Question
Asked Nov 22, 2019
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A 65 kg hunter, standing on frictionless ice, shoots a 30 g bullet at a speed of 700 m/s.  What is the recoil speed of the hunter?

Draw a before and after diagram, define the system that you will use in the problem, write Newton's second law  in terms of momentum, solve for the final answer symbolically, and finally plug in.

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Expert Answer

Step 1

Consider the mass of the hunter be mh, the mass of the bullet be mb, the initial speed of the hunter be uh, the initial speed of the bullet be ub, the final speed of the hunter, that is the recoil speed of the hunter, be vh, and the final speed of the bullet be vb.

 

Consider the diagram of the hunter-bullet system before and after fire the bullet.

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Vh Before fire After fire

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Step 2

The initial speed of the hunter and the bullet will be zero because the hunter and the bullet is at rest before bullet was fired. For recoil speed, the direction of motion of the bullet will be opposite to that of the hunter. Thus, the final speed of the bullet will be taken as negative.

 

The given values are,

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=65 kg 1 kg т, %3D 30 g %3D 30 g 1000 g = 0.03 kg И, — и, 3 0 =700 m/s

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Step 3

Write the expression for the Newton’s second law of momentum, and solve for the initial momentum of the hunter-bullet s...

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Р. F t P (F)(t) P (0)(t) =0

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