A perfectly (non-head 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 head-on. Calculate the direction of motion for one of the 2 objects after the collision.

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  1. 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.

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