Initial velocity of bullets - collisions

  • Thread starter nrc_8706
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  • #1
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a 11.9g bullet is fired vertically into a 5.49kg block of wood. the bullet gets stuck in the block, and the impact lifts the block 0.11m up. given g=9.8m/s^2. what was the initial velocity of the bullet?


Pi=Pf is this right?
m1Vi1=(m1+m2)Vf
 

Answers and Replies

  • #2
Fermat
Homework Helper
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Yes, that's right. Where Vf is the initial velocity of the block/bullet combination just before it rises up.
You now have to figure out Vf from the motion of the block/bullet.
 
  • #3
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?

in order find Vf of the block/bullet do i use the same formula and that will equal Vf=M1Vio/(m1+m2) if so, i now have two unknowns
 
  • #4
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use energy to find the velocity of the combination
 
  • #5
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ok..........is it m2gh=1/2(m1+m2)Vf^2 ?
 
  • #6
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the bullet gets stuck in the wood, and the equation would probably be:

[tex] KE_i = GPE_f [/tex]
 
  • #7
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is it 1/2m1Vo^2=m2gh ?
 
  • #8
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alright, since the bullet gets stuck in the wood, the block and the bullet would rise together, so you need to use both masses

the equation would be [itex] \frac{1}{2}(m_1 + m_2)(v_i)^2 = (m_1+m_2)(g)(h) [/itex]
 
  • #9
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ok...........now is this correct?


Vi^2=2(m1+m2)*g*h/(m1+m2)

Vi=(2(m1+m2)*g*h/(m1+m2))^.5

but isnt that final velocity?
 
  • #10
Fermat
Homework Helper
872
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The block rises 0.11m.

Using v² - u² = 2as

with u = Vf, v = 0, a = -g, then

Vf² = 2*9.8*0.11 = 2.156
Vf = 1.468 m/s
============

Plug that into your original momentum eqn,

m1Vi1=(m1+m2)Vf

to get v1.
 
  • #11
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its the final velocity after the collision occurs, but also the initial velocity as the block+bullet move upwards. so yeah, you're right in a way.
 
  • #12
71
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Gracias!!!!

thank you so much for your help. i actually worked it out myself but i thought it was such a big number that it couldnt be correct. thanks again for everything
 

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