| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Collision then wall impact |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring two collision analyses with different geometries. Part (a) is standard direct collision mechanics (3 marks), but part (b) requires resolving velocities for oblique wall collision, applying restitution in the normal direction only, then using the kinetic energy condition to form and solve an equation involving e. The multi-stage reasoning, oblique collision geometry, and algebraic manipulation of the energy ratio make this substantially harder than typical A-level questions, though the individual techniques are within Further Maths scope. |
| Spec | 6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks |
|---|---|
| 6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | mu =mw+2mv | B1 |
| v−w= eu | B1 | Restitution with consistent signs |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | B1 | Both correct. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | Perpendicular to plane: y=evsinθ | |
| Parallel to plane: x=vcosθ | B1 | Both |
| Answer | Marks | Guidance |
|---|---|---|
| | M1 | Speed of B |
| Answer | Marks | Guidance |
|---|---|---|
| 2 5 9 | M1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 9 32 2 5 9 | M1 A1 | Relate the two KEs |
| 4 ( 1−2e )2 =( e+1 )2 or 15e2 −18e+3=0 | M1 | Rearrange and simplify to quadratic |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | A1 | Both values |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/31 Cambridge International AS & A Level – Mark Scheme October/November 2020
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2020 Page 5 of 14
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | mu =mw+2mv | B1 | Momentum equation (with m)
v−w= eu | B1 | Restitution with consistent signs
v= u( e+1 )
3
w= u( 1−2e )
3 | B1 | Both correct.
3
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | Perpendicular to plane: y=evsinθ
Parallel to plane: x=vcosθ | B1 | Both
4 2 3 2 2
Speed of B = x2 + y2 = v2( )2 + . (= v)
5 3 5 5
| M1 | Speed of B
KE of B = 1 .2m 4 . u2 ( e+1 )2
2 5 9 | M1 | 1
KE of B in terms of u. and 2m needed
2
KE of A = 1 .m. u2 ( 1−2e )2
2 9
So 1 .m. u2 ( 1−2e )2 = 5 . 1 .2m 4 . u2 ( e+1 )2
2 9 32 2 5 9 | M1 A1 | Relate the two KEs
4 ( 1−2e )2 =( e+1 )2 or 15e2 −18e+3=0 | M1 | Rearrange and simplify to quadratic
1+e=±2 ( 1−2e )
1
e= , 1
5 | A1 | Both values
7
Question | Answer | Marks | GuidanceTwo smooth spheres $A$ and $B$ have equal radii and masses $m$ and $2m$ respectively. Sphere $B$ is at rest on a smooth horizontal floor. Sphere $A$ is moving on the floor with velocity $u$ and collides directly with $B$. The coefficient of restitution between the spheres is $e$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $u$ and $e$, the velocities of $A$ and $B$ after the collision. [3]
\end{enumerate}
Subsequently, $B$ collides with a fixed vertical wall which makes an angle $\theta$ with the direction of motion of $B$, where $\tan\theta = \frac{3}{4}$.
The coefficient of restitution between $B$ and the wall is $\frac{2}{3}$. Immediately after $B$ collides with the wall, the kinetic energy of $B$ is $\frac{5}{27}$ of the kinetic energy of $B$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the possible values of $e$. [7]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q6 [10]}}