How Do Bullet Impact and Spring Constants Determine Oscillation Characteristics?

[Biggidy]

1. Homework Statement

A 4kg block resting on frictionless surface is attached to spring of k=100n/m. A 2g bullet is fired at 100m/s at the block. What is the period and amplitude of oscillations?



2. Homework Equations

1/2mv² = kinetic energy
1/2kx² = elastic energy
2 x pi x square root (mass/k) = period

3. The Attempt at a Solution

No answer choices given, just seeing if my thought process is valid.

First, period. 2 x pi x square root (4/100) = 2 x pi x 2/10 = 4pi/10 = approx. 1.2

Next, amplitude. 1/2 (.002g) (100m/s²) = 1/2 100n/m x²
therefore, x equals out to .45, or 45 cm?



Does this make sense, I think I have the period right, but one of my friends got 1 cm for the amplitude and another one of my friends got 0.83 seconds for the period. Any ideas?


Thanks

 

[aim1732]

There's been a mistake.
You took an inelastic collision for a completely elastic one.Instead conserve momentum to find velocity of bullet-block system and then proceed for momentum.

 

[Biggidy]

That makes great sense. Thanks