Solve Harmonic Problem: Find Bullet Speed Before Impact

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The problem involves calculating the speed of a bullet before impact using the principles of conservation of momentum and spring dynamics. A 25.0g bullet strikes a 0.600kg block attached to a spring with a spring constant of 6.70x103 N/m, resulting in a combined mass of 0.625kg. The correct speed of the bullet before impact is determined to be 557 m/s, achieved by applying the formula V = A√(k/m) and correctly utilizing conservation of momentum principles.

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ceeforcynthia
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Hi! I've been working on this problem for a while, subtracting this and that.. but i can't get the answer which is 557m/s. Help would be greatly appreciated! Thanks in advance ><

1. A 25.0g bullet strikes a .600kg block attached to a fixed horizontal spring whose spring constant is 6.70x10^3 and sets it into vibration with an amplitude of 21.5cm. What is the speed of the bullet before impact if the two objects move together after impact?


2. V=A\sqrt{k/m}



3. I used the equation above, adding the masses together (.625kg). I got the velocity to be 22.26m/s... but now i can't figure out how to get the speed of the bullet.
 
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