Collisions in an Elastic System: Solving for Motion and Energy Loss

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The discussion focuses on analyzing a collision between a 1 kg mass suspended by a 2 m cord and a 2 kg mass on the ground. When the 1 kg mass is released from a horizontal position, it collides elastically with the 2 kg mass, allowing for the application of conservation of momentum and energy principles to determine the final velocities of both masses. In a second scenario, where 25% of energy is lost during the collision, the same principles apply, but adjustments must be made to account for energy loss. The discussion emphasizes the importance of using kinematic equations and energy conservation to fully describe the motion of both blocks post-collision.

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nihaokailan
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A 1 kg mass is attached to a 2 m long piece of very light cord and suspended in such a way that it can swing freely and pass just above the ground. A 2 kg mass is placed on the ground so that the 1 kg mass will collide with it when the cord is certical. The coefficient of kinetic friction between the 2 kg mass and the ground is 0.1.
a) If the 1 kg block is pulled to one side until the cord is horizontal and releaed, the two blocks collide elastically at the bottom. Completely describe the subsequent motion of each block.
b) Repeat (a) assuming that 25% of the energy is lost during the collision.

Please help! It would be greatly appricated!
 
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I won't go into the numerical part but you can conserve energy to obtain speed of collision of suspended block. Then you can apply momentum conservation and energy consvation(yes even in second case) to calculate final velocities after collision. Then use kinematical equations for block on ground and energy conservation and the force equation on the suspended block to completely describe it's motion.
 

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