| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Two wall collisions sequence |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring systematic application of coefficient of restitution in 2D, resolving velocities parallel and perpendicular to walls, and connecting energy loss to restitution. Part (a) is standard technique, part (b) requires geometric insight about the symmetry of rebounds, and part (c) involves simultaneous equations linking energy, angles, and restitution. The multi-step reasoning and need to synthesize several concepts places it well above average difficulty. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | ucosα=vcosβ | M1 |
| eusinα=vsinβ | M1 | |
| Divide: tanβ=etanα | A1 | AG. Must see divide OE. |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | vsinβ=wcosγ (=eusinα) | M1 |
| evcosβ=wsinγ (=eucosα) | M1 | |
| Divide: tanγ=1/tanα: γ=90°−α | *A1 | |
| After second rebound, direction of motion is parallel to initial path. | DB1 |
| Answer | Marks |
|---|---|
| 7(c) | ( ) |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 | Energy expression in terms of u. |
| Answer | Marks |
|---|---|
| 2 9 2 3 | A1 |
| Answer | Marks |
|---|---|
| Part (a) gives tan | M1 |
| So tanα= 3 | A1 |
Question 7:
--- 7(a) ---
7(a) | ucosα=vcosβ | M1
eusinα=vsinβ | M1
Divide: tanβ=etanα | A1 | AG. Must see divide OE.
3
--- 7(b) ---
7(b) | vsinβ=wcosγ (=eusinα) | M1
evcosβ=wsinγ (=eucosα) | M1
Divide: tanγ=1/tanα: γ=90°−α | *A1
After second rebound, direction of motion is parallel to initial path. | DB1
4
--- 7(c) ---
7(c) | ( )
Final KE = 1 m ( eusinα)2 +( eucosα)2 = 1 me2u2
2 2 | M1 | Energy expression in terms of u.
1 1 1 1
So me2u2 = × mu2 giving e=
2 9 2 3 | A1
( 90−α)=etanα
Part (a) gives tan | M1
So tanα= 3 | A1
4\includegraphics{figure_7}
The smooth vertical walls $AB$ and $CB$ are at right angles to each other. A particle $P$ is moving with speed $u$ on a smooth horizontal floor and strikes the wall $CB$ at an angle $\alpha$. It rebounds at an angle $\beta$ to the wall $CB$. The particle then strikes the wall $AB$ and rebounds at an angle $\gamma$ to that wall (see diagram). The coefficient of restitution between each wall and $P$ is $e$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \beta = e \tan \alpha$. [3]
\item Express $\gamma$ in terms of $\alpha$ and explain what this result means about the final direction of motion of $P$. [4]
\end{enumerate}
As a result of the two impacts the particle loses $\frac{8}{9}$ of its initial kinetic energy.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Given that $\alpha + \beta = 90°$, find the value of $e$ and the value of $\tan \alpha$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q7 [11]}}