1990 Paper 2 Q12

Year: 1990
Paper: 2
Question Number: 12

Course: UFM Mechanics
Section: Momentum and Collisions 2

Difficulty: 1600.0 Banger: 1484.0

Problem

A straight staircase consists of \(N\) smooth horizontal stairs each of height \(h\). A particle slides over the top stair at speed \(U\), with velocity perpendicular to the edge of the stair, and then falls down the staircase, bouncing once on every stair. The coefficient of restitution between the particle and each stair is \(e\), where \(e<1\). Show that the horizontal distance \(d_{n}\) travelled between the \(n\)th and \((n+1)\)th bounces is given by \[ d_{n}=U\left(\frac{2h}{g}\right)^{\frac{1}{2}}\left(e\alpha_{n}+\alpha_{n+1}\right), \] where \({\displaystyle \alpha_{n}=\left(\frac{1-e^{2n}}{1-e^{2}}\right)^{\frac{1}{2}}}\). If \(N\) is very large, show that \(U\) must satisfy \[ U=\left(\frac{L^{2}g}{2h}\right)^{\frac{1}{2}}\left(\frac{1-e}{1+e}\right)^{\frac{1}{2}}, \] where \(L\) is the horizontal distance between the edges of successive stairs.

No solution available for this problem.

Rating Information

Difficulty Rating: 1600.0

Difficulty Comparisons: 0

Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source
A straight staircase consists of $N$ smooth horizontal stairs each
of height $h$. A particle slides over the top stair at speed $U$,
with velocity perpendicular to the edge of the stair, and then falls
down the staircase, bouncing once on every stair. The coefficient
of restitution between the particle and each stair is $e$, where
$e<1$. Show that the horizontal distance $d_{n}$ travelled between
the $n$th and $(n+1)$th bounces is given by 
\[
d_{n}=U\left(\frac{2h}{g}\right)^{\frac{1}{2}}\left(e\alpha_{n}+\alpha_{n+1}\right),
\]
where ${\displaystyle \alpha_{n}=\left(\frac{1-e^{2n}}{1-e^{2}}\right)^{\frac{1}{2}}}$. 

If $N$ is very large, show that $U$ must satisfy
\[
U=\left(\frac{L^{2}g}{2h}\right)^{\frac{1}{2}}\left(\frac{1-e}{1+e}\right)^{\frac{1}{2}},
\]
where $L$ is the horizontal distance between the edges of successive
stairs.