Calculating angle of ball after collision

[Felicity Wang]

Homework Statement

The white ball in the figure has a speed of 1.81 m/s and the yellow ball is at rest prior to an elastic glancing collision. After the collision the white ball has a speed of 1.37 m/s. To the nearest tenth of a degree, measured counterclockwise from east, what angle does it scatter at if the yellow ball is scattered at 280degrees?

Homework Equations

mv1=mv1'sin(@)+mv2'sin(@)

The Attempt at a Solution

(2kg x 1.81m/s) = (2kg x 1.37) sin(@) +(1kg) v2' sin(280)

v2':
1/2mv1^2 = 1/2mv1'^2 + 1/2mv2'^2
v2'= 1.67m/s

The answer is 37 degrees but I am not sure how to progress from here.

 

[JeremyG]

I would recommend drawing a free body diagram of the two balls as well as vector triangles to get a better grasp of the situation here. For one, do not apply the equations blindly. The very first step here:

(2kg x 1.81m/s) = (2kg x 1.37) sin(@) +(1kg) v2' sin(280)

is meant to be a statement of Conservation of Linear Momentum in the horizontal direction, yes?

What then, is the direction of the momentum vector you are considering when you use sin(@)?

 

[haruspex]

Something odd about this question...
It appears to be overspecified. There are only two unknowns, the final speed of the yellow ball and the final direction of the white ball. But since it is elastic there are three available equations: energy, x-momentum and y-momentum.
The component of the white ball's momentum orthogonal to the yellow ball's departure direction should be conserved. But |1.81*sin(280) |=1.78 > 1.37, a contradiction.