CAIE Further Paper 3 2021 June — Question 7 9 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeTime when specific condition met
DifficultyChallenging +1.2 This is a standard projectiles question requiring systematic application of SUVAT equations. Part (a) involves setting up a quadratic for time, using the difference of roots formula (a moderately sophisticated step), and solving for H. Part (b) requires finding the total time of flight then working backwards one second. While it has multiple steps and requires careful algebra, the techniques are all standard for Further Maths mechanics with no novel insight required. The 9 marks and two-part structure place it above average difficulty but well within expected Further Maths territory.
Spec3.02d Constant acceleration: SUVAT formulae3.02i Projectile motion: constant acceleration model
A particle \(P\) is projected from a point \(O\) on a horizontal plane and moves freely under gravity. The initial velocity of \(P\) is 100 ms\(^{-1}\) at an angle \(\theta\) above the horizontal, where \(\tan\theta = \frac{4}{3}\). The two times at which \(P\)'s height above the plane is \(H\) m differ by 10 s.
  1. Find the value of \(H\). [5]
  1. Find the magnitude and direction of the velocity of \(P\) one second before it strikes the plane. [4]
Question 7:
AnswerMarks Guidance
7(a)At greatest height 0=100sinθ− gt M1
t = 8A1
Therefore times at height H are t=3 (and t =13)B1
1
Substitute into H =100sinθt− gt2
AnswerMarks
2M1
H =195A1
Alternative method to question 7(a)
1
↑H =100sinθt− gt2
AnswerMarks
2M1
And H =100sinθ( t+10 )− 1 g ( t+10 )2
AnswerMarks
2A1
Subtract: 1000sinθ= 1 g ( 20t+100 )
AnswerMarks Guidance
2M1
t=3B1
H =195A1
QuestionAnswer Marks
7(a)Alternative method to question 7(a)
1
↑H =100sinθt− gt2
AnswerMarks
2B1
(100sinθ)2−2gH
Difference between roots =
1
g
AnswerMarks
2M1 A1
Equate to 10 and rearrange to find HM1
H =195A1
5
AnswerMarks Guidance
7(b)Time to required point = 15 s B1
( )
↑v=100sinθ−10×15 =−70
AnswerMarks Guidance
→v=100cosθ=60B1 Both components.
Magnitude = 92.2B1
Angle below horizontal = tan -1 (70/60) = 49.4°B1
4
Question 7:
--- 7(a) ---
7(a) | At greatest height 0=100sinθ− gt | M1
t = 8 | A1
Therefore times at height H are t=3 (and t =13) | B1
1
Substitute into H =100sinθt− gt2
2 | M1
H =195 | A1
Alternative method to question 7(a)
1
↑H =100sinθt− gt2
2 | M1
And H =100sinθ( t+10 )− 1 g ( t+10 )2
2 | A1
Subtract: 1000sinθ= 1 g ( 20t+100 )
2 | M1
t=3 | B1
H =195 | A1
Question | Answer | Marks | Guidance
7(a) | Alternative method to question 7(a)
1
↑H =100sinθt− gt2
2 | B1
(100sinθ)2−2gH
Difference between roots =
1
g
2 | M1 A1
Equate to 10 and rearrange to find H | M1
H =195 | A1
5
--- 7(b) ---
7(b) | Time to required point = 15 s | B1
( )
↑v=100sinθ−10×15 =−70
→v=100cosθ=60 | B1 | Both components.
Magnitude = 92.2 | B1
Angle below horizontal = tan -1 (70/60) = 49.4° | B1
4
A particle $P$ is projected from a point $O$ on a horizontal plane and moves freely under gravity. The initial velocity of $P$ is 100 ms$^{-1}$ at an angle $\theta$ above the horizontal, where $\tan\theta = \frac{4}{3}$. The two times at which $P$'s height above the plane is $H$ m differ by 10 s.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $H$. [5]
\end{enumerate}

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the magnitude and direction of the velocity of $P$ one second before it strikes the plane. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q7 [9]}}
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