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1987 Paper 1 Q12
D: 1500.0 B: 1500.0
momentum and collisions coefficient of restitution projectile bouncing particle steps kinematics repeated motion

\(\,\) \vspace{-1cm}

TikZ diagram
A particle is placed at the edge of the top step of a flight of steps. Each step is of width \(2d\) and height \(h\). The particle is kicked horizontally perpendicular to the edge of the top step. On its first and second bounces it lands exactly in the middle of each of the first and second steps from the top. Find the coefficient of restitution between the particle and the steps. Determine whether it is possible for the particle to continue bouncing down the steps, hitting the middle of each successive step.

Solution: Considering the horizontal component, this will be constant as there are no forces acting in that direction. The first step will take the particle \(t = \sqrt{\frac{2h}g}\) to reach. At which point it will be travelling with speed \(v = \sqrt{2gh} \) (by energy considerations, \(mgh = \frac12 mv^2\)). To reach the second step must take twice as long (since the ball has to travel \(2d\) horizontally, rather than \(d\)). Since \(t = 2\sqrt{\frac{2h}g}\) we must have that: \begin{align*} && s &= ut + \frac12 gt^2 \\ \Rightarrow && h &= u 2\sqrt{\frac{2h}g} + \frac12 g \frac{8h}g \\ \Rightarrow && u &= -\frac{3}{2} h\sqrt{\frac{g}{2h}} \\ &&&= -\frac{3}{2\sqrt{2}} \sqrt{gh} \end{align*} Therefore, using Newton's experimental law, we must have that \(e = \frac{\frac{3}{2 \sqrt{2}} \sqrt{gh}}{\sqrt{2} \sqrt{gh}} = \frac{3}{4}\). Again by conservation of energy \(mgh + \frac12 \frac{9}{8} mgh = \frac12 mv^2 \Rightarrow v = \frac{5}{2\sqrt{2}} \sqrt{gh}\) when it lands on the next step. Therefore we would need the coefficient of restitution for the second (and subsequent steps) to be: \(\displaystyle \frac{\frac{3}{2\sqrt{2}} \sqrt{gh}}{\frac{5}{2\sqrt{2}} \sqrt{gh}} = \frac35\)

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1987 Paper 2 Q13
D: 1500.0 B: 1500.0
momentum and collisions coefficient of restitution geometry reflection parallel paths rectangular table snooker/billiards locus

Ice snooker is played on a rectangular horizontal table, of length \(L\) and width \(B\), on which a small disc (the puck) slides without friction. The table is bounded by smooth vertical walls (the cushions) and the coefficient of restitution between the puck and any cushion is \(e\). If the puck is hit so that it bounces off two adjacent cushions, show that its final path (after two bounces) is parallel to its original path. The puck rests against the cushion at a point which divides the side of length \(L\) in the ratio \(z:1\). Show that it is possible, whatever \(z\), to hit the puck so that it bounces off the three other cushions in succession clockwise and returns to the spot at which it started. By considering these paths as \(z\) varies, explain briefly why there are two different ways in which, starting at any point away from the cushions, it is possible to perform a shot in which the puck bounces off all four cushions in succession clockwise and returns to its starting point.

Solution:

TikZ diagram
The puck sets off at some velocity \(\displaystyle \binom{u_x}{u_y}\), after the first bounce off the wall parallel to the \(y\)-axis, it has velocity \(\displaystyle \binom{-eu_x}{u_y}\). After it bounces off the wall parallel to the \(x\)-axis, it has velocity \(\displaystyle \binom{-eu_x}{-eu_y}\) which is clearly parallel to the original velocity.
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If the puck bounces off 3 walls and returns to the same point the shape formed must be a parallelogram. We need to hit the point on the opposite side which is in a ratio of \(1:z\), but this must be possible if we aim towards the side further away.
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For a fixed path, as \(z\) increases we generate more parallelograms which cross ours (on two of the legs) twice. As they move the full length it will cover the full leg of the parallogram. Similarly going the other way will cover the other leg of the parallelogram. Therefore from every point there are two circuits round the table

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