Posted by **ashley --- HELP PLEASEE!!!** on Saturday, May 7, 2011 at 8:04pm.

A 3.1-kilogram gun initially at rest is free to move. When a 0.015-kilogram bullet leaves the gun with a speed of 500. meters per second, what is the speed of the gun?

- physics -
**drwls**, Saturday, May 7, 2011 at 8:29pm
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

- physics -
**Stephanie**, Monday, December 5, 2011 at 10:32pm
7.5 m/s

- physics -
**Joey Fourier**, Wednesday, January 1, 2014 at 6:12pm
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)

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