1998 Paper 1 Q9

Year: 1998
Paper: 1
Question Number: 9

Course: UFM Mechanics
Section: Simple Harmonic Motion

Difficulty: 1500.0 Banger: 1484.0

SHM simple pendulum Newton's experimental law coefficient of restitution collision conservation of momentum successive collisions equal pendulums period

Problem

Two small spheres \(A\) and \(B\) of equal mass \(m\) are suspended in contact by two light inextensible strings of equal length so that the strings are vertical and the line of centres is horizontal. The coefficient of restitution between the spheres is \(e\). The sphere \(A\) is drawn aside through a very small distance in the plane of the strings and allowed to fall back and collide with the other sphere \(B\), its speed on impact being \(u\). Explain briefly why the succeeding collisions will all occur at the lowest point. (Hint: Consider the periods of the two pendulums involved.) Show that the speed of sphere \(A\) immediately after the second impact is \(\frac{1}{2}u(1+e^{2})\) and find the speed, then, of sphere \(B\).

No solution available for this problem.

Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1484.0

Banger Comparisons: 1

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Problem source

Two small spheres $A$ and $B$ of equal mass $m$
are suspended in contact by two light inextensible strings
of equal length so that
the strings are vertical and
the line of centres is horizontal.
The coefficient of restitution between the spheres is $e$.
The sphere $A$ is
drawn aside through a very small distance
in the plane of the strings
and allowed
to fall back and collide with the other sphere $B$, its speed
on impact being $u$.
Explain briefly why the succeeding collisions will all occur
at the lowest point. (Hint: Consider the periods of the
two pendulums involved.)
Show that the speed of sphere $A$
immediately after the second impact is 
$\frac{1}{2}u(1+e^{2})$
and find the speed, then, of sphere $B$.

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