Question

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?

Answer 1

what is meant by "muzzle velocity"?

is it the velocity of the bullet relative to the ground or relative to the gun?

Answer 2

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

Answer 3

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?

Answer 4

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?

Answer 5

Careful with units. The bullet is 3.5 GRAMS. The gun is 5.5 KILOGRAMS.

Correct approach, however.

Answer 6

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

Answer 7

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

Answer 8

and where does the 2012.5 come from?

Answer 9

The 2012.5 came from the 3.50 x 575.

Answer 10

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

Answer 11

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.