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

[endeavor]

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...

 

[christianjb]

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.

 

[m2003]

I would approach this problem by first acknowledging that the given information is not enough to accurately determine the speed of the bullet. The equation provided is used to calculate the kinetic energy of the block after the impact, but does not take into account the energy lost due to friction and other factors.

To accurately calculate the speed of the bullet, we would need to know the initial position and velocity of the block, as well as the coefficient of restitution (a measure of how much energy is lost during the impact) between the bullet and the block.

Without this information, it is not possible to accurately calculate the speed of the bullet. However, based on the given information, a speed of 10.5 m/s does seem reasonable and falls within the typical range for bullet speeds. It is important to note that the speed of a bullet can vary greatly depending on factors such as the type of gun and ammunition used, as well as external factors like air resistance.