Stored Spring Energy Calculation

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SUMMARY

The discussion focuses on calculating the initial velocities and energy stored in a compressed spring when two trolleys, A and B, are released. Trolley A, with a mass of 0.70 kg, moves left at an initial velocity of 4.0 m/s, while Trolley B, with a mass of 0.80 kg, achieves an initial velocity of 3.5 m/s based on momentum conservation principles. The initial energy stored in the spring is calculated using the kinetic energy formula, resulting in a corrected value of 10.5 J after addressing a calculation error in the initial attempt.

PREREQUISITES
  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of conservation of momentum
  • Familiarity with kinetic energy calculations (KE = 1/2mv^2)
  • Basic principles of elastic potential energy
NEXT STEPS
  • Study the principles of conservation of momentum in elastic collisions
  • Learn about energy transfer between potential and kinetic forms
  • Explore advanced topics in spring mechanics and energy storage
  • Review examples of similar physics problems involving multiple bodies
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Students studying physics, educators teaching mechanics, and anyone interested in understanding energy transfer and momentum in dynamic systems.

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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