Conservation of momentum in a bullet-block-spring system

[AL115]

According to the first equation, the final potential energy is equal to the initial kinetic energy of the block. So that means that Vblok is the instantaneous speed of the block right before it moves to the right and compress the spring, right? But doesn't the second equation (The initial total momentum = the final total momentum) tell us that the moment where the block has the speed Vbolk = 1.5 m/s is the same moment where the bullet have its final speed (100m/s) after it emerges from the the block? Does that mean that the block started moving to the right after the bullet emerged from it? How? Wouldn't the force that is applied by the bullet on the block move the block before the bullet emerges?
I am very sorry if I was unclear.

 

[A.T.]

Does hat mean that the block started moving to the right after the bullet emerged from it? How? Wouldn't the force that is applied by the bullet on the block move the block before the bullet emerges?

Do the details of what happens while the bullet is in the block change the final quantities the exercise asks for?

 

[AL115]

Do the details of what happens while the bullet is in the block change the final quantities the exercise asks for?

I don't know...

 

[A.T.]

I don't know...

Details that are not provided in the question are usually not needed for the answer.

 

[AL115]

Details that are not provided in the question are usually not needed for the answer.

Yeah that is correct. But my question is that did the block move after the bullet emerged from it?
mV1 + MV2 = mV3 + MV4 where m = mass of the bullet = 5g , M = mass of the block = 1kg, V1 = the initial speed of the bullet = 400 m/s, V2 = the initial speed of the block = 0 m/s, V3 = the final speed of the bullet after it emerges from the block = 100 m/s , V4 = the final speed of block after the bullet emerges from the block = 1.5 m/s. Because V3 and V4 happened in the same moment... right?
1/2(m)(V)^2 = 1/2(k)(x)^2 tell us that block started to move with initial speed V = V4 = 1.5 m/s. Which imply that the block started to move after the bullet emerged from it. But how come?

 

[A.T.]

Which imply that the block started to move after the bullet emerged from it. But how come?

I think the simplifying assumption here is that the duration of the bullet-block interaction is very short, so the movement of the block during the interaction is negligible.

 

[AL115]

I think the simplifying assumption here is that the duration of the bullet-block interaction is very short, so the movement of the block during the interaction is negligible.

So if we try to describe this mathematically, we can consider that the moment where the bullet makes contact with the block (lets call it t1) and the moment where the bullet emerges from the block (lets call it t2) the same since t2 - t1 is close to zero? That makes sense. Thank you very much.