physics question?

Question

A ball moving at 13.9 m/s makes an off-center elastic collision with another ball of equal mass that is initially at rest. The incoming ball is deected at an angle of 30.1deg from its original direction of motion. Calculate the velocity of the initially moving ball after the collision.

Answer 1

the ball mass is m

The center of mass velocity is constant.

In the x
13.9*m=m*cos(30.1)*v1+m*cos(θ)*v2

where θ is the angle of the deflected ball
It will be below the horizontal, so in the y

0=m*sin(30.1)*v1-m*sin(θ)*v2

Since the collision was elastic, energy is conserved
.5*m*13.9^2=.5*m*v1^2+.5*m*v2^2

solve for v1

The next several steps are me doing algebra. The key is to get the equations where I can apply the identity
sin^2+cos^2=1

13.9-cos(30.1)*v1=cos(θ)*v2
sin(30.1)*v1=sin(θ)*v2
13.9^2-v1^2=v2^2


sin^2(30.1)*v1^2=
sin^2(θ)*(13.9^2-v1^2)

(13.9-cos(30.1)*v1)^2=
cos^2(θ)*(13.9^2-v1^2)

add
sin^2(30.1)*v1^2+
(13.9-cos(30.1)*v1)^2=
(sin^2(θ)+cos^2(θ))*(13.9^2-v1^2)

note that (sin^2(θ)+cos^2(θ))=1

sin^2(30.1)*v1^2+
(13.9-cos(30.1)*v1)^2
=13.9^2-v1^2

sin^2(30.1)*v1^2+13.9^2-
2*13.9*cos(30.1)*v1+
cos^2(30.1)*v1^2
=13.9^2-v1^2

again sin^2+cos^2=1

v1=13.9*cos(30.1)

v1=12.03 m/s

j