Initial velocity of bullets - collisions

[nrc_8706]

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

 

[Fermat]

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.

 

[nrc_8706]

?

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

 

[andrewchang]

use energy to find the velocity of the combination

 

[nrc_8706]

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

 

[andrewchang]

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

KE_i = GPE_f

 

[nrc_8706]

is it 1/2m1Vo^2=m2gh ?

 

[andrewchang]

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 \frac{1}{2}(m_1 + m_2)(v_i)^2 = (m_1+m_2)(g)(h)

 

[nrc_8706]

ok...now is this correct?


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

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

but isn't that final velocity?

 

[Fermat]

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.

 

[andrewchang]

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.

 

[nrc_8706]

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 couldn't be correct. thanks again for everything