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

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To calculate the bullet's impact speed, the kinetic energy of the bullet is equated to the spring's potential energy after the bullet strikes the block. The spring constant is given as 2.25 x 10^3 N/m, and the amplitude of vibration is 12.4 cm. The relevant formula is E = 0.5 k A^2 = 0.5 m v^2. Using the combined mass of the bullet and block for the final velocity calculation yields a speed of 10.5 m/s, which raises concerns about being low for a bullet's speed. The discussion confirms that this is an energy conservation problem, where the initial kinetic energy equals the final spring energy.
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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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