A railroad boxcar rolls on a track at 2.90 m/s toward two identical coupled boxcars, which are rolling in the same direction as the first, b

Question

A railroad boxcar rolls on a track at 2.90 m/s toward two identical coupled boxcars, which are rolling in the same direction as the first, but at a speed of 1.20 m/s. The first reaches the second two and all couple together. The mass of each is 3.05 ✕ 104 kg.(a)What is the speed (in m/s) of the three coupled cars after the first couples with the other two? (Round your answer to at least two decimal places.)Incorrect: Your answer is incorrect.What is the momentum of the two coupled cars? What is the momentum of the first car in terms of its mass and initial speed? Note all cars are initially traveling in the same direction. Apply conservation of momentum to find the final speed. m/s(b)Find the (absolute value of the) amount of kinetic energy (in J) converted to other forms during the collision.J

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Kiệt Gia 5 years 2021-08-18T21:00:49+00:00 1 Answers 217 views 0

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  1. Sapo1
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    2021-08-18T21:01:50+00:00
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    Answer:

    momentum of the coupled cars V =  1.77 m/s

    kinetic energy coverted to other forms during the collision ΔK.E = -2.892×10⁴J

    Explanation:

    given

    m₁ =3.05 × 10⁴kg

    u₁ =2.90m/s

    m₂=6.10× 10⁴kg

    u₂=1.20m/s

    using law of conservation of momentum

    m₁u₁ + m₂u₂ = (m₁ + m₂) V

    3.05 × 10⁴ ×2.90 + 6.10× 10⁴× 1.20 = (9.15×10⁴)V

    V =  1.617×10⁵/9.15×10⁴

    V = 1.77m/s

    K.E =1/2mV²

    ΔK.E = K.E(final) – K.E(initial)

    ΔK.E = ¹/₂ × 9.15×10⁴ ×(1.77)² –  ¹/₂ ×3.05 × 10⁴ × (2.90)² -¹/₂ × 6.10× 10⁴× (1.20)²

    ΔK.E = ¹/₂ × (28.67-25.65-8.784) ×10⁴

    ΔK.E = -2.892×10⁴J

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