Double cheking momentum problem

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To find the bullet's speed as it leaves the gun, the conservation of momentum principle is applied. The equation used is m1V2 + m2v2ii = m1v1i + m2v2i, where m1 is the gun and m2 is the bullet. After simplifying, the bullet's speed is calculated as 66.6 m/s, derived from the gun's recoil speed of 2.5 m/s and its mass of 5.6 kg. The solution appears correct based on the calculations provided. The discussion confirms the accuracy of the derived bullet speed.
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Homework Statement


A bullet of mass 0.21 kg is fired from a gun of mass 5.6 kg. If the guns recoil speed is 2.5 m/s, what speed must the bullet have as it leaves the gun?


Homework Equations



m1V2 + m2v2ii = m1v1i + m2v2i
m1 = gun, m2 = bullett

The Attempt at a Solution



So,
The terms on the right side drop out.
m1V2 + m2v2ii = m1v1i + m2v2i
m1V2 + m2v2ii = 0
m1V2 = -m2v2ii
divide by m2
m1V2/-m2 = v2ii

(5.6 kg x 2.5 m/s) / (0.21 kg)
-66.6
Since it asks for speed 66.6 m/s
Is this correct I don't have a solution guide thanks.
 
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