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23 May, 09:58

A 30-g bullet is fired with a horizontal velocity of 460 m/s and becomes embedded in block B which has a mass of 3 kg. After the impact, block B slides on 30-kg carrier C until it impacts the end of the carrier. Knowing the impact between B and C is perfectly plastic and the coefficient of kinetic friction between B and C is 0.2, determine the energy loss due to friction and the impacts. The energy lost due to friction is J. The impact loss due to AB impacting the carrier is J. The impact loss at first impact is J.

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  1. 23 May, 13:19
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    energy loss due to friction and the impacts = 2.97 J; The impact loss due to AB impacting the carrier is = 25.72J; The impact loss at first impact is 6,316.64J

    Explanation:

    First find the velocity of the bullet after the first impact using

    M1V1 + 0 = (M1 + M2) v'

    Where M1 is the mass of the bullet

    M2 is the mass of the block B

    M3 is the mass of the carrier

    v' is the velocity

    v' = M1V1 / (M1 + M2)

    v' = (30 * 10^-3 kg) (460 m/s) / (30 * 10^-3 kg + 3 kg)

    v' = 13.8/3.03

    v' = 4.55m/s

    Also calculate final velocity of the carrier v2'

    v2' = M1V1 / (M1 + M2 + M3)

    v2' = (30 * 10 kg) (460 m/s) / (30 * 10 kg + 3 kg + 30kg)

    v2' = 0.42m/s

    Now to calculate energy loss due to friction

    Normal force

    N = W1 + W2 = (m1 + m2) g

    Where W1 and W2 is the weight of the bullet and block respectively

    g is gravitational acceleration for taken as 9.81m/s

    = (0.030 kg + 3 kg) (9.81 m/s) 29.724 N

    Friction force = coefficient of kinetic * normal force

    Where coefficient of kinetic = 0.2

    Ff = (0.2) (29.724) = 5.945 N

    Now

    Energy loss due to friction = frictional force * distance

    Assume distance is 0.5 m.

    Energy loss due to friction = 5.945 N * 0.5 m

    = 2.97J

    Kinetic energy of block with embedded bullet immediately after first impact:

    1/2 * (m1 + m2) (v') ^2

    1/2 * (30 * 10^-3 kg + 3 kg) (4.55m/s) ^2

    = 1/2 * (3.03kg) * (4.55m/s) ^2

    = 31.36 J

    Final kinetic energy of bullet, Block, and Carrier together

    1/2 * (m1 + m2 + m3) (v2') ^2

    1/2 * (30 * 10^-3 kg + 3 kg + 30kg) (0.42m/s) ^2

    1/2 * (33.03kg) * (0.42m/s) ^2

    = 2.91 J

    Therefore

    Loss due to friction and stopping impact = Kinetic energy of block with embedded bullet immediately after first impact - Final kinetic energy of bullet, Block, and Carrier together

    = 31.36 J - 2.91 J

    = 28.69 J

    Impact loss due to AB impacting the carrier = loss due to friction - energy due to friction

    28.69J - 2.97J

    =25.72J

    Initial kinetic energy of system ABC = 1/2 (m1vo)

    =1/2 (0.030 kg) (460 m/s) ^2 = 6,348J

    Therefore

    Impact loss at first impact = Initial kinetic energy of system ABC - Kinetic energy of block with embedded bullet immediately after first impact:

    = 6,348J - 31.36 J

    = 6,316.64J
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