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

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0what is meant by "muzzle velocity"? is it the velocity of the bullet relative to the ground or relative to the gun?

mtimko
 one year ago
Best ResponseYou've already chosen the best response.0Unfortunately, I'm not sure. I would assume to the ground

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0i 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?

mtimko
 one year ago
Best ResponseYou've already chosen the best response.03.50 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Careful with units. The bullet is 3.5 GRAMS. The gun is 5.5 KILOGRAMS. Correct approach, however.

mtimko
 one year ago
Best ResponseYou've already chosen the best response.0Does it matter if I change units of the bullet or the gun?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0done some research: "Muzzle velocity is the speed a projectile has at the moment it leaves the muzzle of the gun."

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0and where does the 2012.5 come from?

mtimko
 one year ago
Best ResponseYou've already chosen the best response.0The 2012.5 came from the 3.50 x 575.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can change the units of either. You're dividing the two masses, so the units will cancel, provided they are the same units.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Your 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.
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