Challenging +1.8 This is a challenging projectiles problem requiring students to set up velocity component equations at two different times, apply the perpendicularity condition (dot product = 0), and solve algebraically. While the mechanics concepts are standard A-level, the problem requires sustained multi-step reasoning with non-trivial algebra involving the specific angle and time values, placing it well above average difficulty but not at the extreme end for Further Maths.
A particle \(P\) is projected from a point \(O\) on a horizontal plane and moves freely under gravity. Its initial speed is \(u\) ms\(^{-1}\) and its angle of projection is \(\sin^{-1}(\frac{3}{5})\) above the horizontal. At time 8 s after projection, \(P\) is at the point \(A\). At time 32 s after projection, \(P\) is at the point \(B\). The direction of motion of \(P\) at \(B\) is perpendicular to its direction of motion at \(A\).
Find the value of \(u\). [7]
Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
Answer
Marks
Guidance
5
At A: ↑usinθ−8g →ucosθ
M1
usinθ−8g
tanα=
Answer
Marks
Guidance
ucosθ
A1
At B: ↑usinθ−32g →ucosθ
M1
Both.
usinθ−32g
tanβ=
Answer
Marks
ucosθ
A1
usinθ−8g usinθ−32g
× =−1
Answer
Marks
Guidance
ucosθ ucosθ
B1
Perpendicular directions, so tanα×tanβ=−1.
u2 −320u+25600=0
M1
Simplify to a quadratic in u.
u=160
A1
7
Answer
Marks
Guidance
Question
Answer
Marks
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
5 | At A: ↑usinθ−8g →ucosθ | M1 | Both.
usinθ−8g
tanα=
ucosθ | A1
At B: ↑usinθ−32g →ucosθ | M1 | Both.
usinθ−32g
tanβ=
ucosθ | A1
usinθ−8g usinθ−32g
× =−1
ucosθ ucosθ | B1 | Perpendicular directions, so tanα×tanβ=−1.
u2 −320u+25600=0 | M1 | Simplify to a quadratic in u.
u=160 | A1
7
Question | Answer | Marks | Guidance
A particle $P$ is projected from a point $O$ on a horizontal plane and moves freely under gravity. Its initial speed is $u$ ms$^{-1}$ and its angle of projection is $\sin^{-1}(\frac{3}{5})$ above the horizontal. At time 8 s after projection, $P$ is at the point $A$. At time 32 s after projection, $P$ is at the point $B$. The direction of motion of $P$ at $B$ is perpendicular to its direction of motion at $A$.
Find the value of $u$. [7]
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q5 [7]}}