A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially h

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A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of 170 kgkg and is traveling east with a velocity of magnitude 4.60 m/sm/s. Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.

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Dâu 5 years 2021-09-05T01:55:03+00:00 1 Answers 509 views 1

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  1. minhkhang
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    2021-09-05T01:56:41+00:00
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    This question is not complete, the complete question is;

    A railroad handcar is moving along straight, frictionless tracks with negligible air resistance.

    In the following cases, the car initially has a total mass (car and contents) of 170 kg and is traveling east with a velocity of magnitude 4.60 m/s.

    Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.

    Part A

    An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car’s initial velocity.

    Part B

    An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.

    Answer:  

    Part A) the final velocity of the car is  4.6 m/s

    Part B) the final velocity of the car is 5.28 m/s

    Explanation:

    Given the data in the question;

    Total mass (m₁+m₂) = 170 kg

    velocity of magnitude Vx = 4.60 m/s

    PART A)

    An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car’s initial velocity,

    i.e

    m₂ = 22.0 kg

    so m₁ = 170 – 22 = 148 kg

    so, we apply conservation of momentum

    since the object thrown out of the car, it has nothing to do with the car’s velocity.

    (m₁+m₂)Vx = m₁Vx₁ + m₂Vx₂

    we substitute

    (170)4.60 = 148Vx₁ + 22(4.60)

    782 = 148Vx₁ + 101.2

    148Vx₁ = 782 – 101.2

    148Vx₁ = 680.8

    Vx₁ = 680.8 / 148

    Vx₁ = 4.6 m/s

    Therefore, the final velocity of the car is  4.6 m/s

    Part B)

    An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.

    Vx = V(m₁+m₂) / ((m₁+m₂) – m₁)

    we substitute

    Vx = 4.60(170) / ((170) – 22)

    Vx = 782 / 148

    Vx = 5.28 m/s

    Therefore, the final velocity of the car is 5.28 m/s

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