What is the spring constant of the spring?

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In summary, the pellet gun fires a bullet into a stationary block of wood that is attached to a spring. The block and bullet enter into simple harmonic motion with amplitude = 11cm. The bullet was initially traveling at 650 m/s before hitting the block of wood. After the collision, the bullet and block reach a new velocity of 1.297 m/s.
  • #1
JumpinJohny
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Homework Statement


A pellet gun fires a bullet into a stationary block of wood that is attached to a spring on a frictionless surface. When the bullet enters the wood, it reamins inside, and the bullet and block enter into simple harmonic motion with amplitude = 11cm. The bullet ( m = 5g) was initially traveling at 650 m/s before hitting the block of wood ( m = 2.5kg).

A: What is the spring constant of the spring?
B: What is the total energy of the system after the collision?(I know how to solve this one, I just need A to solve it.)
C: What is the maximum acceleration of the bullet/block system once it begins its oscillation?
D:Where will the bullet and block reach 0 velocity?(I think this is at the equilibrium position, but I'm not sure how to solve for it)

We know:
the amplitude of the block-and-bullet is .11m
the mass of the bullet is .005kg
the mass of the block is 2.5kg
the mass of the bullet-and-block is 2.505kg
the velocity of the bullet before it hits the block is 650 m/s.

Homework Equations


I know for B that you use the equation for total energy, kinetic energy + potential energy.

The Attempt at a Solution


My attempt at A; it wasn't so good. I tried using F = -kx, but that of course didn't work since we don't know the acceleration to solve for F. Then I considered kx=mg, but that doesn't work either because that only works for a spring that is suspended. So I'm not really sure where to go with this :eek:
 
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  • #2
Hint: What's the speed of "block+bullet" immediately after the collision?
 
  • #3
:yuck:I really don't know Al.
 
  • #4
Hint: What's conserved in any collision?
 
  • #5
Errrrr it's either energy or momentum. I think momentum.
 
  • #6
Good. Use a conservation law to figure out the speed of "block+bullet" after they collide.
 
  • #7
Alright. p=mv, so then .005*650=3.25; so 3.25/2.505= 1.297.
So 1.297 would be the new velocity?
 
  • #8
Is that correct?
 
  • #9
JumpinJohny said:
Is that correct?
Perfect!

Now see if you can make use of that speed to solve part A (and part B, for that matter).
 

1. What is the "Bullet-in-Block with Spring" experiment?

The "Bullet-in-Block with Spring" experiment is a physics demonstration that shows the principles of conservation of energy and momentum. It involves a block attached to a spring and a bullet fired into it, causing the block to move.

2. How does the experiment demonstrate conservation of energy?

The experiment demonstrates conservation of energy by showing that the initial potential energy of the compressed spring is converted into kinetic energy as the block moves after being hit by the bullet. The total energy remains constant throughout the experiment.

3. What is the purpose of the bullet in this experiment?

The bullet is used to introduce an external force on the block, causing it to move. This allows us to observe the transfer of energy and momentum from the bullet to the block.

4. What factors affect the motion of the block in this experiment?

The motion of the block is affected by the mass and velocity of the bullet, the mass of the block, and the stiffness of the spring. Air resistance and friction may also play a role in the motion of the block.

5. What are some real-world applications of the "Bullet-in-Block with Spring" experiment?

This experiment can be used to understand the principles of energy and momentum in collisions, which have many real-world applications. It can also be used to study the behavior of springs and the effects of external forces on objects.

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