Bullet through block and spring

  • #1
dkgojackets
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Homework Statement



A bullet of mass 9 g moving with initial speed 300 m/s passes through a block of mass 4 kg, initially at rest on a frictionless, horizontal surface. The block is connected to a spring of force constant 904 N/m. If the block moves 2 cm to the right after the bullet passed through it, find the speed at which the bullet emerges from the block and the magnitude of the energy lost in the collision.

Homework Equations



Conservation of momentum.

The Attempt at a Solution



Well initial mometum is just (.009)(300), and it must equal (.009)(Vf)=(4)(Vblk)

to solve for the block velocity immediately after the bullet passes, I used Vf = sqrt(2ad). I said acceleration was (904)(.02)/(4)...(may be where I went wrong). I took that velocity and plugged it back into the original momentum equation and got velocity of the bullet to be 111 m/s, which was wrong.
 

Answers and Replies

  • #3
AngeloG
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First draw a basic idea of how this looks; a good force body diagram or a diagram on how this works. Drawing out what happens can get you far when conceptualizing what happens and going about problems.

After you do this, we can start working on the problem :).

But here's a few things to think about:

There's a few things to deal with this such as momentum [bullet through the block] and energy [spring energy].
 
Last edited:
  • #4
dkgojackets
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This is an energy question. There have been many similar threads. From the two links below I'm sure you can work out what you need to do for this question.

https://www.physicsforums.com/showthread.php?t=146628&page=3&highlight=bullet+in+block

https://www.physicsforums.com/showthread.php?t=148204&highlight=bullet+in+block

I used conservation of energy (initial KE of bullet = final KE of bullet + spring potential), but I got almost the exact same speed for the bullet. Not even .1 m/s less.
 
  • #5
Doc Al
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I used conservation of energy (initial KE of bullet = final KE of bullet + spring potential), but I got almost the exact same speed for the bullet. Not even .1 m/s less.
Kinetic energy is not conserved during the collision. This problem involves both conservation of momentum and conservation of energy--separately. (See the links that Kurdt provided.)
 
  • #6
Kurdt
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If you show the work you have done we can determine where you are going wrong.

EDIT: Doc Al beat me
 
  • #7
AngeloG
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Kinetic energy is not conserved during the collision. This problem involves both conservation of momentum and conservation of energy--separately. (See the links that Kurdt provided.)
To go along with this.

The reason why you can't use conservation of energy is due to the fact that there maybe thermal energy caused by this or sound and other various forms of energy. Energy is not conserved. Conservation of energy only works with perfectly elastic or a complete inelastic collision.

As I stated in my post there are things at work here =). Spring energy being one and the momentum.
 
  • #8
Hootenanny
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  • #9
dkgojackets
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I get that momentum is conserved during the collision, and mechanical energy after. But why would conservation of energy not work if I have just one unknown in the equation?

.5(.009)(300^2) = .5(.009)(v^2) + .5(904)(.02^2) is my energy conservation equation.
 
  • #10
Hootenanny
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I get that momentum is conserved during the collision, and mechanical energy after. But why would conservation of energy not work if I have just one unknown in the equation?

.5(.009)(300^2) = .5(.009)(v^2) + .5(904)(.02^2) is my energy conservation equation.
What about thermal energy? Sound...?
 
  • #11
dkgojackets
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I got it now. The bullet into the block gives the block a kinetic energy, which is immediately transferred into the spring. I know the spring potential energy and used it to solve for the velocity of the block immediately after the collision, which I put into conservation of momentum equation.
 

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