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A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 650 m/s . Part A What is the recoil speed of the hunter
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A 70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 650 m/s . Part A What is the recoil speed of the hunter

User SomeKoder
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1 Answer

5 votes
5 votes

Answer:

Step-by-step explanation:

momentum is conserved. Initial momentum was zero, so final total momentum must also be zero

0.042(650) + 70v = 0

v = -0.39 m/s

|v| = 0.39 m/s

User Paul Armstrong
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