Energy conservation: Bullet-block collision

  • Thread starter n.hirsch1
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  • #1
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Homework Statement


A 2.30 g bullet embeds itself in a 1.6 kg block, which is attached to a spring of force constant 770 N/m.
A) If the maximum compression of the spring is 5.40 cm, find the initial speed of the bullet.


Homework Equations


Step one: convert everything to kg and m
Step two: 1/2 mvi^2 = 1/2 kA^2


The Attempt at a Solution


I first tried to solve the above equation for vi, and then plug in the rest since m, k, and A are given, but this gave me 29.1 m/s and that is not the correct answer. I am not sure where else to go with this, this is the only equation I know that deals with initial velocity!
 

Answers and Replies

  • #2
Delphi51
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Your conservation of energy equation should be correct for the compression stage, so you can find the initial speed of the bullet/block combination that way (much less than 29 m/s). Use conservation of momentum in the bullet-block collision stage.
 
  • #3
mgb_phys
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Sounds like the correct method, check your calculations

1/2 mvi^2 = 1/2 kA^2

0.0023 v^2 = 770 * 0.054
v = sqrt (770 * 0.054/0.0023) = 134m/s
 
  • #4
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134 m/s isn't correct either. I don't understand what is wrong with this equation, everything is given.
I can't use conservation of momentum for inelastic collisions because I don't have a final velocity.
 
  • #5
mgb_phys
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You do have the final velocity - it says maximum compression of the spring = block is stationary
 
  • #6
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mgb_phys: I must be looking at the wrong equation. for conservation of momentum, if the final velocity of the system is zero (which it is), and it is an inelastic collision, then the equation is :
vf = m1vi +m2vi / m1 + m2
and m2vi is equal to zero since its at rest, so
mvi = (m + M) vf
and if the whole system goes to rest, then everything is zero!!
 
  • #7
mgb_phys
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You can use conservation of momentum to find the initial velocity of bullet + block and then use conservation of energy to find the kinetic to spring energy but this seems a little pointless if you are assuming elastic collisions then the ke of the bullet becomes the spring energy.
 
  • #8
Delphi51
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For the spring compressing:
1/2 mvi^2 = 1/2 kA^2
vi = sqrt(kA²/m) = sqrt(770*.054²/1.623) = 1.176 m/s

Now you know the speed of the bullet + block after the collision.
Use conservation of momentum to work back to the speed of the bullet before the collision. This collision cannot be elastic because the bullet embeds itself into the block with very much friction!
 

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