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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: /0 on April 28, 2008, 08:38:17 pm

Title: Spring and collision
Post by: /0 on April 28, 2008, 08:38:17 pm
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!

Title: Re: Spring and collision
Post by: Mao on April 28, 2008, 08:46:28 pm
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).
Title: Re: Spring and collision
Post by: /0 on April 28, 2008, 08:48:10 pm
Well good luck on that Mao! You can move this topic I guess
Title: Re: Spring and collision
Post by: enwiabe on April 28, 2008, 09:04:49 pm
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.
Title: Re: Spring and collision
Post by: enwiabe on April 28, 2008, 09:06:59 pm
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. :P
Title: Re: Spring and collision
Post by: cara.mel on April 28, 2008, 09:09:11 pm
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.
Title: Re: Spring and collision
Post by: enwiabe on April 28, 2008, 09:11:24 pm
Is the mass of block 2 .6 kg?
Title: Re: Spring and collision
Post by: enwiabe on April 28, 2008, 09:13:35 pm
wouldn't you use conservation of momentum, not conservation of energy?
Title: Re: Spring and collision
Post by: /0 on April 28, 2008, 09:13:50 pm
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
Title: Re: Spring and collision
Post by: dcc on April 28, 2008, 09:15:29 pm


note i did this with my ruler, measuring the relative lengths of h & d
Title: Re: Spring and collision
Post by: /0 on April 28, 2008, 09:16:50 pm
This is what I've got to so far:







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)
Title: Re: Spring and collision
Post by: cara.mel on April 28, 2008, 09:20:02 pm
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 :(
Title: Re: Spring and collision
Post by: Neobeo on April 28, 2008, 09:50:49 pm
Step 1: Mass of block 2
Using formula for simple harmonic motion for mass on a spring:




Step 2: Collision of blocks
Using elastic collision formulas:






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

Initial Energy:


Final Energy:


where

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:



Title: Re: Spring and collision
Post by: /0 on April 28, 2008, 10:22:32 pm
Wow neobeo! Thanks a heap!

I just don't understand how you got those momentum formulas though ? 
Title: Re: Spring and collision
Post by: Neobeo on April 28, 2008, 10:30:05 pm
I just don't understand how you got those momentum formulas though ? 

Copied and pasted from the textbook :P. In any case http://en.wikipedia.org/wiki/Elastic_collision should be a good starting point, letting u2 = 0.
Title: Re: Spring and collision
Post by: /0 on April 28, 2008, 11:03:45 pm
Also how did you get

?
Title: Re: Spring and collision
Post by: Neobeo on April 28, 2008, 11:51:49 pm
Also how did you get

?

Well, by first principles you could derive for the spring:



and then
Title: Re: Spring and collision
Post by: /0 on May 01, 2008, 01:36:25 am
Sorry for replying so late, but I checked in the book and it says





I'm guessing this is if the block is being stretched at t = 0, is this right?

When the block starts at rest, does that mean

?

Also, which direction do you usually choose to be 'negative'? The direction in the compression of the spring, or the direction stretching the spring?

Thanks again :p