1998 Paper 1 Q11

Year: 1998
Paper: 1
Question Number: 11

Course: LFM Pure and Mechanics
Section: Pulley systems

Difficulty: 1500.0 Banger: 1502.5

pulley systems Newton's second law constant acceleration optimisation calculus connected particles vertical motion ore mine

Problem

Hank's Gold Mine has a very long vertical shaft of height \(l\). A light chain of length \(l\) passes over a small smooth light fixed pulley at the top of the shaft. To one end of the chain is attached a bucket \(A\) of negligible mass and to the other a bucket \(B\) of mass \(m\). The system is used to raise ore from the mine as follows. When bucket \(A\) is at the top it is filled with mass \(2m\) of water and bucket \(B\) is filled with mass \(\lambda m\) of ore, where \(0<\lambda<1\). The buckets are then released, so that bucket \(A\) descends and bucket \(B\) ascends. When bucket \(B\) reaches the top both buckets are emptied and released, so that bucket \(B\) descends and bucket \(A\) ascends. The time to fill and empty the buckets is negligible. Find the time taken from the moment bucket \(A\) is released at the top until the first time it reaches the top again. This process goes on for a very long time. Show that, if the greatest amount of ore is to be raised in that time, then \(\lambda\) must satisfy the condition \(\mathrm{f}'(\lambda)=0\) where \[\mathrm{f}(\lambda)=\frac{\lambda(1-\lambda)^{1/2}} {(1-\lambda)^{1/2}+(3+\lambda)^{1/2}}.\]

No solution available for this problem.

Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1502.5

Banger Comparisons: 4

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Problem source

Hank's Gold Mine has a very long vertical shaft
of height $l$.
A light chain of length $l$
passes over a 
small smooth
light fixed pulley at the top of the shaft.
To one end of the chain is attached a bucket $A$ of
negligible mass and to the other a bucket $B$ of mass
$m$. The system is used to raise ore from the mine as follows.
When bucket $A$ is at the top it is filled with mass $2m$
of water and bucket $B$ is filled with mass $\lambda m$
of ore, where $0<\lambda<1$. The buckets are then released,
so that bucket $A$ descends and bucket $B$ ascends. 
When bucket $B$
reaches the top both buckets are emptied and released,
so that bucket $B$ descends and bucket $A$ ascends. The time to
fill and empty the buckets is negligible. Find the
time taken from the moment bucket $A$ is released at the top
until the first time it reaches the top again.
This process
goes on for a very long time. Show that, if the greatest
amount of ore is to be raised in that time, then
$\lambda$ must satisfy the condition $\mathrm{f}'(\lambda)=0$
where
\[\mathrm{f}(\lambda)=\frac{\lambda(1-\lambda)^{1/2}}
{(1-\lambda)^{1/2}+(3+\lambda)^{1/2}}.\]

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