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.
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.