Bullet in Block- 2D Projectiles w/ Motion

[physicsluv]

Homework Statement

A(n) 9 g bullet is fired into a 396 g block
that is initially at rest at the edge of a frictionless table of height 0.9 m. The bullet remains
in the block, and after impact the block lands
2.3 m from the bottom of the table.
The acceleration of gravity is 9.8 m/s.

Homework Equations

ΔX = Vi(t) + (1/2)(a)(t^2)

m(Vf-Vi) = -m(Vf-Vi)

The Attempt at a Solution

I separated the equation into vertical and horizontal parts.

Horizontal: ΔX=2.3, a = 0 m/s/s
Vertical: ΔY= .9, a= -9.8 m/s/s, Vi = 0 m/s (because it is launched horizontally).

Using the kinematic equation ΔX = Vi(t) + (1/2)(a)(t^2) to solve for the vertical time:
t = 0.4285714286 seconds

Since the vertical time is equal to the horizontal time, I plugged this time into the same kinematic equation to solve for the horizontal Vi:
Vi = 5.366666667 m/s

Now, I know you need to use the collision equation to solve for the initial velocity in the bullet, but I do not know how to set it up...
I have this so far

9(5.366666667-vi)=-____?_____

Thank you for the help :)

 

[Barakn]

m(Vf-Vi) = -m(Vf-Vi) That is not the collision equation I am familiar with. With just a single m, it suggests a collision involving a single object, an oxymoron if I've ever heard one. Try m₁v_1i+m₂v_2i=m₁v_1f+m₂v_2f
You should then be able to quickly simplify one side by equating one velocity with another, and it will all fall out from there.

 

[physicsluv]

m(Vf-Vi) = -m(Vf-Vi) That is not the collision equation I am familiar with. With just a single m, it suggests a collision involving a single object, an oxymoron if I've ever heard one. Try m₁v_1i+m₂v_2i=m₁v_1f+m₂v_2f
You should then be able to quickly simplify one side by equating one velocity with another, and it will all fall out from there.

What would m2, v2i, and v2f be?

 

[Zula110100100]

m2 would be mass of the block, v2i is the initial velocity of the block, and v2f is the final velocity of the block

 

[physicsluv]

I got it, thanks!