Stored Spring Energy Calculation

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Homework Help Overview

The problem involves two trolleys on a horizontal track connected by a compressed spring. Upon release, the trolleys separate, and the task is to calculate the initial velocity of one trolley and the energy stored in the spring. The subject area includes concepts of momentum and energy transfer.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum to find the velocity of trolley A. There is also a focus on the conversion of elastic potential energy to kinetic energy, with attempts to calculate the total energy involved.

Discussion Status

Some participants have provided guidance on the calculations, noting a mistake in the original energy calculation. There is a recognition of the correct application of momentum conservation, and a clarification of the energy formula used has been discussed.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the depth of exploration into the concepts involved. There is an emphasis on checking methods and understanding rather than providing direct solutions.

lemon
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Hi - Would someone check my method here please?
Thank you

Homework Statement



Two trolleys A and B, of mass 0.70kg and 0.80kg respectively, are on a horizontal track and held together by a compressed spring. When the spring is released the trolleys separate freely and A moves to the left with an initial velocity of 4.0m/s.
Calculate:
a) the initial velocity of A
b) the energy initially stored in the spring



Homework Equations



F=ma



The Attempt at a Solution



a) (0.70x4.0)=(0.80xV)
v=(0.70x4.0)/(0.80)
v=3.5m/s

b) This is a transfer of elastic potential energy to kinetic energy -
Energy=1/2mv^2
mv=p(momentum)
1/2(0.70x4.0)^2 + 1/2(0.80x3.5)^2
=0.0392 + 3.92
=3.9592 J
 
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lemon said:
a) (0.70x4.0)=(0.80xV)
v=(0.70x4.0)/(0.80)
v=3.5m/s
This is good. You used conservation of momentum.

b) This is a transfer of elastic potential energy to kinetic energy -
Energy=1/2mv^2
mv=p(momentum)
1/2(0.70x4.0)^2 + 1/2(0.80x3.5)^2
=0.0392 + 3.92
=3.9592 J
Your mistake here was squaring mv, instead of just v.
KE = 1/2mv^2, but you calculated 1/2(mv)^2. Not the same!
 
ahh of course:
silly boy!

(1/2x0.7x4.0^2) + (1/2x0.8x3.5^2) = 10.5 J
 
Now you've got it. :smile:
 
Thanks Doc
 

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