- #1

spark8819

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## 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

pi=pf

KEi=KEf[/B]

## 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!