CAIE Further Paper 3 2024 June — Question 3 5 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypePerpendicular velocity directions
DifficultyChallenging +1.2 This is a standard projectiles problem requiring resolution of velocity components and use of the perpendicularity condition. While it involves setting up equations for velocity at two times and applying the dot product condition, the method is straightforward once the setup is recognized. The 5-mark allocation and single-answer nature make it moderately challenging but not exceptional for Further Maths.
Spec1.10h Vectors in kinematics: uniform acceleration in vector form3.02i Projectile motion: constant acceleration model

At time \(t = 0\) seconds, a particle \(P\) is projected with speed \(u\) m s\(^{-1}\) at an angle \(60°\) above the horizontal from a point \(O\). In the subsequent motion \(P\) moves freely under gravity. The direction of motion of \(P\) when \(t = 5\) is perpendicular to its direction of motion when \(t = 15\). Find the value of \(u\). [5]

Question 3:
AnswerMarks Guidance
3usin605g, and ucos60 , or usin6015g , and ucos60 B1
usin605g
If  is direction of velocity at t 5, tan
AnswerMarks Guidance
ucos60M1* Accept equivalent for t15.
usin605g usin6015g
For perpendicular directions,   1
AnswerMarks Guidance
ucos60 ucos60M1dep Multiply two expressions involving relevant
velocities and equate to – 1.
Simplify: 3u2 75g2 10 3ug1u2 0, u2 100 3u75000
AnswerMarks Guidance
4 4M1 Simplify to quadratic in u (may see g ).
u5 3gA1 OE. Accept 50 3 or 86.6.
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 3:
3 | usin605g, and ucos60 , or usin6015g , and ucos60 | B1
usin605g
If  is direction of velocity at t 5, tan
ucos60 | M1* | Accept equivalent for t15.
usin605g usin6015g
For perpendicular directions,   1
ucos60 ucos60 | M1dep | Multiply two expressions involving relevant
velocities and equate to – 1.
Simplify: 3u2 75g2 10 3ug1u2 0, u2 100 3u75000
4 4 | M1 | Simplify to quadratic in u (may see g ).
u5 3g | A1 | OE. Accept 50 3 or 86.6.
5
Question | Answer | Marks | Guidance
At time $t = 0$ seconds, a particle $P$ is projected with speed $u$ m s$^{-1}$ at an angle $60°$ above the horizontal from a point $O$. In the subsequent motion $P$ moves freely under gravity. The direction of motion of $P$ when $t = 5$ is perpendicular to its direction of motion when $t = 15$.

Find the value of $u$. [5]

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q3 [5]}}