Stored Spring Energy Calculation

AI Thread Summary
The discussion focuses on calculating the initial velocities and energy stored in a spring for two trolleys after release. The initial velocity of trolley A is correctly calculated as 4.0 m/s, while the velocity of trolley B is found to be 3.5 m/s using conservation of momentum. However, a mistake was made in calculating the kinetic energy, where the user initially squared the momentum instead of the velocity. After correcting the formula to use the kinetic energy equation, the total energy stored in the spring is determined to be 10.5 J. The conversation highlights the importance of applying the correct physics equations in energy calculations.
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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