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14 September, 07:06

A block of mass 2 kg is traveling in the positive direction at 3 m/s. Another block of mass 1.5 kg, traveling in the same direction at 4 m/s, collides elastically with the first block. Find the final velocities of the blocks. How much kinetic energy did the system lose

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  1. 14 September, 07:21
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    a. The final velocity of the block of mass 2 kg is 3 m/s or 3.86 m/s. The final velocity of the block of mass 1.5 kg is 4 m/s or 2.86 m/s b. The kinetic energy change is 0 J or - 12.235 J. Since the collision is elastic, we choose ΔK = 0

    Explanation:

    From principle of conservation of momentum,

    momentum before impact = momentum after impact

    Let m₁ = 2 kg, m₂ = 1.5 kg and v₁ = 3 m/s, v₂ = 4 m/s represent the masses and initial velocities of the first and second blocks of mass respectively. Let v₃ and v₄ be the final velocities of the blocks. So,

    m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄

    (2 * 3 + 1.5 * 4) = 2v₃ + 1.5v₄

    6 + 6 = 2v₃ + 1.5v₄

    12 = 2v₃ + 1.5v₄

    2v₃ + 1.5v₄ = 12 (1)

    Since the collision is elastic, kinetic energy is conserved. So

    1/2m₁v₁² + 1/2m₂v₂² = 1/2m₁v₃² + 1/2m₂v₄²

    1/2 * 2 * 3² + 1/2 * 1.5 * 4² = 1/2 * 2v₃² + 1/2 * 1.5v₄²

    9 + 12 = v₃² + 0.75v₄²

    21 = v₃² + 0.75v₄²

    v₃² + 0.75v₄² = 21 (2)

    From (1) v₃ = 6 - 0.75v₄ (3). Substituting v₃ into (2)

    (6 - 0.75v₄) ² + 0.75v₄² = 21

    36 - 9v₄ + 0.5625v₄² + 0.75v₄² = 21

    36 - 9v₄ + 1.3125v₄² - 21 = 0

    1.3125v₄² - 9v₄ + 15 = 0

    Using the quadratic formula,

    v₄ = [ - (-9) ± √[ (-9) ² - 4 * 1.3125 * 15]] / (2 * 1.3125)

    = [9 ± √[81 - 78.75]]/2.625

    = [9 ± √2.25]/2.625

    = [9 ± 1.5]/2.625

    = [9 + 1.5]/2.625 or [9 - 1.5]/2.625

    = 10.5/2.625 or 7.5/2.625

    = 4 m/s or 2.86 m/s

    Substititing v₄ into (3)

    v₃ = 6 - 0.75v₄ = 6 - 0.75 * 4 = 6 - 3 = 3 m/s

    or

    v₃ = 6 - 0.75v₄ = 6 - 0.75 * 2.86 = 6 - 2.145 = 3.855 m/s ≅ 3.86 m/s

    b. The kinetic energy change ΔK = K₂ - K₁

    K₁ = initial kinetic energy of the two blocks = 1/2m₁v₁² + 1/2m₂v₂²

    = 1/2 * 2 * 3² + 1/2 * 1.5 * 4² = 9 + 12 = 21 J

    K₂ = final kinetic energy of the two blocks = 1/2m₁v₃² + 1/2m₂v₄². Using v = 3 m/s and v = 4 m/s

    = 1/2 * 2 * 3² + 1/2 * 1.5 * 4² = 9 + 12 = 21 J.

    ΔK = K₂ - K₁ = 21 - 21 = 0

    Using v = 3.86 m/s and v = 2.86 m/s

    K₂ = 1/2 * 2 * 3.86² + 1/2 * 1.5 * 2.86² = 14.8996 - 6.1347 = 8.7649 J ≅ 8.765 J

    ΔK = K₂ - K₁ = 8.765 - 21 = - 12.235 J

    Since the collision is elastic, we choose ΔK = 0
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