CAIE Further Paper 3 2024 June — Question 1 6 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
DifficultyChallenging +1.8 This is an advanced mechanics problem requiring simultaneous application of conservation of momentum (in two directions), Newton's law of restitution, and the kinetic energy condition. While the concepts are standard for Further Maths mechanics, the problem requires careful algebraic manipulation of multiple equations with trigonometric components and the insight to use the equal kinetic energy condition effectively. The 6-mark allocation and multi-constraint nature place it above average difficulty, though it follows a recognizable problem-solving framework for oblique collisions.
Spec6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts

\includegraphics{figure_1} Two smooth uniform spheres \(A\) and \(B\) of equal radii have masses \(m\) and \(5m\) respectively. Sphere \(A\) is moving on a smooth horizontal surface with speed \(u\) when it collides with sphere \(B\) which is at rest on the surface. Immediately before the collision, \(A\)'s direction of motion makes an angle of \(\theta\) with the line of centres. After the collision, the kinetic energies of \(A\) and \(B\) are equal. The coefficient of restitution between the spheres is \(\frac{1}{3}\). Find the value of \(\tan\theta\). [6]

Question 1:
AnswerMarks Guidance
1Along line of centres, PCLM: 5mv mv mucos
B AM1 Must include correct masses.
NEL: v v  1 ucos
AnswerMarks Guidance
B A 2M1 Signs consistent with PCLM equation.
u u
v  cos, v  cos
AnswerMarks
B 4 A 4A1
Perpendicular to line of centres: speed of A is usinB1
1   u  2 usin2  1  u  2
m    cos   5m  cos 
2   4   2 4 
AnswerMarks Guidance
 M1 Equate final kinetic energies, 3 terms, correct
masses.
cos2  4 , cos 2 , tan 1
AnswerMarks
5 5 2A1
6
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
1 | Along line of centres, PCLM: 5mv mv mucos
B A | M1 | Must include correct masses.
NEL: v v  1 ucos
B A 2 | M1 | Signs consistent with PCLM equation.
u u
v  cos, v  cos
B 4 A 4 | A1
Perpendicular to line of centres: speed of A is usin | B1
1   u  2 usin2  1  u  2
m    cos   5m  cos 
2   4   2 4 
  | M1 | Equate final kinetic energies, 3 terms, correct
masses.
cos2  4 , cos 2 , tan 1
5 5 2 | A1
6
Question | Answer | Marks | Guidance
\includegraphics{figure_1}

Two smooth uniform spheres $A$ and $B$ of equal radii have masses $m$ and $5m$ respectively. Sphere $A$ is moving on a smooth horizontal surface with speed $u$ when it collides with sphere $B$ which is at rest on the surface. Immediately before the collision, $A$'s direction of motion makes an angle of $\theta$ with the line of centres. After the collision, the kinetic energies of $A$ and $B$ are equal. The coefficient of restitution between the spheres is $\frac{1}{3}$.

Find the value of $\tan\theta$. [6]

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q1 [6]}}