Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails while movin

Question

Aaron Agin nodded off while driving home from play practice this past Sunday evening. His 1500-kg car hit a series of guardrails while moving at 19.8 m/s. The first guard rail delivered a resistive impulse of 5700 N•s. The second guard rail pushed against his car with a force of 79000 N for 0.12 seconds. The third guard rail collision lowered the car’s velocity by 3.2 m/s. Determine the final velocity of the car.

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Kim Cúc 4 years 2021-08-10T17:04:37+00:00 1 Answers 120 views 0

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  1. Nem
    0
    2021-08-10T17:06:02+00:00
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    Answer: 6.48m/s

    Explanation:

    First, we know that Impulse = change in momentum

    Initial velocity, u = 19.8m/s

    Let,

    Velocity after first collision = x m/s

    Velocity after second collision = y m/s

    Also, we know that

    Impulse = m(v – u). But then, the question said, the guard rail delivered a “resistive” impulse. Thus, our impulse would be m(u – v).

    5700 = 1500(19.8 – x)

    5700 = 29700 – 1500x

    1500x = 29700 – 5700

    1500x = 24000

    x = 24000/1500

    x = 16m/s

    Also, at the second guard rail. impulse = ft, so that

    Impulse = 79000 * 0.12

    Impulse = 9480

    This makes us have

    Impulse = m(x – y)

    9480 = 1500(16 -y)

    9480 = 24000 – 1500y

    1500y = 24000 – 9480

    1500y = 14520

    y = 14520 / 1500

    y = 9.68

    Then, the velocity decreases by 3.2, so that the final velocity of the car is

    9.68 – 3.2 = 6.48m/s

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