1991 Paper 1 Q11

Year: 1991
Paper: 1
Question Number: 11

Course: UFM Mechanics
Section: Momentum and Collisions 1

Difficulty: 1500.0 Banger: 1500.1

Problem

A piledriver consists of a weight of mass \(M\) connected to a lighter counterweight of mass \(m\) by a light inextensible string passing over a smooth light fixed pulley. By considerations of energy or otherwise, show that if the weights are released from rest, and move vertically, then as long as the string remains taut and no collisions occur, the weights experience a constant acceleration of magnitude \[ g\left(\frac{M-m}{M+m}\right). \] Initially the weight is held vertically above the pile, and is released from rest. During the subsequent motion both weights move vertically and the only collisions are between the weight and the pile. Treating the pile as fixed and the collisions as completely inelastic, show that, if just before a collision the counterweight is moving with speed \(v\), then just before the next collision it will be moving with speed \(mv/\left(M+m\right)\). {[}You may assume that when the string becomes taut, the momentum lost by one weight equals that gained by the other.{]} Further show that the times between successive collisions with the pile form a geometric progression. Show that the total time before the weight finally comes to rest is three times the time from the start to the first impact.

No solution available for this problem.

Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1500.1

Banger Comparisons: 2

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Problem source
A piledriver consists of a weight of mass $M$ connected to a lighter
counterweight of mass $m$ by a light inextensible string passing
over a smooth light fixed pulley. By considerations of energy or otherwise,
show that if the weights are released from rest, and move vertically,
then as long as the string remains taut and no collisions occur, the
weights experience a constant acceleration of magnitude 
\[
g\left(\frac{M-m}{M+m}\right).
\]
Initially the weight is held vertically above the pile, and is released
from rest. During the subsequent motion both weights move vertically
and the only collisions are between the weight and the pile. Treating
the pile as fixed and the collisions as completely inelastic, show
that, if just before a collision the counterweight is moving with
speed $v$, then just before the next collision it will be moving
with speed $mv/\left(M+m\right)$. {[}You may assume that when the
string becomes taut, the momentum lost by one weight equals that gained
by the other.{]}

Further show that the times between successive collisions with the
pile form a geometric progression. Show that the total time before
the weight finally comes to rest is three times the time from the
start to the first impact.