Question

A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball as shown. 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.

So far have figured out that the y component is v1=v1'costheta1+v2'costheta2 and the x compenent is v1'sintheta1=v2'sintheta2. I tried squaring them (conservation of KE and all masses and factors cancel) but I have no idea how to proceed. I'm completely stuck. Help would be much appreciated.

Answer 1

I don't know which is the x, or y component. But you do know the momentum off original axis adds to zero...so one ball component off oxis is equal and opposite to the the other ball. I think that is what you mean by v1'sintheta1=v2'sintheta2.

If that is so, then you have two equations, three unknowns (v1', v2', and theta2)

put in the cosine and sine of 30 into the equations then stop.


V1^2=v1^2 + v2^2 from energy.
Note that
a) v1^2=(v1'cosTheta1)^2 + (v1'sinTheta1)^2
and
b) v2'^2=(v2'cosTheta2)^2 + (v2'sinTheta2)^2

You then have
c) v1=v1'costheta1+v2'costheta2
but V1^2=v1'^2 + v2'^2
so put in c) xquared for V1, and for the right side, put in a) and b). Have a lot of scratch paper, it will solve.

you will know both ' velocityies, and angle2