Help with a Simple Harmonic Motion problem

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SUMMARY

The discussion centers on calculating the speed of a bullet before it impacts a block attached to a spring. The bullet has a mass of 25.0g and strikes a 0.600kg block, resulting in a combined mass of 0.625kg. The spring constant is 6.70x103 N/m, and the amplitude of vibration is 21.5cm. Using the formula V = A√(k/m), the calculated velocity after impact is 22.26m/s, but the correct speed of the bullet before impact is established as 557m/s through conservation of momentum principles.

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  • Knowledge of spring constants and their implications
  • Ability to manipulate equations involving mass and velocity
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ceeforcynthia
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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[tex]\sqrt{k/m}[/tex]
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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ceeforcynthia said:
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[tex]\sqrt{k/m}[/tex]



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.

Use conservation of momentum. momentum of the bullet before the impact = momentum of the block plus bullet moving together at 22.26 m/s
 

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