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A 12g bullet strikes a 5.20kg block standing on a frictionless surface. The bullet embeds itself in the block and the tinal v
Using information from question 2 above. What was the kinetic energy lost by the system (the bullet and the block)?
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Answer #1

•012 (a) Using conservation of momentum •012kg x2 = 5.212 kgx 1.9 m/sec V = 5.212x1.7m/sec =738.4 m/ see 3) Kolimbia = 18.012

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