| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Oblique collision of spheres |
| Difficulty | Challenging +1.8 This is an advanced 2D collision problem requiring conservation of momentum in two perpendicular directions plus Newton's restitution law. The constraint that B moves perpendicular to the line of centres after collision provides the key to eliminating unknowns and deriving the given relationship. While systematic, it requires careful vector decomposition, algebraic manipulation across three equations, and is non-trivial compared to standard momentum questions. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or |
| Answer | Marks |
|---|---|
| 5(a) | Let w be speed of A along line of centres after collision |
| ← mw=−mucosα+musinα | M1 |
| w−0=e(ucosα+usinα) | M1 |
| Answer | Marks |
|---|---|
| sinα( u−eu )=cosα( u+eu ) | M1 |
| Answer | Marks |
|---|---|
| 1−e | A1 |
| Answer | Marks |
|---|---|
| 5(b) | 1 |
| Answer | Marks |
|---|---|
| 3 | B1 |
| Answer | Marks |
|---|---|
| 3 5 5 5 | M1 |
| Answer | Marks |
|---|---|
| Speed = w2 | M1 |
| Answer | Marks |
|---|---|
| 5 5 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | Let w be speed of A along line of centres after collision
← mw=−mucosα+musinα | M1
w−0=e(ucosα+usinα) | M1
Rearrange:
sinα( u−eu )=cosα( u+eu ) | M1
1+e
tanα= . AG
1−e | A1
4
--- 5(b) ---
5(b) | 1
tanα=2e=
3 | B1
1 1 2 u
w= u + =
3 5 5 5 | M1
+( usinα)2
Speed = w2 | M1
u2 4u2
= + =u
5 5 | A1
4
Question | Answer | Marks\includegraphics{figure_5}
Two uniform smooth spheres $A$ and $B$ of equal radii each have mass $m$. The two spheres are each moving with speed $u$ on a horizontal surface when they collide. Immediately before the collision $A$'s direction of motion makes an angle of $\alpha°$ with the line of centres, and $B$'s direction of motion is perpendicular to that of $A$ (see diagram). The coefficient of restitution between the spheres is $e$.
Immediately after the collision, $B$ moves in a direction at right angles to the line of centres.
\begin{enumerate}[label=(\alph*)]
\item Show that $\tan \alpha = \frac{1+e}{1-e}$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q5 [4]}}