A 1 145.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving

Question

A 1 145.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east. What is the velocity of the truck right after the collision?(I found this) =20.826 (b) What is the change in mechanical energy of the car?

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Helga 5 years 2021-08-18T05:10:23+00:00 1 Answers 14 views 0

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  1. Euphemia
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    2021-08-18T05:11:58+00:00
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    Answer:

    (a)v2 = 20.590m/s

    (b) ΔE = 116,150J

    Explanation:

    Let m1 = mass of the car = 1145kg

    m2 = mass of the truck = 9700kg

    u1 = 25.000m/s = initial velocity of the car

    u2 = 20.000m/s = initial velocity of the truck

    v1 = 20.000m/s = final velocity of the car

    v2 = ? = final velocity of the truck

    From the principle of the conservation of momentum, total momentum before collision equals total momentum after collision. We sum the momentum of both vehicles before and after the collision. For the car the momentum before and after the collision are m1u1 and m1v1 while for the truck they are m2u2 and m2v2

    So,

    m1u1 + m2u2 = m1v1 + m2v2

    1145×25 + 9700×20 = 1145×20 + 9700v2

    28,625 + 194,000 = 22,900 + 9700v2

    222,625 = 22,900 +9700v2

    9700v2 = 222625 –22900

    9700v2 = 199,725

    v2= 199725/9700 = 20.59m/s

    (b) ΔE = 1/2× m2× (v2²– v1²)

    = 1/2×9700×(20.590² – 20.000²)

    = 116,148J

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