Bullet Impact Speed Calculation: 2.25x10^3 N/m

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The discussion centers on calculating the speed of a bullet impacting a block attached to a spring with a spring constant of 2.25 x 10^3 N/m. A 0.0125 kg bullet strikes a 0.300 kg block, causing it to vibrate with an amplitude of 12.4 cm. The correct approach involves using the conservation of energy principle, where the initial kinetic energy of the bullet equals the spring's potential energy after impact. The calculated speed of the bullet is 10.5 m/s, which is confirmed as accurate despite initial concerns about its velocity being low for a bullet.

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A 0.0125 kg bullet strikes a 0.300 kg block attached to a fixed horizontal spring whose spring constant is 2.25 * 10^3 N/m and sets it into vibration with an amplitude of 12.4 cm. What was the speed of the bullet if the two objects move together after impact?

E = .5 k A2 = .5 m v2
Do I use m = mass of bullet + mass of block, or just the mass of the bullet?
Using the former, I get v = 10.5 m/s. Is this right? It just seems kind of slow for a bullet...
 
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This is an energy conservation problem

Initially the only energy is the kinetic energy of the bullet.

Finally, the only energy is the spring energy (potential+kinetic).

Those two energies are the same.
 

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