A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally inelastic and

Question

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally inelastic and takes place over an interval of 0.203 s. Assume that no brakes are applied during the collision and the car strikes the rear of the truck. Neglect the friction between the vehicles and the ground.
What is the average x component of the acceleration of the car during the collision?
What is the average x component of the acceleration of the truck during the collision?

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Đan Thu 6 months 2021-08-10T01:58:00+00:00 1 Answers 0 views 0

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    2021-08-10T01:59:35+00:00

    Answer:

    a) Acceleration of the car is given as

    a_{car} = -21 m/s^2

    b) Acceleration of the truck is given as

    a_{truck} = 10.15 m/s^2

    Explanation:

    As we know that there is no external force in the direction of motion of truck and car

    So here we can say that the momentum of the system before and after collision must be conserved

    So here we will have

    m_1v_1 + m_2v_2 = (m_1 + m_2)v

    now we have

    1400 (6.32) + 2900(0) = (1400 + 2900) v

    v = 2.06 m/s

    a) For acceleration of car we know that it is rate of change in velocity of car

    so we have

    a_{car} = \frac{v_f - v_i}{t}

    a_{car} = \frac{2.06 - 6.32}{0.203}

    a_{car} = -21 m/s^2

    b) For acceleration of truck we will find the rate of change in velocity of the truck

    so we have

    a_{truck} = \frac{v_f - v_i}{t}

    a_{truck} = \frac{2.06 - 0}{0.203}

    a_{truck} = 10.15 m/s^2

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