CAIE Further Paper 3 2023 November — Question 1 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImpulse and momentum (advanced)
TypeOblique collision of spheres
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring conservation of momentum along the line of centres, Newton's experimental law, perpendicular component conservation, and the kinetic energy condition. It involves multiple simultaneous equations with trigonometry, but follows a standard oblique collision framework with clear signposting through the parts. The algebraic manipulation is substantial but methodical rather than requiring novel insight.
Spec6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts
\includegraphics{figure_1} Two uniform smooth spheres \(A\) and \(B\) of equal radii have masses \(m\) and \(2m\) respectively. The two spheres are moving with equal speeds \(u\) on a smooth horizontal surface when they collide. Immediately before the collision, \(A\)'s direction of motion makes an angle of \(60°\) with the line of centres, and \(B\)'s direction of motion makes an angle \(\theta\) with the line of centres (see diagram). The coefficient of restitution between the spheres is \(e\). After the collision, the component of the velocity of \(A\) along the line of centres is \(v\) and \(B\) moves perpendicular to the line of centres. Sphere \(A\) now has twice as much kinetic energy as sphere \(B\).
  1. Show that \(v = \frac{1}{2}u(4\cos\theta - 1)\). [1]
  2. Find the value of \(\cos\theta\). [4]
  3. Find the value of \(e\). [2]
Question 1:
AnswerMarks
1(a)PCLM: mv=−mucos60+2mucos
1 1
v=− u+2ucos=v= u(4cos−1)
AnswerMarks Guidance
2 2B1 First line must be seen.
AG
1
AnswerMarks
1(b)1 ( )
KE of A = m v2 +(usin60)2
AnswerMarks
2B1
KE of A = 2  KE of B, so
1 ( ) 1
m v2 +(usin60)2 =2 2m(usin)2
AnswerMarks
2 2M1
2
1  3
 u(4cos−1)  + u2 =4u2(sin)2
2  4
AnswerMarks Guidance
8cos2−2cos−3=0M1 Use result of (a) and rearrange.
Obtain 3-term quadratic.
3 1 3
cos= , − but angle is acute, so cos=
AnswerMarks
4 2 4A1
4
AnswerMarks Guidance
1(c)NEL: v=e(ucos60+ucos) M1
1 3 
v=u, u=e u+ u
 
2 4 
4
e=
AnswerMarks
5A1
2
AnswerMarks Guidance
QuestionAnswer Marks 
Question 1:
--- 1(a) ---
1(a) | PCLM: mv=−mucos60+2mucos
1 1
v=− u+2ucos=v= u(4cos−1)
2 2 | B1 | First line must be seen.
AG
1
--- 1(b) ---
1(b) | 1 ( )
KE of A = m v2 +(usin60)2
2 | B1
KE of A = 2  KE of B, so
1 ( ) 1
m v2 +(usin60)2 =2 2m(usin)2
2 2 | M1
2
1  3
 u(4cos−1)  + u2 =4u2(sin)2
2  4
8cos2−2cos−3=0 | M1 | Use result of (a) and rearrange.
Obtain 3-term quadratic.
3 1 3
cos= , − but angle is acute, so cos=
4 2 4 | A1
4
--- 1(c) ---
1(c) | NEL: v=e(ucos60+ucos) | M1
1 3 
v=u, u=e u+ u
 
2 4 
4
e=
5 | A1
2
Question | Answer | Marks | Guidance
\includegraphics{figure_1}

Two uniform smooth spheres $A$ and $B$ of equal radii have masses $m$ and $2m$ respectively. The two spheres are moving with equal speeds $u$ on a smooth horizontal surface when they collide. Immediately before the collision, $A$'s direction of motion makes an angle of $60°$ with the line of centres, and $B$'s direction of motion makes an angle $\theta$ with the line of centres (see diagram). The coefficient of restitution between the spheres is $e$.

After the collision, the component of the velocity of $A$ along the line of centres is $v$ and $B$ moves perpendicular to the line of centres. Sphere $A$ now has twice as much kinetic energy as sphere $B$.

\begin{enumerate}[label=(\alph*)]
\item Show that $v = \frac{1}{2}u(4\cos\theta - 1)$. [1]

\item Find the value of $\cos\theta$. [4]

\item Find the value of $e$. [2]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q1 [7]}}
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