Answer:
The final speed for the mass 2m is 
and the final speed for the mass 9m is 
.
The angle at which the particle 9m is scattered is 
with respect to the - y axis.
Explanation:
In an elastic collision the total linear momentum and the total kinetic energy is conserved.
<u>Conservation of linear momentum:</u>
Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.
The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m
In the x axis before the collision we have
and after the collision we have that
In the y axis before the collision 
after the collision we have that
so
then
<u>Conservation of kinetic energy:</u>
so
![\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B11%7D%7B2%7D%5C%20m%5C%20v_%7Bi%7D%20%5E%7B2%7D%20%3D%5Cfrac%7B1%7D%7B2%7D%20%5C%209m%5C%20%5B%28%5Cfrac%7B7%7D%7B9%7D%29%20%5E%7B2%7D%5C%20v_%7Bi%7D%20%5E%7B2%7D%2B%20%28%5Cfrac%7B2%7D%7B9%7D%29%20%5E%7B2%7D%5C%20v_%7B2y%7D%20%5E%7B2%7D%5D%2B%20m%5C%20v_%7B2y%7D%20%5E%7B2%7D)
Putting in one side of the equation each speed we get
We know that the particle 2m travels in the -y axis because it was stated in the question.
Now we can get the y component of the speed of the 9m particle:
the magnitude of the final speed of the particle 9m is
The tangent that the speed of the particle 9m makes with the -y axis is
As a vector the speed of the particle 9m is:
As a vector the speed of the particle 2m is:
