CAIE Further Paper 3 2024 November — Question 7 6 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProjectiles
TypeProjectile with bounce or impact
DifficultyChallenging +1.2 This is a multi-step projectiles problem involving coefficient of restitution and energy/kinematics after a bounce. It requires setting up equations for the rebound velocity using e, then applying projectile motion formulas to find maximum height, but follows standard mechanics procedures without requiring novel geometric insight or proof techniques.
Spec6.03k Newton's experimental law: direct impact
In its subsequent motion, the greatest height reached by \(P\) above \(A\) is \(\frac{3}{10}\) of the vertical height of \(A\) above the horizontal plane.
  1. Find the value of \(e\). [6]
Question 7:
AnswerMarks
7(a)When P strikes plane, velocity is →ucos,
3 3
Before impact: parallel to inclined plane ucos, perpendicular to plane usin
AnswerMarks Guidance
5 5M1 3
u
5
3 3
After impact: components ucos (parallel) and eusin (perpendicular)
AnswerMarks
5 5A1
3 3
Since velocity is vertical after impact, tan= ucos/ eusin
AnswerMarks Guidance
5 5M1
tan=1/etan, e tan2 =1A1 AG
4
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
7(b)(usin)2 8u2
Greatest height of P before impact: H = =
AnswerMarks Guidance
2g 25gM1A1 Note: alternative methods.
3
After impact, vertical speed of P is u (cos)2 +e2(sin)2
AnswerMarks
5M1
3
Use V2 =U2 +2as to greatest height, equal to H
16
9 ( ) 3
u2 (cos)2 +e2(sin)2 =2g H
AnswerMarks
25 16M1
1 e 1
Use part (a): tan= , cos= , sin=
e 1+e 1+e
AnswerMarks
Substitute to find eM1
1
3e2 +2e−1=0, e=
AnswerMarks
3A1
6
Question 7:
--- 7(a) ---
7(a) | When P strikes plane, velocity is →ucos,
3 3
Before impact: parallel to inclined plane ucos, perpendicular to plane usin
5 5 | M1 | 3
u
5
3 3
After impact: components ucos (parallel) and eusin (perpendicular)
5 5 | A1
3 3
Since velocity is vertical after impact, tan= ucos/ eusin
5 5 | M1
tan=1/etan, e tan2 =1 | A1 | AG
4
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | (usin)2 8u2
Greatest height of P before impact: H = =
2g 25g | M1A1 | Note: alternative methods.
3
After impact, vertical speed of P is u (cos)2 +e2(sin)2
5 | M1
3
Use V2 =U2 +2as to greatest height, equal to H
16
9 ( ) 3
u2 (cos)2 +e2(sin)2 =2g H
25 16 | M1
1 e 1
Use part (a): tan= , cos= , sin=
e 1+e 1+e
Substitute to find e | M1
1
3e2 +2e−1=0, e=
3 | A1
6
In its subsequent motion, the greatest height reached by $P$ above $A$ is $\frac{3}{10}$ of the vertical height of $A$ above the horizontal plane.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $e$. [6]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q7 [6]}}
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Q1 5 Q2 5 Q3 6 Q3 2 Q4 4 Q4 3 Q5 4 Q5 3 Q6 3 Q6 3 Q6 2 Q7 4 Q7 6