## Elastic collisions

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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hi doub! welcome to pf!
 Quote by doub 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)
 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?

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## Elastic collisions

hi doub!

(try using the X2 button just above the Reply box )
 Quote by doub 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)
 yeah I'm totally lost now
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor 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
 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
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor 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
 So, cos(theta) = v1 - (v1'cos30)/v2' and sin(theta) = (-va'sin30)/v2' where do the sin2theta come from?
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor uhh? square both equations!
 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
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor ok, now add … θ will miraculously disappear! pzzaaam!
 so if we add are we left with; (v12 -v12'cos302) + (-v12'sin302) / 2v2'2 ?

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