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Momentum of a Bullet Contacting a Block

  • Thread starter LastXdeth
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  • #1
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Homework Statement


A bullet with a mass of 4.0 g and a speed of 650 m/s is fired at a block of wood with a mass of 0.095 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 23 m/s.

(a) What is the speed of the bullet when it exits the block?


(b) Is energy conserved in this collision? (yes or no)

Homework Equations


(m1)(v1o)+(m2)(v2o)=(m1)(v1)+(m2)(v2)


The Attempt at a Solution



(4000)(650)+(0.095)(0)=(4000)(v1)+(0.095)(23)

v1≈650.00 (Apparently this is wrong.)
 
Last edited:

Answers and Replies

  • #2
lewando
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Your mass units need attention.
 
  • #3
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Your mass units need attention.
I don't see any mass unit that needs attention. The only thing I did with mass was convert m1 (4.0 g) to the SI unit kg.
 
  • #4
lewando
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4g is not 4000kg. :smile:
 
  • #5
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4g is not 4000kg. :smile:
Aww crap! Dang, I feel so ridiculous for reversing it up!
 
  • #6
lewando
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We all have our moments. That would be some big bullet. No wonder it didn't slow down.
 
  • #7
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With the correct unit conversions and your same formula, I got 296.94
 
  • #8
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With the correct unit conversions and your same formula, I got 296.94
I got 103.75. :confused:
 
  • #9
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MV+MV = MV+MV
.004kg*650m/s + .095kg*0 = .004kg*VF + .095kg*23
2.6 = .004*VF + 2.185
.0415 = .004VF
Vf= 103.75

0.0 my bad, man sorry about that
 
  • #10
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MV+MV = MV+MV
.004kg*650m/s + .095kg*0 = .004kg*VF + .095kg*23
2.6 = .004*VF + 2.185
.0415 = .004VF
Vf= 103.75

0.0 my bad, man sorry about that
No problem! Thanks for confirming!
 

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