A block falling onto a spring (SHM related question)

In summary, the question is asking if the explanations for the answer make sense and whether mechanical energy is conserved before and after the collision. The answer relies on the ideality of the spring, which has negligible mass and can be ignored. The usual way to approach this problem is by using conservation of energy, where the initial height of the spring is considered to be at x=0.
  • #1
Tea_Aficionado
14
6
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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  • #2
An usual way is to use conservation of energy, i.e.
[tex]\frac{1}{2}mv^2+\frac{1}{2}kx^2+mgx=const.[/tex]
where x =0 at initial end's height of the spring.
 
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FAQ: A block falling onto a spring (SHM related question)

How does the mass of the block affect the frequency of the spring's oscillations?

The mass of the block does not affect the frequency of the spring's oscillations. The frequency of a spring's oscillations is determined by the spring constant and the mass of the spring itself, not the mass of the object attached to it.

Can the amplitude of the spring's oscillations be changed by altering the mass of the block?

Yes, the amplitude of the spring's oscillations can be changed by altering the mass of the block. A heavier block will cause the spring to stretch further and have a larger amplitude, while a lighter block will cause the spring to stretch less and have a smaller amplitude.

What is the relationship between the spring constant and the frequency of oscillations?

The spring constant and the frequency of oscillations have a direct relationship. As the spring constant increases, the frequency of oscillations also increases. This means that a stiffer spring will have a higher frequency of oscillations compared to a less stiff spring.

How does the initial position of the block affect the period of oscillations?

The initial position of the block does not affect the period of oscillations. The period of oscillations is solely determined by the spring constant and the mass of the spring itself. The initial position of the block only affects the amplitude of the oscillations.

Can the spring's oscillations continue indefinitely with a block attached?

No, the spring's oscillations cannot continue indefinitely with a block attached. The energy of the system will eventually dissipate due to friction and air resistance, causing the oscillations to decrease in amplitude and eventually come to a stop.

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