[jackyjoy123 (OP)]

Hello,

This question is to be considered in space away from any effecting masses.  I am trying to find a formula to figure out the loss of energy when an object weighing x amount traveling at an x amount of speed in one starting direction makes a gradual turn of 90 degrees.  What is the force applied to its opposition and what speed is lost?  Imagine a track in space and an object was traveling down this track at a set speed.  When the object reached the curve, how fast would the track want to start moving and how much will the object want to slow down?  For ease of formula imagine the track can only travel in the objects original direction, so the object would want to turn and the track would go straight.

thanks
jackyjoy

[Halc]

This question is to be considered in space away from any effecting masses.

Any reaction mass (rocket exhaust, track, poles, etc.) is an effecting mass so to speak.  I think you mean no gravity in consideration.

I am trying to find a formula to figure out the loss of energy when an object weighing x

I am presuming here 'massing m' since it has no weight except whatever centripetal force you are applying to it. I changed it to 'm' since you're also using x for speed, which is best expressed as 'v'.

What is the force applied to its opposition and what speed is lost?

Any force you like. The higher the force, the less time it will take to make the turn. If you want a nice circular trajectory, the force is always going to be perpendicular to the velocity at any given time.
So for instance, imagine there's a handy bar pole off to the side. The mass m snakes out a grappling hook and snags the bar pole, effecting the nice neat turn. If the pole is well grounded (masses much more than m), then you have your nice turn without any expenditure of energy.
Likewise, a car moving at 100 km/hr can make a 90° turn on the highway without any expenditure of energy. Any reduction in speed is due to friction, not to the force required to turn it. The road is the track of which you speak.

Imagine a track in space and an object was traveling down this track at a set speed.  When the object reached the curve, how fast would the track want to start moving and how much will the object want to slow down?

Assuming the track was initially stationary in the frame in which m is moving at v, then just compute the component momentum of each object and apply the conservation of momentum law.  If the track is well balanced, it can be done without any resulting rotation, so it all just depends on the ratio of mass of object to track. If rotation results, then it gets more complicated.

For ease of formula imagine the track can only travel in the objects original direction, so the object would want to turn and the track would go straight.

OK, so the track then absorbs all the original momentum, and new lateral momentum is applied to the object from outside the system. So all you have to do is compute the kinetic energy transferred to the track and subtract that from the object.
Remember that the track is moving at the end, so it bends more than 90° if the object riding it is to have straight lateral motion.  So if the track masses the same as m, then the track gets all the momentum, the track shape is a U-turn, and the original mass m ends up stationary at the end.
If, like our car on the highway, the track masses a lot more than the object, then the object doesn't lose significant speed at all as it turns.

If all you want is exit velocity, just treat it as an elastic collision with the track. The result is the same as the gradual turn.
Momentum is mv.  Energy is mv²/2
The rest is a little bit of algebra.