CAIE Further Paper 3 2024 November — Question 3 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
DifficultyChallenging +1.8 This is an advanced 2D collision problem requiring careful decomposition of velocities along and perpendicular to the line of centres, application of both momentum conservation and the coefficient of restitution equation, and multi-step reasoning across three parts. While the techniques are standard for Further Mechanics, the geometric setup with perpendicular initial velocities and the constraint that B moves perpendicular to the line of centres after collision requires careful analysis and algebraic manipulation beyond typical A-level questions.
Spec6.03i Coefficient of restitution: e6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
\includegraphics{figure_3} The diagram shows two identical smooth uniform spheres \(A\) and \(B\) of equal radii and each of mass \(m\). The two spheres are moving on a smooth horizontal surface when they collide with speeds \(2u\) and \(3u\) respectively. Immediately before the collision, \(A\)'s direction of motion makes an angle \(\theta\) with the line of centres and \(B\)'s direction of motion is perpendicular to that of \(A\). After the collision, \(B\) moves perpendicular to the line of centres. The coefficient of restitution between the spheres is \(\frac{1}{3}\).
  1. Find the value of \(\tan \theta\). [3]
  2. Find the total loss of kinetic energy as a result of the collision. [2]
  3. Find, in degrees, the angle through which the direction of motion of \(A\) is deflected as a result of the collision. [2]
Question 3:
AnswerMarks Guidance
3(a)PCLM along line of centres: −mv=m2ucos−m3usin B1
accept positive or negative v.
If velocity of B after collision is included, it must
be equated to zero before this mark is awarded.
AnswerMarks Guidance
NEL: v=eu(3sin+2cos)M1 Must have plus sign on RHS,
accept positive or negative v (sign of v does not
need to be consistent with PCLM equation)
If velocity of B after collision is included, it must
be equated to zero before this mark is awarded.
4
Eliminate v: 6sin=8cos, tan=
AnswerMarks Guidance
3A1 Correct work only, except possibly missing m.
3
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks
3(b)Only change in KE is along line of centres
1 ( ) 1
Loss = m (2ucos)2 +(3usin)2 − mv2
AnswerMarks
2 2M1
 2
1 36 144 12 6 72
mu2 + − −  = mu2
 
2 25 25  5 5 25
AnswerMarks Guidance
 A1 (Note that tan= 2 leads to final answer 36mu2)
3 13
2
Alternative method for question 3(b)
Alternative method, using total KE
1 1
Loss in KE = [ m(2u)2 + m(3u)2 ]
2 2
1 1 1 
− mv2 + m(2usin)2 + m(3ucos)2
 
AnswerMarks Guidance
2 2 2 M1 Or equivalent, with all necessary terms present
13 181 72
mu2 − mu2 = mu2
AnswerMarks
2 50 25A1
2
AnswerMarks
3(c)6 8
[Components of velocity of A after collision are  u  u so] angle between
5 5
AnswerMarks Guidance
line of centres and A’s direction is .M1
Angle of deflection = 180 – 2tan-1 4/3 = 73.7A1FT FT their answer to part (a)
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | PCLM along line of centres: −mv=m2ucos−m3usin | B1 | Must include m, must have minus sign on RHS,
accept positive or negative v.
If velocity of B after collision is included, it must
be equated to zero before this mark is awarded.
NEL: v=eu(3sin+2cos) | M1 | Must have plus sign on RHS,
accept positive or negative v (sign of v does not
need to be consistent with PCLM equation)
If velocity of B after collision is included, it must
be equated to zero before this mark is awarded.
4
Eliminate v: 6sin=8cos, tan=
3 | A1 | Correct work only, except possibly missing m.
3
Question | Answer | Marks | Guidance
--- 3(b) ---
3(b) | Only change in KE is along line of centres
1 ( ) 1
Loss = m (2ucos)2 +(3usin)2 − mv2
2 2 | M1
 2
1 36 144 12 6 72
mu2 + − −  = mu2
 
2 25 25  5 5 25
  | A1 | (Note that tan= 2 leads to final answer 36mu2)
3 13
2
Alternative method for question 3(b)
Alternative method, using total KE
1 1
Loss in KE = [ m(2u)2 + m(3u)2 ]
2 2
1 1 1 
− mv2 + m(2usin)2 + m(3ucos)2
 
2 2 2  | M1 | Or equivalent, with all necessary terms present
13 181 72
mu2 − mu2 = mu2
2 50 25 | A1
2
--- 3(c) ---
3(c) | 6 8
[Components of velocity of A after collision are  u  u so] angle between
5 5
line of centres and A’s direction is . | M1
Angle of deflection = 180 – 2tan-1 4/3 = 73.7 | A1FT | FT their answer to part (a)
2
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

The diagram shows two identical smooth uniform spheres $A$ and $B$ of equal radii and each of mass $m$. The two spheres are moving on a smooth horizontal surface when they collide with speeds $2u$ and $3u$ respectively. Immediately before the collision, $A$'s direction of motion makes an angle $\theta$ with the line of centres and $B$'s direction of motion is perpendicular to that of $A$. After the collision, $B$ moves perpendicular to the line of centres. The coefficient of restitution between the spheres is $\frac{1}{3}$.

\begin{enumerate}[label=(\alph*)]
\item Find the value of $\tan \theta$. [3]
\item Find the total loss of kinetic energy as a result of the collision. [2]
\item Find, in degrees, the angle through which the direction of motion of $A$ is deflected as a result of the collision. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q3 [7]}}
This paper (7 questions)
View full paper
Q1 3 Q2 4 Q3 7 Q4 7 Q5 8 Q6 10 Q7 11