A block falling onto a spring (SHM related question)

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SUMMARY

The discussion centers on the mechanics of a block falling onto a spring, specifically addressing the conservation of mechanical energy during the collision. The participant confirmed that the spring is ideal, meaning its mass is negligible, which simplifies the analysis. The conservation of energy equation, 1/2 mv^2 + 1/2 kx^2 + mgx = const., is utilized to explain the system's behavior, indicating that energy is conserved regardless of whether the collision is elastic or inelastic.

PREREQUISITES
  • Understanding of simple harmonic motion (SHM)
  • Familiarity with conservation of mechanical energy principles
  • Knowledge of spring constants and their role in energy storage
  • Basic mechanics of elastic and inelastic collisions
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  • Study the principles of simple harmonic motion (SHM) in detail
  • Learn about the conservation of mechanical energy in different types of collisions
  • Explore the mathematical modeling of spring systems using Hooke's Law
  • Investigate the effects of non-ideal springs on energy conservation
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Physics students, mechanical engineers, and anyone interested in the dynamics of collisions and energy conservation in spring systems.

Tea_Aficionado
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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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