A 2.0 kg mass is attached to one end of a spring

In summary, the conversation is about a physics homework problem involving a 2.0 kg mass, a 4.0 kg mass, and a spring with a spring constant of 100 N/m on a horizontal frictionless surface. The masses are observed to be moving in opposite directions with different speeds, and there is a debate on whether the spring is compressed or not at the time of the observation. The correct answer is D, based on the conservation of momentum. The initial momentum of the system is not 0, indicating that something was initially moving. The ratios of the speeds and masses of the two weights also support this answer.
  • #1
ctv1337
I am currently taking physics right now and i had the following question on a homework assignment. I have the right answer but i don't believe its correct. Can anyone help me prove him wrong or give me a better explanation of why his answer is correct?

-------here is the problem-----------
A 2.0 kg mass is attached to one end of a spring with a spring constant of 100 N/m and a 4.0 kg mass is attached to the other end. The masses are placed on a horizontal frictionless surface and set into motion. At one instant the 2.0 kg mass is observed to be traveling to the right with a speed of 0.50 m/s and the 4.0 kg mass is observed to be traveling to the left with a speed of 0.30 m/s. Since the only forces acting are the force of gravity, the normal force of the surface, and the force of the spring, we conclude that:
A) the spring is compressed at the time of the observation
B) the spring is not compressed at the time of observation
C) the motion was started with the masses at rest
D) the motion was started with at least one of masses moving
E) the motion was started by compressing the spring \

he said the answer was D can anyone help?
a picture is attached to help see both ways

https://webspace.utexas.edu/pourkc/PHYS/spring.jpg
 

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  • #2
No, he is correct.

Compare the ratios of the speeds and masses of the two weights.
 
  • #3
A)As we don't know the natural length of the spring and the length at the observation, we can't say spring is compressed or extended.
B) same as above.
C)If so, masses stay still.
E)Starting by compressing and stretching both can make the same motion. We can't say so.
Remaining D) is correct.
 
  • #4
I think it has to do with the conservation of momentum. The momentum of the system is
m1*v1 - m2*v2 (the signs can be changed if you change the reference system) = 2*0.5 - 4*0.3 = -0.2 N/s
So the initial momentum of the system is not 0 => something was initially moving...
 

1. What is the relationship between the mass and the spring in this scenario?

The mass and the spring have an inverse relationship. This means that as the mass increases, the spring will stretch more and have a lower spring constant. As the mass decreases, the spring will stretch less and have a higher spring constant.

2. How does the spring constant affect the behavior of the spring?

The spring constant, also known as the stiffness or force constant, determines how much force is required to stretch or compress the spring by a certain distance. A higher spring constant means the spring is stiffer and requires more force to stretch or compress, while a lower spring constant means the spring is more flexible and requires less force.

3. What is the significance of the 2.0 kg mass in this scenario?

The 2.0 kg mass is the object that is attached to the spring and will cause the spring to stretch or compress. It is used to demonstrate the relationship between the mass and the spring in this scenario.

4. How does the displacement of the mass affect the spring?

The displacement, or distance, of the mass from its equilibrium position will determine the amount of stretch or compression of the spring. The greater the displacement, the greater the force exerted by the spring to return the mass to its equilibrium position.

5. Is there a limit to how much the spring can stretch or compress?

Yes, there is a limit to how much the spring can stretch or compress, known as the elastic limit. Once the elastic limit is exceeded, the spring will no longer return to its original shape and will be permanently deformed.

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