5051525354555657585961A 0.12kg bullet is initially traveling 120m/s when it strikes a stationary 4.3kg block.After the collision the block (with the bullet embedded inside) swings up as shown.Neglect drag and the initial height of the bullet above the reference level.KEGPEunitunitunitbullet (time 1)block* (time 2)block* (time 3)h* The bullet is embedded in the block at these times.How fast was the block traveling after the impact?m/sHow much heat was generated during the impact?To what height (h) did the block swing?unitcheck answerscannot be determined

1) bullet (time 1) momentum, P = mv0 = 0.12 ×120 = 14.4 kg m/s the kinetic energy of the bullet, K =…

Answered: A 0.12kg bullet is initially traveling…

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