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Write the principle of conservation of linear momentum.

Physics

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Answer

16.8 m/s to the east\text{16.8 m/s to the east}

49680 J\text{49680 J}

Step-by-step solution

4 steps
  1. Write the principle of conservation of linear momentum.

    In an isolated system, the total linear momentum remains constant.

  2. Calculate the velocity of the truck after the collision.

    Let the car be cc and the truck be tt. Let the initial velocities be viv_i and final velocities be vfv_f.
    Assume the eastern direction is positive.
    Given:
    Mass of car, mc=1200m_c = 1200 kg
    Initial velocity of car, vci=+25v_{ci} = +25 m/s
    Mass of truck, mt=6000m_t = 6000 kg
    Initial velocity of truck, vti=+15v_{ti} = +15 m/s
    Final velocity of car, vcf=+16v_{cf} = +16 m/s

    According to the principle of conservation of linear momentum:

    mcvci+mtvti=mcvcf+mtvtfm_c v_{ci} + m_t v_{ti} = m_c v_{cf} + m_t v_{tf}

    Substitute the known values:

    (1200 kg)(25 m/s)+(6000 kg)(15 m/s)=(1200 kg)(16 m/s)+(6000 kg)vtf(1200 \text{ kg})(25 \text{ m/s}) + (6000 \text{ kg})(15 \text{ m/s}) = (1200 \text{ kg})(16 \text{ m/s}) + (6000 \text{ kg}) v_{tf} 30000 kg⋅m/s+90000 kg⋅m/s=19200 kg⋅m/s+(6000 kg)vtf30000 \text{ kg}\cdot\text{m/s} + 90000 \text{ kg}\cdot\text{m/s} = 19200 \text{ kg}\cdot\text{m/s} + (6000 \text{ kg}) v_{tf} 120000 kg⋅m/s=19200 kg⋅m/s+(6000 kg)vtf120000 \text{ kg}\cdot\text{m/s} = 19200 \text{ kg}\cdot\text{m/s} + (6000 \text{ kg}) v_{tf} 120000−19200=6000vtf120000 - 19200 = 6000 v_{tf} 100800=6000vtf100800 = 6000 v_{tf} vtf=1008006000v_{tf} = \frac{100800}{6000} vtf=16.8 m/sv_{tf} = 16.8 \text{ m/s}

    The velocity is positive, indicating it is in the eastern direction.
    The velocity of the truck after the collision is 16.8 m/s to the east\boxed{\text{16.8 m/s to the east}}.

  3. Calculate the mechanical energy lost during the car-truck collision.

    First, calculate the total initial kinetic energy (KEiKE_i):

    KEi=12mcvci2+12mtvti2KE_i = \frac{1}{2} m_c v_{ci}^2 + \frac{1}{2} m_t v_{ti}^2 KEi=12(1200 kg)(25 m/s)2+12(6000 kg)(15 m/s)2KE_i = \frac{1}{2} (1200 \text{ kg}) (25 \text{ m/s})^2 + \frac{1}{2} (6000 \text{ kg}) (15 \text{ m/s})^2 KEi=12(1200)(625)+12(6000)(225)KE_i = \frac{1}{2} (1200)(625) + \frac{1}{2} (6000)(225) KEi=375000 J+675000 JKE_i = 375000 \text{ J} + 675000 \text{ J} KEi=1050000 JKE_i = 1050000 \text{ J}

    Next, calculate the total final kinetic energy (KEfKE_f):

    KEf=12mcvcf2+12mtvtf2KE_f = \frac{1}{2} m_c v_{cf}^2 + \frac{1}{2} m_t v_{tf}^2

    Using vtf=16.8v_{tf} = 16.8 m/s from Step 2:

    KEf=12(1200 kg)(16 m/s)2+12(6000 kg)(16.8 m/s)2KE_f = \frac{1}{2} (1200 \text{ kg}) (16 \text{ m/s})^2 + \frac{1}{2} (6000 \text{ kg}) (16.8 \text{ m/s})^2 KEf=12(1200)(256)+12(6000)(282.24)KE_f = \frac{1}{2} (1200)(256) + \frac{1}{2} (6000)(282.24) KEf=153600 J+846720 JKE_f = 153600 \text{ J} + 846720 \text{ J} KEf=1000320 JKE_f = 1000320 \text{ J}

    Finally, calculate the mechanical energy lost:

    Energy lost=KEi−KEf\text{Energy lost} = KE_i - KE_f Energy lost=1050000 J−1000320 J\text{Energy lost} = 1050000 \text{ J} - 1000320 \text{ J} Energy lost=49680 J\text{Energy lost} = 49680 \text{ J}

    The mechanical energy lost during the collision is 49680 J\boxed{\text{49680 J}}.

  4. Explain why mechanical energy is lost.

    Mechanical energy is lost in an inelastic collision because it is converted into other forms of energy, such as heat, sound, and potential energy of deformation (due to the crumpling of the car and truck).

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