Momentum of a Bullet Contacting a Block

[LastXdeth]

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.)

 

[lewando]

Your mass units need attention.

 

[LastXdeth]

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]

4g is not 4000kg.

 

[LastXdeth]

4g is not 4000kg.

Aww crap! Dang, I feel so ridiculous for reversing it up!

 

[lewando]

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

 

[PotentialE]

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

 

[LastXdeth]

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

I got 103.75.

 

[PotentialE]

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

 

[LastXdeth]

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!