Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the same weight,

Question

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the same weight, has been pushed by dockworkers and hits car B with a velocity of 1.5 m/s. Knowing that the coefficient of restitution is 0.8 between B and C, and it is 0.5 between A and B, determine the velocity of each car after all collisions have taken place

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RobertKer 2 months 2021-08-06T14:13:13+00:00 1 Answers 1 views 0

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    2021-08-06T14:14:56+00:00

    Answer:

    Explanation:

    Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

    velocity of approach = 1.5 – 0 = 1.5

    velocity of separation = v₁ + v₂

    coefficient of restitution = velocity of separation / velocity of approach

    .8 = v₁ + v₂ / 1.5

    v₁ + v₂ = 1.2

    applying law of conservation of momentum

    m x 1.5 + 0 = mv₂ – mv₁

    1.5 = v₂ – v₁

    adding two equation

    2 v ₂= 2.7

    v₂ = 1.35 m /s

    v₁ = – .15 m / s

    During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

    For second collision ,

    coefficient of restitution = velocity of separation / velocity of approach

    .5 = v₃ + v₄ / 1.35

    v₃ + v₄ = .675

    applying law of conservation of momentum

    m x 1.35 + 0 = mv₄ – mv₃

    1.35 = v₄ – v₃

    adding two equation

    2 v ₄= 2.025

    v₄ = 1.0125 m /s

    v₃ = – 0 .3375  m / s

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