- #1

Chandasouk

- 165

- 0

What was the initial speed of the bullet?

Okay, first off, I know the block is sliding after the collision so that is internal energy.

[tex]\Delta[/tex]U

_{internal}= Friction Force* Distance

=(12.7106N)(.180)(.230m) = 0.52621884J

From here, i do not know what to do.

Here was what I attempted:

KE=[tex]\Delta[/tex]U

_{internal}

so

.5(1.297kg)v^2 = 0.52621884J

Velocity of the block was 0.9007m/s

Then I used Conservation of Momentum

Pi = Pf

.007kgV = (1.297kg)(.9007m/s)

V=166.9m/s

I don't know, seems slow for a bullet.