A block falling onto a spring (SHM related question)

AI Thread Summary
The discussion centers on the conservation of mechanical energy in a scenario where a block falls onto a spring. The contributor confirms that the mechanical energy remains conserved before and after the collision, regardless of whether the collision is elastic or inelastic, due to the ideal nature of the spring. They emphasize that the spring's negligible mass allows it to be ignored in calculations. The conservation of energy equation is referenced to illustrate the concept, highlighting the relationship between kinetic energy, spring potential energy, and gravitational potential energy. Overall, the explanations provided are aimed at clarifying the principles of simple harmonic motion in this context.
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Homework Statement
A block of mass m = 2.0kg is dropped from a height h = 0.45m, above an uncompressed spring as shown here. The spring has an elastic constant k = 200 N/m and negligible mass. The block strikes the end of the spring and sticks to it.

a. Determine the speed of the block at the instant it hits the end of the spring
b. Determine the force in the spring when the block reaches the equilibrium position
c. Determine the distance that the spring is compressed at the equilibrium position
d. Determine the speed of the block at the equilibrium position
e. Determine the resulting amplitude of the oscillation that ensues
f. Is the speed of the block a maximum at the equilibrium position, explain
g. Determine the period of the SHM that ensues
Relevant Equations
Please see below
EDIT for clarity: I solved the question, just asking for if the explanations make sense and if the mechanical energy is considered to be conserved before and after the collision due to reasons listed below the photo.

IMG_2531.jpeg

I hope this image is readable (grr, scanner is janky).

I'm guessing the answer (to the question at the bottom of the paper) is again, due to ideality and therefore conservation of mechanical energy. And as haru has pointed out earlier, it doesn't matter whether the collision was elastic or inelastic because the spring is considered to be ideal, and therefore has negligible mass and should be ignored.

(Also: do the answers' explanations make sense?)
 
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An usual way is to use conservation of energy, i.e.
\frac{1}{2}mv^2+\frac{1}{2}kx^2+mgx=const.
where x =0 at initial end's height of the spring.
 
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