| 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 challenging two-collision mechanics problem requiring careful geometric analysis of velocity components, application of restitution formulas at two walls, and working backwards from the final speed. It demands strong spatial reasoning with the 60° wall angle and systematic resolution of velocities, placing it well above average difficulty but within reach of well-prepared Further Maths students. |
| 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 | Find components of speed after collision at \(E\) |
| \(\sqrt{\{(v/\sqrt{2})^2 + (\frac{3}{4}v/\sqrt{2})^2\}} = \frac{1}{4}u\) | M1 | Relate \(v\) to \(u\), or \(v^2\) to \(u^2\) |
| \((5/4\sqrt{2})\,v = \frac{1}{4}u\) | 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) | Relate angle \(\beta\) after collision to \(u\), \(v\) |
| \(\tan\beta = \frac{3}{4}\) *or* \(\beta = 36.9°\) | (A1) | Find \(\tan\beta\) or \(\beta\) |
| \(\frac{1}{4}u \times (4/5) = v/\sqrt{2}\) | (M1) | Eliminate \(\beta\) |
| \(v = (\sqrt{2}/5)\,u\) | (A1) | A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| \(v\cos 75° = u\cos\alpha\) | M1 | Relate components of speed // to wall after collision at \(D\) |
| \(\cos\alpha = (\sqrt{2}/5)\cos 75°\) \([= 0.0732]\) | A1 | Find \(\cos\alpha\) |
| \(\alpha = 85.8°\) *or* \(1.50\) rads | A1 | Find \(\alpha\) |
| \(v\sin 75° = eu\sin\alpha\) | M1 | Relate components of speed \(\perp\) to wall after collision at \(D\) |
| \(e = (\sqrt{2}/5)\sin 75° / \sin\alpha\) | A1 | |
| *or* \(= \tan 75° / \tan\alpha = 0.274\) |
# Question 2:
## Part (i) — EITHER method
| $v\cos 45°$ // to wall *and* $\frac{3}{4}v\sin 45°$ $\perp$ to wall | M1 A1 | Find components of speed after collision at $E$ |
|---|---|---|
| $\sqrt{\{(v/\sqrt{2})^2 + (\frac{3}{4}v/\sqrt{2})^2\}} = \frac{1}{4}u$ | M1 | Relate $v$ to $u$, or $v^2$ to $u^2$ |
| $(5/4\sqrt{2})\,v = \frac{1}{4}u$ | 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) | Relate angle $\beta$ after collision to $u$, $v$ |
|---|---|---|
| $\tan\beta = \frac{3}{4}$ *or* $\beta = 36.9°$ | (A1) | Find $\tan\beta$ or $\beta$ |
| $\frac{1}{4}u \times (4/5) = v/\sqrt{2}$ | (M1) | Eliminate $\beta$ |
| $v = (\sqrt{2}/5)\,u$ | (A1) | A.G. |
## Part (ii)
| $v\cos 75° = u\cos\alpha$ | M1 | Relate components of speed // to wall after collision at $D$ |
|---|---|---|
| $\cos\alpha = (\sqrt{2}/5)\cos 75°$ $[= 0.0732]$ | A1 | Find $\cos\alpha$ |
| $\alpha = 85.8°$ *or* $1.50$ rads | A1 | Find $\alpha$ |
| $v\sin 75° = eu\sin\alpha$ | M1 | Relate components of speed $\perp$ to wall after collision at $D$ |
| $e = (\sqrt{2}/5)\sin 75° / \sin\alpha$ | A1 | |
| *or* $= \tan 75° / \tan\alpha = 0.274$ | | |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{62d0d8cb-8f8c-4298-9705-71a735a9a4e7-2_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]}}