# Elastic collisions

by doub
Tags: collisions, elastic
 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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P: 26,148
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)
 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?
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P: 26,148
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)
 P: 15 yeah I'm totally lost now
 Sci Advisor HW Helper Thanks P: 26,148 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
 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
 Sci Advisor HW Helper Thanks P: 26,148 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
 P: 15 So, cos(theta) = v1 - (v1'cos30)/v2' and sin(theta) = (-va'sin30)/v2' where do the sin2theta come from?
 Sci Advisor HW Helper Thanks P: 26,148 uhh? square both equations!
 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
 Sci Advisor HW Helper Thanks P: 26,148 ok, now add … θ will miraculously disappear! pzzaaam!
 P: 15 so if we add are we left with; (v12 -v12'cos302) + (-v12'sin302) / 2v2'2 ?
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P: 26,148
 Quote by doub 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 by doub 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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