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28 July, 11:12

A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B which has a mass of 3 kg. After the impact, block B slides on a 30-kg carrier C until it impacts the end of the carrier. Knowing the impact between B and C is perfectly plastic determine (a) velocity of the bullet and B after the first impact, (b) the final velocity of the carrier

(Distance between C and B is 0.5 m)

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  1. 28 July, 11:34
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    a.) 4.46 m/s

    b.) 0.41 m/s

    Explanation:

    a) Given that the mass M of the bullet = 30g = 30/1000 = 0.03 kg

    Velocity V = 450 m/s

    From conservative of linear momentum,

    Sum of momentum before impact = Sum of momentum after impact

    0.03 * 450 = (0.03 + 3) * v₂

    v₂ = 13.5/3.03 = 4.4554 m/s

    Therefore the velocity of the bullet and B after the first impact = 4.46 m/s approximately

    (b) To calculate the velocity of the carrier, you will consider the conservation of linear momentum again.

    (m₁ + m₂) * v₂ = (m₁ + m₂ + m₃) * v₃

    Where:

    Mass of the carrier m₃ = 30 kg

    Substitute all the parameters into the formula

    3.03*4.4554 = (3.03 + 30) * v₃

    v₃ = 13.5 / 33.03 = 0.40872 m/s

    Therefore the velocity of the carrier is 0.41 m/s approximately.
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