A thin block of soft wood with a mass of 0.080 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a

A thin block of soft wood with a mass of 0.080 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 599 m/s at a block of wood and passes completely through it. The speed of the block is 22 m/s immediately after the bullet exits the block.

(a) Determine the speed of the bullet as it exits the block. (m/s)

Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

KEi = J

KEf = J

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  1. Answer:

    (a) 222.126 m/s.

    (b) KEi = 837.8 J, KEf = 308.81 J

    Explanation:

    (a)

    From the law of conservation of momentum,

    Total momentum before collision = Total momentum after collision

    mu+m’u’ = mv+m’v’……………. equation 1

    Where m = mass of block, m’ = mass of bullet, u = initial velocity of block, u’ = initial velocity of bullet, v = final velocity of block, v’ = final velocity of bullet.

    make v’ the subject of the equation,

    v’ = [(mu+m’u’)-mv]/m’……………….. Equation 2

    Given: m = 0.08 kg, m’ = 4.67 g = 0.00467 kg, u = 0 m/s (at rest), u’ = 599 m/s, v = 22 m/s.

    Substitute into equation 2

    v’ = [(0.08×0)+(0.00467×599)-(0.08×22)]/0.00467

    v’ = (2.79733-1.76)/0.00467

    v’ = 1.03733/0.00467

    v’ = 222.126 m/s.

    Hence the speed at which the bullet exit the block = 222.126 m/s.

    (b)

    Initial kinetic energy of the system

    KEi = 1/2mu² + 1/2m’u’² = 1/2(0.08×0²) + 1/2(0.00467×599²)

    KEi = 0+837.8 = 837.8 J.

    Final kinetic energy of the system

    KEf = 1/2mv² + 1/2m’v’²

    KEf = 1/2(0.08×22²) + 1/2(0.00467×222.126²)

    KEf = 19.36+115.21 = 308.81 J

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