Grade 12Mechanics

A small bucket of mass M kg is attached to a long inextensible cord of length L m. The bucket is released when the cord is in horizontal position. At its lowest point the bucket scoops up m kg of water and swings.up to a height h. The height h is??

Shuvam Shukla

13 Years agoGrade 12

2 Answers

Akash Kumar Dutta

Approved Tutor Answer13 Years ago

Dear Shuvam,

Velocity before scooping=root(2gL)
Applying conservation of momentum,
new velocity= (M/M+m).root(2gL)
the bucket stops when velocity = 0
=>u^2=2gh
=>2gh= [ (M/M+m)^2 ]. 2gL
=> h= [ (M/M+m)^2 ].L (ANS)

Regards.

ruhi

7 Years ago

at the lowest position v=root 2gL

now, by using law of conservation of momentum

MV-(M+m)V^'=0

Mv=(M+m)V'

V'=(M/M+m)root2gL

h=V'²/2g = (M/M+m)²L

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A small bucket of mass M kg is attached to a long inextensible cord of length L m. The bucket is released when the cord is in horizontal position. At its lowest point the bucket scoops up m kg of water and swings.up to a height h. The height h is??

Shuvam Shukla

2 Answers

Akash Kumar Dutta

ruhi

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