Posted by Alex on .
A perfectly (nonhead on) elastic collision occurs between a 0.25kg object moving east at 4m/s and a stationary 0.3kg object. Following the collision the 0.25kg object is moving at 2m/s.
a) What is the speed of the 0.3kg object after the collision?
I did:
What formula do you use?
I used m1v1^2 = m1v1'^2 + m2v2'^2
and got 3.16m/s but I'm not sure if that's right.
b) You must select suitable directions of motion for the 2 objects after this collision, which is not headon. Calculate the direction of motion for one of the 2 objects after the collision.

Physics 
bobpursley,
The conservation of energy applies, you did it correctly in a).
For b), you have to use the result in a), and then set up a coordinate system E,N. The N components of both velocities have to add to zero, and the E components have to add to 4m/s. Use trig to get the components of each, assume some angle N of E, and S of E, theta1 and theta2. The algebra is a bit messy, but it works out.