Rotational Collision of Sliding Boys: Velocity and Kinetic Energy Analysis

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SUMMARY

The discussion focuses on the physics of a perfectly inelastic collision between two boys, Jacob and Ethan, sliding towards each other on a frictionless surface. Jacob, with a mass of 45 kg, moves at 8 m/s, while Ethan, weighing 31 kg, slides at 11 m/s. After their collision, the combined velocity is calculated to be 9.34 m/s, with 1.66 x 10^(-4) of their initial kinetic energy remaining as mechanical energy. The second part of the analysis involves determining the moment of inertia as they rotate about their common center of mass after locking arms.

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Two boys are sliding towards each other on a frictionless, ice-covered parking lot. Jacob, mass 45kg, is sliding towards the right at 8m/s and Ethan, mass 31kg, is sliding towards the left at 11m/s along the same line. When they meet, they grab each other and hang on.
(a) What is their velocity immediately after?
(b) What fraction of their initial KE is still mechanical energy after their collision?

That was so much fun that the boys decided to repeat the collision with the same original velocities this time along parallel lines 1.20m apart. At the closest approach they lock arms and start rotating about their common center of mass. Model the boys as particles and their arms as a cord that does not stretch.
(c) Find the velocity of their center of mass.
(d) Find the angular velocity.
(e)What fraction of their initial KE is still mechanical energy after they link arms?My answers to part (a) and (b) are 9.34m/s and 1.66 x 10^(-4) respectively.
For the second part of the question, I need the moment of inertia of the system of boys. How can I find it? Any help?
 
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hey are your answers in (a) and (b) correct? it's perfectly inelastic collision right?
shouldn't it be 45x8 + 31x(-11) =(45+31)(Y), where Y is +0.25?
 

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