A bullet with mass 0.04000 kg, traveling at a speed of 310.0 m/s, strikes a stationary block of wood having mass 6.960 kg. The bullet embeds

Question

A bullet with mass 0.04000 kg, traveling at a speed of 310.0 m/s, strikes a stationary block of wood having mass 6.960 kg. The bullet embeds into the block. The speed, in m/s, of the bullet-plus-wood combination immediately after the collision is

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Thu Thảo 2 weeks 2021-07-22T06:55:11+00:00 1 Answers 1 views 0

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    2021-07-22T06:56:15+00:00

    Answer:

    1.77 m/s

    Explanation:

    From the law of conservation of momentum,

    Total momentum before collision = Total momentum after collision

    mu+m’u’ = V(m+m’)………………………. Equation 1

    Where m = mass of the bullet, m’ = mass of the block, u = initial velocity of the bullet, u’ = initial velocity of the block, V = Velocity of the bullet-wood combination immediately after collision

    make V the subject of the equation

    V = (mu+m’u’)/(m+m’)……………… Equation 2

    Given: m = 0.04 kg, m’ = 6.96 kg, u = 310 m/s, u’ = 0 m/s (stationary)

    Substitute into equation 2

    V = (0.04×310+6.96×0)/(0.04+6.96)

    V = 12.4/7

    V = 1.77 m/s

    Hence the speed, in m/s, of the bullet-plus-wood combination immediately after the collision is 1.77 m/s

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