50sinA1
Use in result from part (a) to form equation in u or sinM1
4
u2 1600, u40 and sin
AnswerMarks Guidance
5A1 Both.
Alternative method for question 6(b)
3
ucos usin
AnswerMarks
4M1
4 4
tan or sin
AnswerMarks
3 5A1
3 4
 ucos u50
AnswerMarks Guidance
4 5M1 A1 Allow sign error.
4
u40 and sin
AnswerMarks Guidance
5A1 Both seen.
QuestionAnswer Marks
6(b)Alternative method for question 6(b)
3
ucos usin
AnswerMarks
4M1
4 4
tan or sin
AnswerMarks
3 5A1
Use in result from part (a) to form equation in uM1
 1000
u40 and
 
AnswerMarks
 7 A1
4
u40 only and sin
AnswerMarks Guidance
5A1 Both seen.
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 6:
--- 6(a) ---
6(a) | Components of velocity are ucos, usingt | B1
2
ucos2 usingt2  3 
 u
 
4  | M1 | Square and add components of velocity and
2
3 
equate to  u .
4 
7
u2 100usin25000
16 | A1 | AG
At least one correct line of working seen.
3
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | Let  be angle of direction of motion with horizontal at t5, then
usin5g
tan
ucos | B1 | Either way up.
 usin5g
tan tan1, so tan   1
 ucos  | M1 | Must be – 1 not +1.
FT their expression for tan.
u50sin | A1
Use in result from part (a) to form equation in u or sin | M1
4
u2 1600, u40 and sin
5 | A1 | Both.
Alternative method for question 6(b)
3
ucos usin
4 | M1
4 4
tan or sin
3 5 | A1
3 4
 ucos u50
4 5 | M1 A1 | Allow sign error.
4
u40 and sin
5 | A1 | Both seen.
Question | Answer | Marks | Guidance
6(b) | Alternative method for question 6(b)
3
ucos usin
4 | M1
4 4
tan or sin
3 5 | A1
Use in result from part (a) to form equation in u | M1
 1000
u40 and
 
 7  | A1
4
u40 only and sin
5 | A1 | Both seen.
5
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 $O$ and moves freely under gravity. After 5 seconds the speed of $P$ is $\frac{3}{4}u$.

\begin{enumerate}[label=(\alph*)]
\item Show that $\frac{7}{16}u^2 - 100u\sin\theta + 2500 = 0$. [3]

\item It is given that the velocity of $P$ after 5 seconds is perpendicular to the initial velocity.

Find, in either order, the value of $u$ and the value of $\sin\theta$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q6 [8]}}
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