Elastic collisions


by doub
Tags: collisions, elastic
doub
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#1
Dec17-12, 12:19 PM
P: 15
1. The problem statement, all variables and given/known data

A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball. If the first ball moves away with an angle 30 degrees to the original path, determine the speed of the first ball after the collision, and the speed and direction of the second ball after the collision.

2. Relevant equations

v_1 = V_1cos30 + v_2cos(theta) for the movement in the "x" direction

and

0 = v_1cos30 + v_2cos(theta) for movement in the "y" directions


3. The attempt at a solution

Played with this for hours but to me it does not seem like there is enough information. I feel like I am missing at velocity for after the collision.

Thanks
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tiny-tim
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#2
Dec17-12, 12:25 PM
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hi doub! welcome to pf!
Quote Quote by doub View Post
A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically
an elastic collision is defined as one in which energy is conserved (as well as momentum, which is always conserved in collisions)
doub
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#3
Dec17-12, 12:46 PM
P: 15
Right,

this is the best answer I got however I do not feel anywhere near confident.

3.0 m/s = v_1'cos30 + v_2'cos(theta) = 2.598 m/s + v_2'cos(theta) = 3.0 m/s - 2.598 m/s
so v_2'cos(theta) = 0.402 m/s in the "x" direction

0 = v_1'sin30 + v_2'sin(theta) = 1.299 m/s + v_2'sin(theta). so v_2'sin(theta) = -1.299 m/s

tan^-1 = v_2'cos(theta)/v_2'sin(theta) = -1.299/0402 = -72 degrees

and using sqrt(v_2'sin(theta)^2 + v_2'cos(theta)^2 = 1.36 m/s

am I anywhere in the ballpark at least?

tiny-tim
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#4
Dec17-12, 01:24 PM
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Elastic collisions


hi doub!

(try using the X2 button just above the Reply box )
Quote Quote by doub View Post
3.0 m/s = v_1'cos30 + v_2'cos(theta) = 2.598 m/s + v_2'cos(theta) = 3.0 m/s - 2.598 m/s
no, v1' isn't 3.0, it's unknown!

try the momentum equations again (and you'll need an energy equation also)
doub
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#5
Dec17-12, 01:38 PM
P: 15
yeah I'm totally lost now
tiny-tim
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#6
Dec17-12, 01:57 PM
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start again, with v1' v2' and θ as your three variables

(you have three equations: x, y, and energy, so that should be solvable )
show us what you get
doub
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#7
Dec17-12, 02:02 PM
P: 15
Ok,

The equations I have gotten are

x --> v_1 = v_1'cos30 + v_2'cos(theta)

y --> 0 = v_1'sin30 + v_2'sin(theta)

Energy --> v_1^2 = v_1'^2 + v_2'^2
tiny-tim
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#8
Dec17-12, 02:11 PM
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fine so far

now fiddle about with the first two equations so that cosθ and sinθ are on their own, then use cos2θ + sin2θ = 1 to eliminate θ
doub
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#9
Dec17-12, 02:25 PM
P: 15
So,

cos(theta) = v1 - (v1'cos30)/v2'

and

sin(theta) = (-va'sin30)/v2'

where do the sin2theta come from?
tiny-tim
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#10
Dec17-12, 02:33 PM
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uhh?
square both equations!
doub
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#11
Dec17-12, 02:45 PM
P: 15
I am just not seeing this...

cos2(θ) = (v12 -v12'cos302)/v2'2

sin2(θ) = (-v12'sin302)/v2'2

thanks very much for helping btw
tiny-tim
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#12
Dec17-12, 02:53 PM
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ok, now add
θ will miraculously disappear!
pzzaaam!
doub
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#13
Dec17-12, 03:11 PM
P: 15
so if we add are we left with;

(v12 -v12'cos302) + (-v12'sin302) / 2v2'2

?
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#14
Dec17-12, 04:05 PM
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Quote Quote by doub View Post
cos2(θ) = (v12 -v12'cos302)/v2'2

sin2(θ) = (-v12'sin302)/v2'2
actually, the RHS of that first equation should be (v1 -v1'cos30)2/v2'2
Quote Quote by doub View Post
so if we add are we left with;

(v12 -v12'cos302) + (-v12'sin302) / 2v2'2

?
how did you get that?

where has the = sign gone?


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