## mtimko one year ago A 3.50-g bullet is fired from a 5.50-kg gun with a muzzle velocity of 575 m/s. What is the speed of the recoil of the gun?

1. IrishBoy123

what is meant by "muzzle velocity"? is it the velocity of the bullet relative to the ground or relative to the gun?

2. mtimko

Unfortunately, I'm not sure. I would assume to the ground

3. IrishBoy123

i would assume to the gun barrel. otherwise you will need to know your speed with respect to ground to answer question. do you have any ideas as to how you might answer this? any equations or laws?

4. mtimko

3.50 g x 575 m/s= 2012.5/ 5.5 Kg= 365.91 I'm not sure about units though. Am I on the right track?

5. anonymous

Careful with units. The bullet is 3.5 GRAMS. The gun is 5.5 KILOGRAMS. Correct approach, however.

6. mtimko

Does it matter if I change units of the bullet or the gun?

7. IrishBoy123

done some research: "Muzzle velocity is the speed a projectile has at the moment it leaves the muzzle of the gun."

8. IrishBoy123

and where does the 2012.5 come from?

9. mtimko

The 2012.5 came from the 3.50 x 575.

10. anonymous

You can change the units of either. You're dividing the two masses, so the units will cancel, provided they are the same units.

11. anonymous

Your notation may be confusing to some. You're using conservation of momentum. Your initial momentum is 0, because everything is at rest. After the bullet is fire, momentum must be conserved. $\vec{\rm{p}}_\rm{b} + \vec{\rm{p}}_\rm{g} = 0 \rightarrow \vec{\rm{p}}_{b} = -\vec{\rm{p}}_\rm{g}$ The negative sign just denotes the opposite direction. So now we have: $m_g\vec{v}_g = -m_b\vec{v}_b \rightarrow \vec{v}_g = -\frac{m_b\vec{v_b}}{m_g}$ And speed is just the absolute value of the velocity vector.