# Bullet in Block- 2D Projectiles w/ Motion

1. Nov 30, 2011

### physicsluv

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

ΔX = Vi(t) + (1/2)(a)(t^2)

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

3. 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 :)

2. Nov 30, 2011

### 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 m1v1i+m2v2i=m1v1f+m2v2f
You should then be able to quickly simplify one side by equating one velocity with another, and it will all fall out from there.

3. Nov 30, 2011

### physicsluv

What would m2, v2i, and v2f be?

4. Nov 30, 2011

### 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

5. Nov 30, 2011

### physicsluv

I got it, thanks!