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January 28, 2020, 01:50:44 pm

### AuthorTopic: Spring and collision  (Read 5398 times) Tweet Share

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#### /0

• Victorian
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##### Spring and collision
« on: April 28, 2008, 08:38:17 pm »
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I don't know if this can be posted here, but there isn't a physics section in the University forums...

The diagram shows block 1 of mass 0.200kg sliding to the right over a frictionless elevated surface at a speed of 8.00m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of spring constant 1208.5N/m. (Assume that the spring does not affect the collision.) After the collision, block 2 oscillates in simple harmonic motion with a period of 0.140s, and block 1 slides off the opposite end of the elevated surface, landing a distance d from the base of that surface after falling height h = 4.90m. What is the value of d?

Thanks!

#### Mao

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##### Re: Spring and collision
« Reply #1 on: April 28, 2008, 08:46:28 pm »
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woah~ no idea...
wherever u got this from, it cannot be part of the VCE course!

try wikipedia
if i dont have an English SAC 2morrow, i'll actually read their section on that harmonic oscillation (under spring).
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• Victorian
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##### Re: Spring and collision
« Reply #2 on: April 28, 2008, 08:48:10 pm »
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Well good luck on that Mao! You can move this topic I guess

#### enwiabe

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##### Re: Spring and collision
« Reply #3 on: April 28, 2008, 09:04:49 pm »
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Lol I can write up a solution to that, but that's first year engineering dynamics. You most DEFINITELY don't need to know how to do it.

Simple harmonic motion is governed by

x = Asin(Wnt) + Bcos(Wnt)

And the differential equation you need to model it is:

md^2x/dt + kx = 0

Rofl seriously that is like... so far off the course it isn't funny.

#### enwiabe

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##### Re: Spring and collision
« Reply #4 on: April 28, 2008, 09:06:59 pm »
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I just started trying to work that out and... I'm stuck. Rofl. I'll keep trying to nut it out but omg this is hard.

#### cara.mel

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##### Re: Spring and collision
« Reply #5 on: April 28, 2008, 09:09:11 pm »
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Don't have time to help but quickly:
Angular frequency = sqrt(k/m) = 2*pi/f, from this get m of block 2.
Then you can get total energy of block 2 from either a^2*k/2 (a = w^2*x) or finding potential energy at the end or kinetic energy at the middle by using the normal KE formula but subbing in the equation for velocity in SHM, I haven't thought about this too much yet.
Then use conservation of energy to get final v of block 1, then youre on home stretch

Note: I cant think atm and I don't guarantee I've rememered all the SHM stuff right, it's at least half-right.

#### enwiabe

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##### Re: Spring and collision
« Reply #6 on: April 28, 2008, 09:11:24 pm »
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Is the mass of block 2 .6 kg?

#### enwiabe

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##### Re: Spring and collision
« Reply #7 on: April 28, 2008, 09:13:35 pm »
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wouldn't you use conservation of momentum, not conservation of energy?

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• Victorian
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##### Re: Spring and collision
« Reply #8 on: April 28, 2008, 09:13:50 pm »
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Is the mass of block 2 .6 kg?

Yeah sorry I didn't read your question correctly, using angular momentum I get block 2 = 0.6kg.

Now I just need to know if the collision is elastic or inelastic
« Last Edit: April 28, 2008, 09:17:42 pm by DivideBy0 »

#### dcc

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##### Re: Spring and collision
« Reply #9 on: April 28, 2008, 09:15:29 pm »
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$d = \boxed{5.6 m}$

note i did this with my ruler, measuring the relative lengths of h & d

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##### Re: Spring and collision
« Reply #10 on: April 28, 2008, 09:16:50 pm »
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This is what I've got to so far:

$\omega = 2\pi \frac{1}{T}=\sqrt{\frac{k}{m_2}}$

$m_2 = 0.6kg$

$a(t) = -{\omega}^2 \times A \cos{(\omega t)}$

Now I've go the acceleration but I'm missing the Amplitude, but somehow I hope to get the force from this.

(sorry for taking so long to get the latex working)
« Last Edit: April 28, 2008, 09:19:09 pm by DivideBy0 »

#### cara.mel

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##### Re: Spring and collision
« Reply #11 on: April 28, 2008, 09:20:02 pm »
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Is the mass of block 2 .6 kg?

Yeah sorry I didn't read your question correctly, using angular momentum I get block 2 = 0.6kg.

Now I just need to know if the collision is elastic or inelastic

Question said elastic.

Sorry I dont have time to help

#### Neobeo

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##### Re: Spring and collision
« Reply #12 on: April 28, 2008, 09:50:49 pm »
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Step 1: Mass of block 2
Using formula for simple harmonic motion for mass on a spring:

$m = k(\frac{T}{2\pi})^2 = 1208.5\times(\frac{0.14}{2\pi})^2 = 0.6 \mbox{ kg}$

Step 2: Collision of blocks
Using elastic collision formulas:

$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 = \frac{0.2-0.6}{0.2+0.6}\times8.0=-4.0\mbox{ ms^{-1}}$

$v_2 = \frac{2m_1}{m_1+m_2}u_1 = \frac{2\times0.2}{0.2+0.6}\times8.0=4.0\mbox{ ms^{-1}}$

Verification step: Conservation of Energy (not required)
Since all the objects are level, (before falling off), we conveniently ignore gravitational potential energy.

Initial Energy:
$E = E_k = \frac{1}{2}m_1 u_1^2 = \frac{1}{2}\times0.2\times8.0^2=6.4\mbox{ J}$

Final Energy:
$E=E_k+E_{spring}=\frac{1}{2}m_1 v_1^2 + \frac{1}{2}k A^2 = \frac{1}{2}\times0.2\times(-4.0)^2 + \frac{1}{2}\times1208.5\times0.08913^2 = 6.4\mbox{ J}$

where $A = \frac{v_2 T}{2\pi} = \frac{4.0\times0.14}{2\pi} = 0.08913$

This step isn't required. It's just to check that energy is conserved to ensure no mistakes anywhere.

Step 3: Trajectory
Using linear motion formulas in the vertical direction:

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times4.9}{9.8}}=1.0\mbox{ s}$

$d=|v_1 t|=|-4.0\times1.0|=4.0\mbox { m}$
« Last Edit: April 28, 2008, 10:13:23 pm by Neobeo »
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#### /0

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##### Re: Spring and collision
« Reply #13 on: April 28, 2008, 10:22:32 pm »
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Wow neobeo! Thanks a heap!

I just don't understand how you got those momentum formulas though ?

#### Neobeo

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##### Re: Spring and collision
« Reply #14 on: April 28, 2008, 10:30:05 pm »
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I just don't understand how you got those momentum formulas though ?

Copied and pasted from the textbook . In any case http://en.wikipedia.org/wiki/Elastic_collision should be a good starting point, letting u2 = 0.
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