| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Ball between two walls, successive rebounds |
| Difficulty | Challenging +1.8 This is a sophisticated two-collision mechanics problem requiring careful geometric analysis of velocity components, application of Newton's restitution law at two walls, and working backwards from the final condition. While the individual concepts (restitution, component resolution) are standard Further Maths topics, the multi-step reasoning through two sequential collisions with geometric constraints and the need to work systematically through the velocity transformations elevates this significantly above routine exercises. However, the question provides clear structure and guidance ('show that'), making it more accessible than open-ended proof problems. |
| Spec | 1.05g Exact trigonometric values: for standard angles6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| \(v\cos 45°\) // to wall and \(\frac{3}{4}v\sin 45°\) \(\perp\) to wall | M1 A1 | |
| \(\sqrt{\{(v/\sqrt{2})^2 + (\frac{3}{4}v/\sqrt{2})^2\}} = \frac{1}{4}u\) | M1 | |
| \((5/4\sqrt{2})\,v = \frac{1}{4}u\) | A1 | |
| \(v = (\sqrt{2}/5)\,u\) | A1 | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{4}u\cos\beta = v\cos 45°\) and \(\frac{1}{4}u\sin\beta = \frac{3}{4}v\sin 45°\) | (M1 A1) | |
| \(\tan\beta = \frac{3}{4}\) or \(\beta = 36.9°\) | (A1) | |
| \(\frac{1}{4}u \times (4/5) = v/\sqrt{2}\) | (M1) | |
| \(v = (\sqrt{2}/5)\,u\) | (A1) | A.G. |
| Answer | Marks |
|---|---|
| \(v\cos 75° = u\cos\alpha\) | M1 |
| \(\cos\alpha = (\sqrt{2}/5)\cos 75°\) [= \(0.0732\)] | A1 |
| \(\alpha = 85.8°\) or \(1.50\) rads | A1 |
| \(v\sin 75° = eu\sin\alpha\) | M1 |
| \(e = (\sqrt{2}/5)\sin 75° / \sin\alpha\) | |
| or \(= \tan 75° / \tan\alpha = 0.274\) | A1 |
# Question 2:
## Part (i) — EITHER method
| $v\cos 45°$ // to wall and $\frac{3}{4}v\sin 45°$ $\perp$ to wall | M1 A1 | |
|---|---|---|
| $\sqrt{\{(v/\sqrt{2})^2 + (\frac{3}{4}v/\sqrt{2})^2\}} = \frac{1}{4}u$ | M1 | |
| $(5/4\sqrt{2})\,v = \frac{1}{4}u$ | A1 | |
| $v = (\sqrt{2}/5)\,u$ | A1 | A.G. |
**OR:**
| $\frac{1}{4}u\cos\beta = v\cos 45°$ and $\frac{1}{4}u\sin\beta = \frac{3}{4}v\sin 45°$ | (M1 A1) | |
|---|---|---|
| $\tan\beta = \frac{3}{4}$ or $\beta = 36.9°$ | (A1) | |
| $\frac{1}{4}u \times (4/5) = v/\sqrt{2}$ | (M1) | |
| $v = (\sqrt{2}/5)\,u$ | (A1) | A.G. |
## Part (ii)
| $v\cos 75° = u\cos\alpha$ | M1 | |
|---|---|---|
| $\cos\alpha = (\sqrt{2}/5)\cos 75°$ [= $0.0732$] | A1 | |
| $\alpha = 85.8°$ or $1.50$ rads | A1 | |
| $v\sin 75° = eu\sin\alpha$ | M1 | |
| $e = (\sqrt{2}/5)\sin 75° / \sin\alpha$ | | |
| or $= \tan 75° / \tan\alpha = 0.274$ | A1 | |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{184020e1-7ff2-4172-8d33-baff963afa76-2_531_760_927_696}
Two smooth vertical walls each with their base on a smooth horizontal surface intersect at an angle of $60 ^ { \circ }$. A small smooth sphere $P$ is moving on the horizontal surface with speed $u$ when it collides with the first vertical wall at the point $D$. The angle between the direction of motion of $P$ and the wall is $\alpha ^ { \circ }$ before the collision and $75 ^ { \circ }$ after the collision. The speed of $P$ after this collision is $v$ and the coefficient of restitution between $P$ and the first wall is $e$. Sphere $P$ then collides with the second vertical wall at the point $E$. The speed of $P$ after this second collision is $\frac { 1 } { 4 } u$ (see diagram). The coefficient of restitution between $P$ and the second wall is $\frac { 3 } { 4 }$.\\
(i) By considering the collision at $E$, show that $v = \frac { \sqrt { } 2 } { 5 } u$.\\
(ii) Find the value of $\alpha$ and the value of $e$.
\hfill \mbox{\textit{CAIE FP2 2016 Q2 [10]}}