- #1

- 71

- 0

Pi=Pf is this right?

m1Vi1=(m1+m2)Vf

- Thread starter nrc_8706
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- #1

- 71

- 0

Pi=Pf is this right?

m1Vi1=(m1+m2)Vf

- #2

Fermat

Homework Helper

- 872

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You now have to figure out Vf from the motion of the block/bullet.

- #3

- 71

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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

- 217

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use energy to find the velocity of the combination

- #5

- 71

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ok..........is it m2gh=1/2(m1+m2)Vf^2 ?

- #6

- 217

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the bullet gets stuck in the wood, and the equation would probably be:

[tex] KE_i = GPE_f [/tex]

[tex] KE_i = GPE_f [/tex]

- #7

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is it 1/2m1Vo^2=m2gh ?

- #8

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the equation would be [itex] \frac{1}{2}(m_1 + m_2)(v_i)^2 = (m_1+m_2)(g)(h) [/itex]

- #9

- 71

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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

- 1

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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- #12

- 71

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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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