Register to reply

Elastic collisions

by doub
Tags: collisions, elastic
Share this thread:
doub
#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
Phys.Org News Partner Science news on Phys.org
NASA team lays plans to observe new worlds
IHEP in China has ambitions for Higgs factory
Spinach could lead to alternative energy more powerful than Popeye
tiny-tim
#2
Dec17-12, 12:25 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
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
#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
#4
Dec17-12, 01:24 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
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
#5
Dec17-12, 01:38 PM
P: 15
yeah I'm totally lost now
tiny-tim
#6
Dec17-12, 01:57 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
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
#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
#8
Dec17-12, 02:11 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
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
#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
#10
Dec17-12, 02:33 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
uhh?
square both equations!
doub
#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
#12
Dec17-12, 02:53 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
ok, now add
θ will miraculously disappear!
pzzaaam!
doub
#13
Dec17-12, 03:11 PM
P: 15
so if we add are we left with;

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

?
tiny-tim
#14
Dec17-12, 04:05 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,160
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?


Register to reply

Related Discussions
Elastic collisions Introductory Physics Homework 1
The elastic collisions of gas Classical Physics 1
Elastic Collisions Introductory Physics Homework 12
Elastic collisions Introductory Physics Homework 1