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

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In summary, two masses, 6 kg and 3 kg, at rest on a frictionless table, are connected by a compressed massless spring and a string. When the string is cut, the 6 kg mass moves to the left at 4 m/s. Using the equation F=kx and assuming a constant spring constant, k, the 3 kg mass is calculated to move to the right at 8 m/s, thus conserving momentum. This can be considered a collision or an explosion, but regardless, momentum is always conserved in the absence of external forces.
  • #1
dnt
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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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  • #2
Conservation of momentum.
 
  • #3
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"?
 
  • #4
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.
 

1. How do you calculate the spring constant for a 6kg mass moving at 4m/s?

The spring constant can be calculated using the formula k = m * v^2 / x, where k is the spring constant, m is the mass, v is the velocity, and x is the displacement of the spring.

2. What is the potential energy of the spring with a 6kg mass moving at 4m/s?

The potential energy of the spring can be calculated using the formula PE = 1/2 * k * x^2, where PE is the potential energy, k is the spring constant, and x is the displacement of the spring.

3. How much work is done on the spring with a 6kg mass moving at 4m/s?

The work done on the spring can be calculated using the formula W = F * x, where W is the work done, F is the force applied to the spring, and x is the displacement of the spring.

4. What is the frequency of the spring with a 6kg mass moving at 4m/s?

The frequency of the spring can be calculated using the formula f = 1 / (2 * pi * sqrt(m/k)), where f is the frequency, m is the mass, and k is the spring constant.

5. How does the displacement of the spring affect the motion of the 6kg mass at 4m/s?

The displacement of the spring affects the motion of the mass by changing the amount of potential energy stored in the spring. As the displacement increases, so does the potential energy, resulting in a larger force and a greater acceleration of the mass.

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