Springs, Simple Harmonic Motion

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A 25.0g bullet strikes a 0.600kg block attached to a spring, initiating simple harmonic motion with an amplitude of 21.5cm. The spring constant is 6.70x10^3, and the combined mass of the bullet and block is 0.625kg. The initial calculations suggest a velocity of 22.26m/s, but the correct speed of the bullet before impact is 557m/s, as determined by equating the kinetic energy of the bullet to the potential energy stored in the spring at maximum deflection. This scenario exemplifies an inelastic collision, where momentum conservation principles also apply. Understanding the relationship between kinetic and potential energy is crucial for solving such problems.
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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.


The answer in the book says that the speed of the bullet is 557m/s.
 
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One determines the maximum spring deflection from the amplitude, and from that the stored energy in the spring at max deflection, which by conservation of energy (and neglecting friction) should equal the initial kinetic energy of the bullet and block.

From the KE of the bullet and block, one can determine the velocity of the combination.

Since the bullet striking the block is an example of an inelastic collision, what about the momentum of the bullet and block?
 
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