- #1

amaryllia

- 8

- 0

## Homework Statement

An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of height 1 m. The bullet remains in the block, and after the impact the block lands 2 m from the bottom of the table. Determine the initial speed of the bullet.

## Homework Equations

Conservation of Momentum (for perfectly inelastic conditions)

m1v1(initial) + m2v2 (initial) = (m1+m2)vf

m1 = mass of bullet (.008 kg)

v1 = velocity of bullet

m2 = mass of block (.250 kg)

v2 = velocity of block (0 m/s)

vf = final velocity of block and bullet right after collision

Conservation of Energy

KE + PE (after collision) = KE + PE (on ground), rewritten as:

1/2mvi^2 + mgyi = 1/2mvf^2 + mgyf

m = mass of bullet + block (.258 kg)

vi = initial velocity of bullet + block after collision

g = 9.8 m/s^2

yi = 1 m

vf = 0 m/s

yf = 0 m

## The Attempt at a Solution

I think I have this problem solved correctly, but wondered if someone could confirm that for me (those darn even-numbered problems at the back of the book without answers!).

First, I use the conservation of momentum and put in my known values, ending up with:

.008v1 = .258vf

Next, I use the conservation of energy and put in the known values, ending up with 4.427 m/s for the initial velocity of the block and bullet, which si the final velocity in my first equation.

I substitute 4.427 m/s into my first equation and get an answer for the initial speed of the bullet as 142.75 m/s.