physics

A 3.1-kilogram gun initially at rest is free to move. When a 0.015-kilogram bullet leaves the gun with a speed of 500. meters per second, what is the speed of the gun?

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  1. This a chance for you to apply the law of conservation of momentum.

    That means the momentum of the gun going backward is equal and opposite to the momentum of the bullet going forward. The two momenta cancel out.

    Mbullet*Vbullet = -Mgun*Vgun

    Solve for Vgun

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    posted by drwls
  2. 7.5 m/s

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  3. pbefore=pafter
    m1v1+m2v2 = m1v’1+m2v’2
    (3.1kg)(0m/s)+(0.015kg)(0m/s) =
    (3.1kg)( v’1)+(0.015kg)(500.m/s)
    0kg∙m/s=(3.1kg)( v’1)+7.5kg∙m/s
    -7.5kg∙m/s=(3.1kg)( v’1)
    v’1=-2.4m/s (its speed is 2.4 m/s)

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  4. u=0
    0+0= 3.1 v1 + 0.015(500)m/s
    -3.1v1= 7.5
    v1= 7.5:(-3.1)

    V1=-2.42 in absolute value then the answer will be 2.42

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  5. u=0
    0+0=3.1V+ 0.015(500)M/S
    -3.1V = 7.5
    V= -2.42M/S OR 2.42m/s in absolute value

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    posted by agahozo
  6. Stephanie is a troll

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    posted by h

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