Solve Spring Question: 6kg Mass Moves 4m/s

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SUMMARY

The discussion centers on a physics problem involving a 6 kg mass and a 3 kg mass on a frictionless table, separated by a compressed spring. Upon releasing the spring, the 6 kg mass moves left at 4 m/s, prompting the question of the velocity of the 3 kg mass. Using the principle of conservation of momentum, the correct calculation shows that the 3 kg mass moves to the right at 8 m/s. The conclusion confirms that the spring acts similarly to a collision, and momentum is conserved in this system.

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Homework Statement



A 6 kg mass and 3 kg mass are at rest on a frictionless table. A massless spring is compressed between them and they are held together with a string. When released (string is cut) the 6 kg mass moves to the left at 4 m/s - what is the velocity of the 3 kg mass as it moves to the right?

Homework Equations



a little unsure - i believe i need to use the F=kx (force of a spring) and assume that k is constant throughout since its the same spring for both.

The Attempt at a Solution



what i did was assume that the force on each mass was the same since each was compressed the same and had the same spring constant, k.

what i then did was assume that if that force moved the 6 kg mass 4 m/s to the left, that it would have an equal reaction to the 2nd mass which was only 3 kg. thinking in terms of momentum (is that correct?) i said the 2nd mass would move to the right at 8 m/s:

(6 kg)(4 m/s) = (3 kg)(x m/s)
x = 8 m/s

is this correct? does my logic work?

thanks.
 
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Conservation of momentum.
 
so then my answer is correct.

m1v1 + m2v2 = m1v1 + m2v2

the first two are 0 because v1=v2=0

then you have:

0 = (6)(4) + (3)(x)

x = -8 (8 m/s in the other direction)

right?

so basically this spring is considered a "collision"?
 
dnt said:
so then my answer is correct.

m1v1 + m2v2 = m1v1 + m2v2

the first two are 0 because v1=v2=0

then you have:

0 = (6)(4) + (3)(x)

x = -8 (8 m/s in the other direction)

right?

so basically this spring is considered a "collision"?
I'd say it was more of an explosion than a collision, but in either case, in the absence of external forces, momentum is always conserved, and your answer is correct, good job.
 

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