CAIE Further Paper 3 2021 June — Question 7 4 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeMaximum range or optimal angle
DifficultyEasy -1.2 This question requires only direct application of standard projectile formulae from the formula sheet, with minimal algebraic manipulation. Part (a) involves substituting y=0 into the trajectory equation and solving for range, while part (b) uses symmetry or differentiation to find maximum height. Both are routine textbook exercises with no problem-solving insight required, making this easier than average despite being Further Maths.
Spec3.02i Projectile motion: constant acceleration model
A particle \(P\) is projected with speed \(u\) at an angle \(\theta\) above the horizontal from a point \(O\) on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of \(P\) from \(O\) at a subsequent time \(t\) are denoted by \(x\) and \(y\) respectively.
  1. Use the equation of the trajectory given in the List of formulae (MF19), together with the condition \(y = 0\), to establish an expression for the range \(R\) in terms of \(u\), \(\theta\) and \(g\). [2]
  2. Deduce an expression for the maximum height \(H\), in terms of \(u\), \(\theta\) and \(g\). [2]
Question 7:
AnswerMarks
7(a)R2
y = 0 in trajectory equation: Rtanθ−g =0
( cosθ)2
AnswerMarks
2u2M1
R=)2u2sinθcosθ
(
only
AnswerMarks Guidance
gA1 Any equivalent single term expression, for example:
u2sin2θ 2u2tanθ
, , at least one intermediate line of
g g sec2θ
working, not just quoting a result.
SC B1 using SUVAT.
2
AnswerMarks
7(b)u2sinθcosθ
x=their
and substitute in trajectory equation.
AnswerMarks Guidance
gM1 Or use SUVAT.
( sinθ)2
u2
H =
AnswerMarks Guidance
2gA1 Single term.
2
AnswerMarks
7(c)4H
Use R= and simplify: tanθ= 3, θ = 60°
AnswerMarks Guidance
3B1 AG
1
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(d)dy gx
=tanθ−
( cosθ)2
AnswerMarks Guidance
dx u2M1 Differentiate with respect to x.
x
tanθ− =±1
used
( cosθ)2
AnswerMarks Guidance
4M1 dy
Use =±1 as limiting case.
dx
AnswerMarks Guidance
x = 3 +1, x = 3 −1A1
3 −1< x < 3 +1A1 Strict inequality, exact values.
Alternative method for question 7(d)
1 dy
y= 3x− x2, = 3−x
AnswerMarks Guidance
2 dxM1 Differentiate with respect to x.
dy
=±1 used
AnswerMarks Guidance
dxM1 dy
Use =±1 as limiting case.
dx
AnswerMarks Guidance
x = 3 +1 , x = 3 −1A1
3 −1< x < 3 +1A1 Strict inequality, exact values.
QuestionAnswer Marks
7(d)Alternative method for question 7(d)
v =±v
When moving at 45° to horizontal,
AnswerMarks Guidance
x yM1 Used, both cases considered.
v = 40 cosθ , v = 40sinθ−10t
x y
1 ( ) 1 ( )
t= 30− 10 , t= 30+ 10
AnswerMarks Guidance
10 10M1
x = 3 +1 , x = 3 −1A1
3 −1< x < 3 +1A1 Strict inequality, exact values.
4
Question 7:
--- 7(a) ---
7(a) | R2
y = 0 in trajectory equation: Rtanθ−g =0
( cosθ)2
2u2 | M1
R=)2u2sinθcosθ
(
only
g | A1 | Any equivalent single term expression, for example:
u2sin2θ 2u2tanθ
, , at least one intermediate line of
g g sec2θ
working, not just quoting a result.
SC B1 using SUVAT.
2
--- 7(b) ---
7(b) | u2sinθcosθ
x=their
and substitute in trajectory equation.
g | M1 | Or use SUVAT.
( sinθ)2
u2
H =
2g | A1 | Single term.
2
--- 7(c) ---
7(c) | 4H
Use R= and simplify: tanθ= 3, θ = 60°
3 | B1 | AG
1
Question | Answer | Marks | Guidance
--- 7(d) ---
7(d) | dy gx
=tanθ−
( cosθ)2
dx u2 | M1 | Differentiate with respect to x.
x
tanθ− =±1
used
( cosθ)2
4 | M1 | dy
Use =±1 as limiting case.
dx
x = 3 +1, x = 3 −1 | A1
3 −1< x < 3 +1 | A1 | Strict inequality, exact values.
Alternative method for question 7(d)
1 dy
y= 3x− x2, = 3−x
2 dx | M1 | Differentiate with respect to x.
dy
=±1 used
dx | M1 | dy
Use =±1 as limiting case.
dx
x = 3 +1 , x = 3 −1 | A1
3 −1< x < 3 +1 | A1 | Strict inequality, exact values.
Question | Answer | Marks | Guidance
7(d) | Alternative method for question 7(d)
v =±v
When moving at 45° to horizontal,
x y | M1 | Used, both cases considered.
v = 40 cosθ , v = 40sinθ−10t
x y
1 ( ) 1 ( )
t= 30− 10 , t= 30+ 10
10 10 | M1
x = 3 +1 , x = 3 −1 | A1
3 −1< x < 3 +1 | A1 | Strict inequality, exact values.
4
A particle $P$ is projected with speed $u$ at an angle $\theta$ above the horizontal from a point $O$ on a horizontal plane and moves freely under gravity. The horizontal and vertical displacements of $P$ from $O$ at a subsequent time $t$ are denoted by $x$ and $y$ respectively.

\begin{enumerate}[label=(\alph*)]
\item Use the equation of the trajectory given in the List of formulae (MF19), together with the condition $y = 0$, to establish an expression for the range $R$ in terms of $u$, $\theta$ and $g$. [2]
\item Deduce an expression for the maximum height $H$, in terms of $u$, $\theta$ and $g$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q7 [4]}}
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