| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, find velocities/angles |
| Difficulty | Challenging +1.8 This is a challenging oblique collision problem requiring resolution of velocities, application of Newton's experimental law along the line of centres, and conservation of momentum perpendicular to it. Part (ii) requires algebraic proof for all values of e, demonstrating that the x-component remains positive. While the techniques are standard for Further Maths mechanics, the oblique nature, the proof element, and the need to work systematically with both components make this significantly harder than routine collision questions. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
$\text{Mom}^m(\rightarrow)$: $mu\cos30 = mw + mx$
$\text{Rest}^n$: $x - w = 0.9\sqrt{3}u/2$
Solve: $w = 0.0433u$
$\text{Mom}^m(\uparrow)$: $mu\sin30 = vm$ so $v = 0.5u$
$\Rightarrow$ direction is $\tan^{-1}(0.5/0.0433) = \mathbf{85.05°}$ to $x$-axis
- M1: C of M equation, needn't have $m$, ignore signs, needs cos and sin
- M1: Restitution equation, ignore signs of LHS
- A1: Correctly obtain $w =$ a.r.t. $0.0433u\ [=\sqrt{3}u/40]$
- B1: Obtain, state, or use $v = u/2$
- A1: Direction, $[85.0°, 85.1°]$ to $x$-axis ($5°$ or $4.9°$ to $y$)
**[5]**
(ii) $u\cos30 = w + x,\quad ue\cos30 = x - w$
$\Rightarrow 2w = u\cos30(1-e)$ but $e \leq 1$
so $w$ cannot be negative
- M1: One general equation
- M1: Second equation, and use $e \leq 1$
- A1: Correctly deduce given conclusion
**[3]**12\\
\includegraphics[max width=\textwidth, alt={}, center]{1a89caec-6da8-4b83-9ffa-efc209ecbc8d-5_205_200_264_497}\\
\includegraphics[max width=\textwidth, alt={}, center]{1a89caec-6da8-4b83-9ffa-efc209ecbc8d-5_284_899_349_753}
A white snooker ball of mass $m$ moves with speed $u$ towards a stationary black snooker ball of the same mass and radius. Taking the $x$-axis to be the line of centres of the two balls at the moment of collision, the direction of motion of the white ball before the collision makes an angle of $30 ^ { \circ }$ with the positive $x$-axis (see diagram).\\
(i) Given that the coefficient of restitution is 0.9 , find the angle made with the $x$-axis by the velocity of the white ball after the collision.\\
(ii) Show that after the collision the white ball cannot have a negative $x$-component of velocity whatever the value of the coefficient of restitution.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q12 [8]}}