# Initial velocity of bullets - collisions

1. Jan 6, 2006

### 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

2. Jan 6, 2006

### 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.

3. Jan 6, 2006

### 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

4. Jan 6, 2006

### andrewchang

use energy to find the velocity of the combination

5. Jan 6, 2006

### nrc_8706

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

6. Jan 6, 2006

### andrewchang

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

$$KE_i = GPE_f$$

7. Jan 6, 2006

### nrc_8706

is it 1/2m1Vo^2=m2gh ?

8. Jan 6, 2006

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

9. Jan 6, 2006

### 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 isnt that final velocity?

10. Jan 6, 2006

### 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.

11. Jan 6, 2006

### 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.

12. Jan 6, 2006

### 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 couldnt be correct. thanks again for everything