On an icy road, a 1100 kg car moving at 55 km/h strikes a 480 kg truck moving in the same direction at 37 km/h . The pair is soon hit from b

Question

On an icy road, a 1100 kg car moving at 55 km/h strikes a 480 kg truck moving in the same direction at 37 km/h . The pair is soon hit from behind by a 1300 kg car speeding at 49 km/h. If all three vehicles stick together, what is the speed ofthe wreckage?

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Vân Khánh 5 years 2021-07-30T07:56:49+00:00 2 Answers 28 views 0

Answers ( )

  1. quangkhai
    0
    2021-07-30T07:57:59+00:00
    Reply

    Answer:

    Speed of the wreckage = 49.29 km/hr

    Explanation:

    This question is solved simply by using the conservation of momentum law.

    The momentum of the three moving bodies are calculated below:

    Momentum of Car 1 : 1100 * 55 = 60500 kg.km/hr

    Momentum of Truck : 480 * 37 = 17760 kg.km/hr

    Momentum of Car 2 : 1300 * 49 = 63700 kg.km/hr

    Total mass of all three vehicles: 1100 + 480 + 1300 = 2880 kg

    The final momentum equals the initial momentum if it is conserved. Thus we have the following equation:

    Final Momentum = Initial Momentum

    Final Velocity * Total mass = Momentum of all three vehicles combined

    Final Velocity * 2880 = 60500 + 17760 + 63700

    Final Velocity = 49.29 km/hr

  2. Tryphena
    0
    2021-07-30T07:58:15+00:00
    Reply

    Answer:

    the speed of the wreckage is 48.39km/hr

    Explanation:

    Given that ,

    Mass of first car, m₁ =1100kg

    Speed of the first car, v₁ = 55km/hr

    Mass of second car, m₂ = 480kg

    Speed of second car, v₂ = 37km/hr

    Mass of third car, m₃ =1300kg

    Speed of third car, v₃ = 47km/hr

    In the first case,

    m_1v_1+m_2v_2=(m_1+m_2)v

    (1100\times 55) + (480\times 37)=(1100+480)v\\\\v =\frac{78260}{1580} \\\\v =49.53km/hr

    In the second case

    (m_1+m_2)v+m_3v_3=(m_1+m_2+m_3)v

    where v is the velocity of the combined system

    (1100+480)\times 49.53=(1300\times 47)=(1100+480+1300)\\\\(78257.4+61100)=2880\\\\v = \frac{139357.4}{2880} \\\\v =48.39km/hr

    Thus , the speed of the wreckage is 48.39km/hr

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