Conservation of moentum and a spring.

AI Thread Summary
The discussion revolves around a physics problem involving the conservation of momentum and energy in a collision between two balls, where one ball is attached to a spring. The first ball, moving at a speed of 2/√3 m/s, collides with a second ball that has three times its mass, resulting in the two balls sticking together and compressing the spring. Participants express confusion over the application of conservation of energy, noting that energy is lost during the collision, which complicates calculations for the spring constant k. They emphasize the need to account for the kinetic energy of the combined mass post-collision and the work done by the spring and friction. The conversation highlights the importance of correctly applying energy principles and understanding the relationship between work and potential energy in this context.
JohnGG
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Homework Statement


We got a ball moving at speed 2/√3 m/s, and it hits another ball that is attached to the string. Ball 2 has 3 times the mass of ball 1. After the crash, ball one stucks in ball 2 and the spring gets .02m shorter at its minimum length. There are frictions with n=1/12 and g=10m/s^2. Find the constant k of the spring.

Homework Equations


ΔP=0=>mu=4mu', u'=u/4, you can't use hooke's law as it isn't in the textbook.
Fy=0<=>N=mg
M=3m, M'=4m

The Attempt at a Solution


I used conservation of energy principle to get 1/2Mu'^2=1/2kx^2-nM'g=>Mu'^2+2nM'g=kx^2=>(Mu'^2+2nM'g)/x^2=k and thn just plug in numbers, but the problem is I ended up with a k as a function of m.
 
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Your equation for the conservation of energy cannot be correct. It has nM'g as an addend, and its dimension is that of force.

Another issue is when one ball gets stuck in another ball in a collision, there is a loss of energy, while you assume it is fully conserved.
 
voko said:
Your equation for the conservation of energy cannot be correct. It has nM'g as an addend, and its dimension is that of force.

Another issue is when one ball gets stuck in another ball in a collision, there is a loss of energy, while you assume it is fully conserved.

What would be the correct equation?

I found that 1/3 of kinetic energy turned into heat in a question I skipped here. How could I use that to find k?
 
The correct equation would have the kinetic energy of the two balls immediately after the collision, and the work of the spring and friction over the distance of maximum compression.
 
voko said:
The correct equation would have the kinetic energy of the two balls immediately after the collision, and the work of the spring and friction over the distance of maximum compression.

1/2M'u'^2=-Tx+?
What is the work of the spring?
 
How is work related with potential energy?
 
voko said:
How is work related with potential energy?

W=-ΔU. But where is the k I search for?
 
In you very first message, you had ##{1 \over 2} kx^2 ##. Where did that come from?
 

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