Elastic collision in 2-dimensions

[spark8819]

Homework Statement

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 angle 30° to the original path, determine:

a. the speed of the first ball after the collision.

b. the speed and direction of the second ball after the collision. 2. Homework Equations
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The Attempt at a Solution

I have attempted this problem multiple times using several variations in the math but always seem to run into the same issue of getting an impossible value for v'a as shown below.
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Momentum conserved in x-axis:

mvax+mvbx=mv'ax+mv'bx

va=v'acos30+v'bcosθ

3=0.866v'a+v'bcosθ (1)

Momentum conserved in y-axis:

mvay+mvby=mv'ay+mv'by

0=v'ay+(-v'by)

v'asin30=v'bsinθ

0.5v'a=v'bsinθ (2)

Kinetic energy conserved:

va^2=v'a^2+v'b^2

9=v'a^2+v'b^2 (3)

Next, I squared equations (1) and (2) and added them together so I could use the trig identity sin^2θ+cos^2θ equals one to get rid of the theta and end up with the following equation:

v'b^2(cos^2θ+sin^2θ) = 0.75v'a^2 - 9 + 0.25v'a^2

v'b^2 = v'a^2 - 9 (4)

Next, I write equation (3) in terms of v'b^2 and substitute it into equation (4) to get the following:

va^2 - v'a^2 = v'a^2 - 9

3^2 - v'a^2 = v'a^2 - 9

18 = 2v'a^2

v'a= 3 m/s

I know this answer can't be right because some of the energy from ball A will be lost after the collision which will cause a reduction in its velocity. Fairly confident in the methodology behind this but I think I am making a mistake in the math somewhere that I am missing.

Any assistance/comments are appreciated!

 

[haruspex]

v'b^2(cos^2θ+sin^2θ) = 0.75v'a^2 - 9 + 0.25v'a^2

Try that step again, maybe taking smaller steps.

 

[spark8819]

Square eq (1):

(3)^2=(0.866v'a+v'bcosθ)^2

9=(0.866v'a+v'bcosθ)(0.866v'a-v'bcosθ)

9=0.75v'a^2-0.866v'av'bcosθ+0.866v'av'bcosθ-v'b^2cos^2θ

9=0.75v'a^2-v'b^2cos^2θ

REARRANGE:

v'b^2cos^2θ=0.75v'a^2-9 (1)

Square eq (2):

(0.5v'a)^2=(v'bsinθ)^2

0.25v'a^2=v'b^2sin^2θ

REARRANGE:

v'b^2sin^2θ=0.25v'a^2 (2)

ADD EQ (1) and (2):

v'b^2cos^2θ+v'b^2sin^2θ=0.75v'a^2-9+0.25v'a^2

Then, factor out v'b^2 and combine like terms on the right side:

v'b^2(cos^2θ+sin^2θ)=v'a^2-9

then I am left with:

v'b^2=v'a^2-9I am having a hard time seeing where I am going wrong here... I feel like its positive/negative error but not sure.

 

[SammyS]

Square eq (1):

(3)^2=(0.866v'a+v'bcosθ)^2

9=(0.866v'a+v'bcosθ)(0.866v'a-v'bcosθ)

Why does the second fctor on the right have a minus sign ?

The right hand side should be (0.866v'a + v'bcosθ)(0.866v'a + v'bcosθ)

... but that won't result in getting rid of θ in the end.

9=0.75v'a^2-0.866v'av'bcosθ+0.866v'av'bcosθ-v'b^2cos^2θ

9=0.75v'a^2-v'b^2cos^2θ

REARRANGE:

v'b^2cos^2θ=0.75v'a^2-9 (1)

Square eq (2):

(0.5v'a)^2=(v'bsinθ)^2

0.25v'a^2=v'b^2sin^2θ

REARRANGE:

v'b^2sin^2θ=0.25v'a^2 (2)

ADD EQ (1) and (2):

v'b^2cos^2θ+v'b^2sin^2θ=0.75v'a^2-9+0.25v'a^2

Then, factor out v'b^2 and combine like terms on the right side:

v'b^2(cos^2θ+sin^2θ)=v'a^2-9

then I am left with:

v'b^2=v'a^2-9I am having a hard time seeing where I am going wrong here... I feel like its positive/negative error but not sure.