In a closed system with no friction, a red sphere of 2.5 kg stands stationary. A blue sphere with a mass of 5.8 kg approaches the first sphe

Question

In a closed system with no friction, a red sphere of 2.5 kg stands stationary. A blue sphere with a mass of 5.8 kg approaches the first sphere with a speed of 4.1 m/s. The two collide. After the collision, the blue sphere begins moving forward with a speed of 1.3 m/s. What is the velocity of the red sphere after the collision? (Show all work)

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Kim Chi 2 months 2021-07-22T00:08:01+00:00 1 Answers 0 views 0

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    2021-07-22T00:09:45+00:00

    Answer:

    m1 = the mass of the blue sphere = 5.8 kg

    m2 = the mass of the red sphere = 2.5 kg

    v1 = initial velocity of the blue sphere before the collision = 4.1 m/s

    v2 = initial velocity of the red sphere before the collision = 0 m/s

    v’1 = final velocity of the blue sphere after the collision = 1.3 m/s

    v’2 = final velocity of the red sphere after the collision = ?

    using conservation of momentum

    m1v1 + m2v2 = m1v’1 + m2v’2

    (5.8) (4.1) + (2.5) (0) = (5.8) (1.3) + (2.5) (v’2)

    23.78 = 7.54 + (2.5) (v’2)

    -7.54    -7.54

    16.24 = (2.5) v’2

    ———   ———–

     2.5         2.5

    v’2 = 6.5

    Explanation:

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