Conservation of momentum in a bullet-block-spring system

In summary: 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.
  • #1
AL115
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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.
 
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  • #2
AL115 said:
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?
 
  • #3
A.T. said:
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...
 
  • #4
AL115 said:
I don't know...
Details that are not provided in the question are usually not needed for the answer.
 
  • #5
A.T. said:
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?
 
  • #6
AL115 said:
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.
 
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  • #7
A.T. said:
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.
 

Related to Conservation of momentum in a bullet-block-spring system

1. What is conservation of momentum in a bullet-block-spring system?

Conservation of momentum in a bullet-block-spring system refers to the principle that the total momentum of the system remains constant before and after a collision. This means that the momentum of the bullet and the block, as well as the momentum of the spring, must all add up to the same value before and after the collision.

2. How does conservation of momentum apply to a bullet-block-spring system?

In a bullet-block-spring system, the bullet and the block have an initial momentum before the collision. When the bullet hits the block, the momentum is transferred from the bullet to the block. The momentum of the block then causes the spring to compress, storing potential energy. After the collision, the total momentum of the system remains the same, but it is now divided between the block's motion and the spring's potential energy.

3. What factors can affect the conservation of momentum in a bullet-block-spring system?

The conservation of momentum in a bullet-block-spring system can be affected by the mass and velocity of the bullet and the block, as well as the stiffness of the spring. The angle and direction of the bullet's impact on the block can also play a role in how much momentum is transferred.

4. How does conservation of momentum impact the motion of a bullet-block-spring system?

The conservation of momentum in a bullet-block-spring system is what allows for the transfer of momentum from the bullet to the block and then to the spring. This transfer of momentum governs the motion of the system, causing the block to move and the spring to compress. It also allows for the potential energy stored in the spring to be released as the spring pushes the block back out.

5. What other laws of physics are involved in a bullet-block-spring system?

In addition to conservation of momentum, a bullet-block-spring system also involves the laws of motion, energy conservation, and Hooke's law. The laws of motion describe how objects move and accelerate, while energy conservation states that energy cannot be created or destroyed, only transferred or transformed. Hooke's law explains the relationship between the force applied to an elastic object, such as a spring, and the resulting displacement.

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