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Given a Schema as in the Image attached. Each of the 2 balls have the velocity = V0 and Mass = M0. The system in the lower part of the image has Velocity = V1 and Mass = M1. The arcs are tubes that are hollow and are 1/4th (or a quadrant) of a circle that are perfectly aligned with the 2 balls.
Let the Radius of each of the external arc (that forms the outer circle) of the tube be R1.
Lets ignore all the complicating factors (gravity, air resistance, friction, heat loss and so on) and assume 100% efficiency. When each of the balls leaves the system simultaneously after traveling through the tubes:
1. What is the resultant velocity of the balls as they exit the Schema (they exit exactly horizontally)?
2. What is the new (vertical) velocity of the Schema when the balls exit the Schema?
3. How does the velocity depend on the radius R1?
2. Approach
If the balls and the schema have a direct head collision all the momentum of the system (balls and schema) would be completely utilized in the vertical direction. But since the balls follow a circular arc a part of momentum is used in vertical direction and the rest is used in horizontal direction:
1. The momentum available in the vertical direction is given by: M1*V1 + 2* M0*V0* sin(45)
2. The momentum available in the horizontal direction is given by: 2* M0*V0*cos(45)