physics
posted by ashley  HELP PLEASEE!!! on .
A 3.1kilogram gun initially at rest is free to move. When a 0.015kilogram bullet leaves the gun with a speed of 500. meters per second, what is the speed of the gun?

This a chance for you to apply the law of conservation of momentum.
That means the momentum of the gun going backward is equal and opposite to the momentum of the bullet going forward. The two momenta cancel out.
Mbullet*Vbullet = Mgun*Vgun
Solve for Vgun 
7.5 m/s

pbefore=pafter
m1v1+m2v2 = m1v’1+m2v’2
(3.1kg)(0m/s)+(0.015kg)(0m/s) =
(3.1kg)( v’1)+(0.015kg)(500.m/s)
0kg∙m/s=(3.1kg)( v’1)+7.5kg∙m/s
7.5kg∙m/s=(3.1kg)( v’1)
v’1=2.4m/s (its speed is 2.4 m/s)