Analyzing Spring Oscillations with Colliding Blocks

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Homework Help Overview

The problem involves analyzing the dynamics of two blocks colliding and subsequently oscillating on a spring. The first block has a mass of 600 grams and is attached to a spring with a spring constant of 22 N/m, while the second block, weighing 220 grams, collides with the first block at a speed of 1.50 m/s. The discussion focuses on determining the amplitude and period of the resulting oscillation after the collision.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the use of conservation of momentum to find the velocity of the combined blocks after the collision. There are attempts to relate kinetic energy to potential energy to determine the amplitude of oscillation. Questions arise about unit consistency and the implications of using different unit systems.

Discussion Status

Several participants have provided insights into the calculations needed to find the amplitude and period of oscillation. There is ongoing clarification regarding the importance of using consistent units, particularly converting grams to kilograms. Some participants have noted potential errors in calculations and are working towards a clearer understanding of the relationships between kinetic and potential energy.

Contextual Notes

Participants are discussing the implications of unit conversions and the necessity of including units in calculations to avoid errors. The problem is framed within the context of homework constraints, emphasizing the need for careful reasoning and verification of steps taken in the calculations.

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Homework Statement


A horizontal spring with a spring constant of 22 N/m has a 600 gram block attached to it and is at rest on a frictionless surface. A second block which has a mass of 220 grams is pushed toward the 600 gram block at a speed of 1.50 m/s. The second block collides with and sticks to the 600 gram block.

a. What is the amplitude of the subsequent oscillation?

b. What is the period of the subsequent oscillation?



Homework Equations


Period = 1/F
A=sqroot(X^2+(V^2/w^2))


The Attempt at a Solution



I tried finding how far the spring compresses but i am not sure how to do this with the collision of the blocks
 
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jjd101 said:

Homework Statement


A horizontal spring with a spring constant of 22 N/m has a 600 gram block attached to it and is at rest on a frictionless surface. A second block which has a mass of 220 grams is pushed toward the 600 gram block at a speed of 1.50 m/s. The second block collides with and sticks to the 600 gram block.

a. What is the amplitude of the subsequent oscillation?

b. What is the period of the subsequent oscillation?



Homework Equations


Period = 1/F
A=sqroot(X^2+(V^2/w^2))


The Attempt at a Solution



I tried finding how far the spring compresses but i am not sure how to do this with the collision of the blocks

Use conservation of momentum to figure out the velocity of the combined block after the collision. Once you know this, you know the initial kinetic energy at the time when the spring first starts to compress. Figure out how far the spring gets compressed by assuming that all of this initial kinetic energy gets converted into elastic potential energy.
 
The speed of the 2 blocks together is .4024m/s KE=66.4 therefore when PE=66.4 the delta x is 2.46, therefore the Amp = 2.46, how do i find the period from this?
 
jjd101 said:
The speed of the 2 blocks together is .4024m/s KE=66.4 therefore when PE=66.4 the delta x is 2.46, therefore the Amp = 2.46, how do i find the period from this?

It's SHM, so by definition, the period is independent of the amplitude. The period only depends on two parameters. For a spring, what are these? (Hint: they have both been given to you in your problem).
 
jjd101 said:
The speed of the 2 blocks together is .4024m/s KE=66.4 therefore when PE=66.4 the delta x is 2.46, therefore the Amp = 2.46,

That's not what I get. Also, 2.46 in what UNITS? Just the number by itself is totally meaningless.
 
T= 2(pie)(sqroot(m/k)) T= 38.4sec?
 
220(1.5) = 330 820(v) = 330 v = .4024 KE = 1/2mv^2 KE = 66.4? 2.46 in m
 
jjd101 said:
T= 2(pie)(sqroot(m/k)) T= 38.4sec?

That's not the answer I'm getting. I know what mistake you're making. You need to pay closer attention to the units of your quantities. What units is m in? What units is k in? Are you consistently using the the same unit system for both? I would recommend converting everything to standard SI (i.e. mks) units.

Oh, and just FYI, the name of the Greek letter is "pi", not pie.
 
ohhhhhh, i was forgetting to change grams to kg? so with units as .82 kg now i get T = 1.213sec?
 
  • #10
jjd101 said:
220(1.5) = 330 820(v) = 330 v = .4024 KE = 1/2mv^2 KE = 66.4? 2.46 in m

You're having the same problem with units here again. Please include units on all quantities in your calculations. I cannot emphasize this enough. I'm not just saying so to be pedantic. It is crucially important for two reasons:

1) Without units, the equations in your computational steps really are meaningless.

2) More importantly, including units helps you to catch calculation mistakes. In a sensible physical equation, the units have to be the same on both sides. If the units aren't the same on both sides of the equation, then it's just nonsense. So, if you're doing some calculations, and you notice that the units don't work out, then you know you've made an error, and you can go back and fix it. So it makes a really really hand way of checking your work.

In particular, if you had included units on all of your quantities above, then you would have caught the mistake that you're making above. :wink:

EDIT: Also, where does the 330 come from? I don't get that.

EDIT 2: Nevermind. 330 (g*m/s) = (220 g)*(1.5 m/s). Now I get it. I was just confused before, because you didn't have any line breaks between your equations, So I though that the 330 was multiplying the 820v.
 
Last edited:
  • #11
i normally do include units it's just tedious on the computer haha, and 330 is the product of the mass of the block at 1.5m/s therefore it is a quantity of the momentum i believe? and i believe my mistake is when i do KE=1/2mv^2 i use grams instead of kilograms, therefore it should be KE=(1/2)(.82g)(.4024m/s)^2 KE = .066 J so when PE = .066J the distance changes to .078meters?
 
  • #12
jjd101 said:
ohhhhhh, i was forgetting to change grams to kg? so with units as .82 kg now i get T = 1.213sec?

Yeah that seems alright now.
 
  • #13
jjd101 said:
i normally do include units it's just tedious on the computer haha, and 330 is the product of the mass of the block at 1.5m/s therefore it is a quantity of the momentum i believe? and i believe my mistake is when i do KE=1/2mv^2 i use grams instead of kilograms, therefore it should be KE=(1/2)(.82g)(.4024m/s)^2 KE = .066 J so when PE = .066J the distance changes to .078meters?

I agree with that. Good work.
 
  • #14
Thank you!
 

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