| 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 restitution laws at two walls, and working backwards from the final condition. While the individual collision mechanics are standard Further Maths content, the geometric setup with angled walls and the need to track velocity components through two sequential collisions with different restitution coefficients requires strong spatial reasoning and systematic algebraic manipulation. This is significantly harder than routine single-collision problems but doesn't require truly novel insights beyond careful application of standard principles. |
| 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(v\cos 45°\) // to wall and \(\frac{3}{4}v\sin 45°\ \perp\) to wall | M1 A1 | Find comps. of speed after colln. 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 colln. 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 |
|---|---|---|
| Answer | Marks | Guidance |
| \(v\cos 75° = u\cos\alpha\) | M1 | Relate comps. of speed // to wall after colln. 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 comps. of speed \(\perp\) to wall after colln. at \(D\) |
| \(e = (\sqrt{2}/5)\sin 75°/\sin\alpha\) or \(= \tan 75°/\tan\alpha = 0.274\) | A1 | Find \(e\) |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v\cos 45°$ // to wall and $\frac{3}{4}v\sin 45°\ \perp$ to wall | M1 A1 | Find comps. of speed after colln. 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 colln. 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)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v\cos 75° = u\cos\alpha$ | M1 | Relate comps. of speed // to wall after colln. 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 comps. of speed $\perp$ to wall after colln. at $D$ |
| $e = (\sqrt{2}/5)\sin 75°/\sin\alpha$ or $= \tan 75°/\tan\alpha = 0.274$ | A1 | Find $e$ |
---2\\
\includegraphics[max width=\textwidth, alt={}, center]{58728f93-bfdb-4f76-a9b9-3a1d1592bfc9-2_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]}}