A ball moving at 13.9 m/s makes an off-center elastic collision with another ball of equal mass that is?

Question

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.


1.20E+01 m/s

1.20E+01 cm/s

1.40E+01 m/s

1.40E+02 m/s

Answer 1

Using conservation of momentum in the direction of motion of the ball prior to collision

m*13.9=m*v1*cos(30.1)+m*v2*cos(θ)
where θ is the angle of the ball that was struck, v1 is the velocity of the initially moving ball after collision and v2 is the velocity of the ball originally at rest after collision

we also know
m*v1*sin(30.1)-m*v2*sin(θ)=0

and
.5*m*13.9^2=.5*m*v1^2+.5*m*v2^2

solve for v1

13.9^2=v1^2+v2^2
v1*sin(30.1)-v2*sin(θ)=0
13.9=v1*cos(30.1)+v2*cos(θ)

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

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

sin^2(θ)+cos^2(θ)=1

v1^2*sin^2(30.1)/v2^2+
(13.9-v1*cos(30.1))^2/v2^2=1

multiply by v2^2
v1^2*sin^2(30.1)+
(13.9-v1*cos(30.1))^2=v2^2

Since
13.9^2-v1^2=v2^2

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

Solve for v1
v1=13.9*cos(30.1)
v1=12.026

v2=6.971

θ=59.9 degrees below the original path of m1.

The answer from the multiple choice is
1.2E+01 m/s

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