# A spring and two blocks with kinetic energy

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1. Mar 3, 2015

### hs764

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^2

3. 1/2(1.6)(3.00)^2 + 1/2(2.1)(-1.74)^2 = 1/2(600)x^2 x = 0.19 m

2. Mar 3, 2015

### BvU

Hello hs, and welcome to PF

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: Mar 3, 2015
3. Mar 3, 2015

### hs764

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?

4. Mar 3, 2015

### BvU

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 ?

5. Mar 3, 2015

### hs764

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....?

6. Mar 3, 2015

### BvU

What's the calculation that gives you 0.173 m ?

7. Mar 4, 2015

### hs764

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.

8. Mar 4, 2015

### BvU

Excellent. (And yes, ahem, it was me who had an error in the calculation, so I was puzzled )

9. Mar 4, 2015