CAIE Further Paper 3 2023 June — Question 2 5 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
DifficultyChallenging +1.2 This is a standard Further Maths mechanics problem requiring application of coefficient of restitution, conservation of momentum parallel to the barrier, and energy loss. While it involves multiple equations and algebraic manipulation across three conditions, the approach is methodical and follows well-established techniques for oblique collisions. The given information (tan θ and energy loss percentage) makes the problem more structured than open-ended, placing it moderately above average difficulty.
Spec6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
\includegraphics{figure_2} A particle \(P\) of mass \(m\) is moving with speed \(u\) on a fixed smooth horizontal surface. It collides at an angle \(\alpha\) with a fixed smooth vertical barrier. After the collision, \(P\) moves at an angle \(\theta\) with the barrier, where \(\tan\theta = \frac{1}{3}\) (see diagram). The coefficient of restitution between \(P\) and the barrier is \(e\). The particle \(P\) loses 20% of its kinetic energy as a result of the collision. Find the value of \(e\). [5]
Question 2:
AnswerMarks Guidance
2Parallel to wall vcosucos
Perpendicular to wall vsineusinM1 Both
1
Dividing, e
AnswerMarks Guidance
2tanA1 AEF
KE reduced by 20%, so 1mu2 cos2e2sin2   41mu2
AnswerMarks Guidance
2 5 2M1 Dimensionally correct equation in u or v, but not
both. Must have either  or , but not both.
Must see 4 on the correct side of the equation.
5
4
Eliminate e: cos
AnswerMarks
5A1
2
e
AnswerMarks Guidance
3A1
QuestionAnswer Marks
2Alternative method for question 2
Parallel to wall vcosucos
AnswerMarks Guidance
Perpendicular to wall vsineusinM1 Both
 5 2 5 5v 2 5v
sin , cos  usin , ucos
5 5 5e 5
AnswerMarks
 A1
4v2 v2
u2 u2cos2u2sin2 
AnswerMarks Guidance
  5 5e2A1 v2 1 
AEF, e.g.  4 
5  e2 
1 4 1 2 v2  1 
mv2   mu2  m 4
AnswerMarks Guidance
2  5 2  5 5   e2  M1 Dimensionally correct equation in v.
Must have either  or , but not both.
4 2
Must see or on the correct side of the
5 5
equation.
2
e
AnswerMarks
3A1
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
2 | Parallel to wall vcosucos
Perpendicular to wall vsineusin | M1 | Both
1
Dividing, e
2tan | A1 | AEF
KE reduced by 20%, so 1mu2 cos2e2sin2   41mu2
2 5 2 | M1 | Dimensionally correct equation in u or v, but not
both. Must have either  or , but not both.
Must see 4 on the correct side of the equation.
5
4
Eliminate e: cos
5 | A1
2
e
3 | A1
Question | Answer | Marks | Guidance
2 | Alternative method for question 2
Parallel to wall vcosucos
Perpendicular to wall vsineusin | M1 | Both
 5 2 5 5v 2 5v
sin , cos  usin , ucos
5 5 5e 5
  | A1
4v2 v2
u2 u2cos2u2sin2 
  5 5e2 | A1 | v2 1 
AEF, e.g.  4 
5  e2 
1 4 1 2 v2  1 
mv2   mu2  m 4
2  5 2  5 5   e2   | M1 | Dimensionally correct equation in v.
Must have either  or , but not both.
4 2
Must see or on the correct side of the
5 5
equation.
2
e
3 | A1
5
Question | Answer | Marks | Guidance
\includegraphics{figure_2}

A particle $P$ of mass $m$ is moving with speed $u$ on a fixed smooth horizontal surface. It collides at an angle $\alpha$ with a fixed smooth vertical barrier. After the collision, $P$ moves at an angle $\theta$ with the barrier, where $\tan\theta = \frac{1}{3}$ (see diagram). The coefficient of restitution between $P$ and the barrier is $e$. The particle $P$ loses 20% of its kinetic energy as a result of the collision.

Find the value of $e$. [5]

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q2 [5]}}
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