# Momentum of a Bullet Contacting a Block

## Homework Statement

A bullet with a mass of 4.0 g and a speed of 650 m/s is fired at a block of wood with a mass of 0.095 kg. The block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 23 m/s.

(a) What is the speed of the bullet when it exits the block?

(b) Is energy conserved in this collision? (yes or no)

## Homework Equations

(m1)(v1o)+(m2)(v2o)=(m1)(v1)+(m2)(v2)

## The Attempt at a Solution

(4000)(650)+(0.095)(0)=(4000)(v1)+(0.095)(23)

v1≈650.00 (Apparently this is wrong.)

Last edited:

## Answers and Replies

lewando
Homework Helper
Gold Member
Your mass units need attention.

Your mass units need attention.
I don't see any mass unit that needs attention. The only thing I did with mass was convert m1 (4.0 g) to the SI unit kg.

lewando
Homework Helper
Gold Member
4g is not 4000kg. 4g is not 4000kg. Aww crap! Dang, I feel so ridiculous for reversing it up!

lewando
Homework Helper
Gold Member
We all have our moments. That would be some big bullet. No wonder it didn't slow down.

With the correct unit conversions and your same formula, I got 296.94

With the correct unit conversions and your same formula, I got 296.94

I got 103.75. MV+MV = MV+MV
.004kg*650m/s + .095kg*0 = .004kg*VF + .095kg*23
2.6 = .004*VF + 2.185
.0415 = .004VF
Vf= 103.75

0.0 my bad, man sorry about that

MV+MV = MV+MV
.004kg*650m/s + .095kg*0 = .004kg*VF + .095kg*23
2.6 = .004*VF + 2.185
.0415 = .004VF
Vf= 103.75

0.0 my bad, man sorry about that

No problem! Thanks for confirming!