| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Topic | Oblique and successive collisions |
| Type | Sphere rebounds off fixed wall obliquely |
| Difficulty | Challenging +1.2 This is a standard oblique collision problem requiring resolution of velocities parallel and perpendicular to the wall, application of the coefficient of restitution formula, and geometric reasoning with angles. Part (i) involves straightforward algebra with given angles, while part (ii) requires working backwards to show a specific result. The problem is more involved than basic mechanics questions but follows well-established techniques for oblique impacts without requiring novel insight. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
(i) Momentum: $u\cos\theta = v\sin\theta$ (Method mark can be awarded in part (ii), if not here) **M1A1**
NEL: $eu\sin\theta = v\cos\theta$ (Method mark can be awarded in part (ii), if not here) **M1A1**
$e\frac{\sin\theta}{\cos\theta} = \frac{\cos\theta}{\sin\theta} \Rightarrow e = \frac{1}{\tan^2\theta}$ **M1**
$\Rightarrow e = \frac{1}{9}$ **A1** [6]
(ii) Momentum: $u\cos\theta = v\cos(135° - \theta) = v\sin(\theta - 45°)$ **B1**
NEL: $\frac{2}{3}u\sin\theta = v\cos(\theta - 45°)$ **B1**
Dividing: $\frac{2}{3}\tan\theta = \frac{\cos\theta\cos45° + \sin\theta\sin45°}{\sin\theta\cos45° - \cos\theta\sin45°} = \frac{1 + \tan\theta}{\tan\theta - 1}$ **DM1**
$\Rightarrow 2(t^2 - t) = 3 + 3t$ (where $t = \tan\theta$) $\Rightarrow 2t^2 - 5t - 3 = 0$ **A1**
$(2t+1)(t-3) = 0 \Rightarrow t = 3$ or $-\frac{1}{2}$. Since $\theta$ is acute, $\theta = \tan^{-1}3$ **DM1A1** [6]12\\
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A small smooth sphere is projected from a point $A$ across a smooth horizontal surface. The sphere strikes a smooth vertical wall at the point $P$. The acute angle between the direction of motion of the sphere and the wall is $\theta$. After the impact, the sphere passes through the point $B$, where angle $A P B = \phi$ (see diagram). The coefficient of restitution between the sphere and the wall is $e$.\\
(i) Given that $\theta = \tan ^ { - 1 } 3$ and $\phi = 90 ^ { \circ }$, find the exact value of $e$.\\
(ii) Given instead that $e = \frac { 2 } { 3 }$ and $\phi = 45 ^ { \circ }$, show that $\theta = \tan ^ { - 1 } 3$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q12 [6]}}