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March 25, 2017

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A bullet of mass 0.01 kg is moving horizontally with a speed of 100 m/s when it hits a block of mass 2 kg that is at rest on a horizontal surface with a coefficient of friction of 0.4. After the collision the bullet becomes embedded in the block.

  • physical - ,

    initial momentum = 0.01 * 100

    momentum immediately after collision = 2.01 Vi

    so
    2.01 Vi = 0.01 * 100
    solve for Vi

    Then you have a force of F = -2.01 * g * 0.4 retarding the motion

    a = F/m = F/2.01 = -0.4 g

    V = Vi -0.4 g t
    x = Vi t -(0.4 g/2)t^2

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