A spring and two blocks with kinetic energy

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Homework Help Overview

The discussion revolves around a problem involving two blocks colliding and compressing a spring. The original poster describes the scenario with specific masses, velocities, and a spring constant, seeking to determine the distance the spring is compressed at a particular instant.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to combine the kinetic energies of both blocks to find the compression of the spring, questioning whether this approach is valid given the two blocks are moving in opposite directions.
  • Some participants question the completeness of the problem statement and the assumptions made regarding the velocities and kinetic energy calculations.
  • Others suggest that the kinetic energy lost during the collision is relevant for determining the spring compression.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the relationship between kinetic energy and spring compression. Some guidance has been offered regarding the calculations, but there is no explicit consensus on the final values or methods used.

Contextual Notes

There is mention of a two-part problem, with some confusion about the initial conditions and the velocities of the blocks at the moment of interest. The discussion highlights potential missing information or misinterpretations in the problem setup.

hs764
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1. I have two blocks, one that is 1.6 kg moving left at 3.00 m/s and the other is 2.1 kg moving right at -1.74 m/s, and they are compressing a block with a spring constant of k=600 N/m. I'm supposed to be finding the distance that the spring is compressed at that instant, and I know that 1/2mv^2=1/2kx^2 but every other problem I've encountered only had one block and kinetic energy from only one direction, so I just want to make sure that I'm doing the right thing by combining the kinetic energy of the two blocks and then using that to solve for x. The answer I came up with is below but I'm not 100% sure that it's correct. 2. 1/2mv^2=1/2kx^23. 1/2(1.6)(3.00)^2 + 1/2(2.1)(-1.74)^2 = 1/2(600)x^2 x = 0.19 m
 
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Hello hs, and welcome to PF :smile:

There is something missing from your problem statement, or something else is wrong. Is this the literal staement ? If not, could you post the complete, exact problem statement ?

Reason I ask is that from the given information (two masses with two velocities) the distance the spring is compressed at that moment can't be determined.
 
Last edited:
Ah, okay. Well it's a two part problem, but I may have gotten the first part wrong as well.

It says, "A block of mass m1 = 1.60 kg moving to the right with a speed of 4.00 m/s on a frictionless horizontal track collides with a spring attached to a second block of mass m2 = 2.10 kg moving to the left with a speed of 2.50 m/s, as in Figure 2. The spring has a spring constant of 600 N/m. At the instant when m1 is moving to the right with a speed of 3.00 m/s, determine (a) the velocity of m2 and (b) the distance x that the spring is compressed."

So for part a) I had worked out that the velocity of m2 was -1.74 m/s. Does that make sense?
 
Yes it does make sense. There is something going on before the velocity of m1 has dropped to 3 m/s: the spring has been compressed.
You get the -1.74 m/s from a momentum balance and it seems fine to me.

In post #1 you convert the remaining kinetic energy to spring energy. So that would result in an answer that gives the further compression of the spring (unitil no kinetic energy is lefet over.

I have the impression part (b) of the exercise is asking for something else. Don't you ?
 
Okay, so it's the kinetic energy that's already been lost that determines the value of x, so that changes x to 0.173 m...?
 
What's the calculation that gives you 0.173 m ?
 
It's the difference in kinetic energy, right? So [1/2(1.6)(4.0)^2 + 1/2(2.1)(-2.5)^2] - [1/2(1.6)(3.0)^2 + 1/2(2.1)(-1.74)^2] = 1/2kx^2, which gave me x = 0.173 m.
 
Excellent. (And yes, ahem, it was me who had an error in the calculation, so I was puzzled o:) )
 
Awesome, thanks for your help!
 

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