Physics - Speeding Bullet?

Question

A bullet with a mass of 6.00g is fired through a 1.25kg block of wood on a frictionless surface. The initial speed of the bullet is 896m/s, and the speed of the bullet after it exits the block is 435m/s. At what speed does the block move after the bullet passes through it?

I need the calculations used as well as the answer - I'm homeschooling my kids, but I don't get it. Thanks!

Answer 1

Conservation of momentum

Momentum = Mass x Velocity

Initial mass x initial velocity = final mass x final velocity

(6.00e-3kg)(896m/s) = (6.00e-3kg)(435m/s) + (1.25kg)(X m/s)

5.376 kgm/s = 2.61kgm/s + (1.25 kg)(Xm/s)

2.766kgm/s = (1.25 kg)(Xm/s)

2.21 m/s = X The block would be moving 2.21m/s

Answer 2

The Bullet travels from 1900 FPS to 3000 FPS so if you're vacationing in one hundred mph (replace the mile to feet and hour to 2d ) if you're capturing it in the option route so subtract the bullet % out of your % and then get the answer.

Answer 3

Conservation of momentum...
m1*v1 = m2*v2
6 g * 896m/s = 6 g * 435m/s + (1250 g * x m/s)
5.37600 m kg / s = 2.61 m kg / s + (1.250 kg * x)
2.76600 m kg / s = 1.250 kg * x
2.76600 / 1.25000 = 2.2128 m/s
or
2.21 m/s with the right no. of significant digits