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3: The law of conservation of linear momentum states that in an isolated system, the total linear momentum remains constant. This means that the total momentum before a collision or interaction is e

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4.3: The law of conservation of linear momentum states that in an isolated system, the total linear momentum remains constant. This means that the total momentum before a collision or interaction is equal to the total momentum after the collision or interaction, provided no external forces act on the system.

4.4:

  1. Define a positive direction and list the given values.

    Let the initial direction of the car be positive.
    Mass of car (mcm_c) = 1000 kg1000 \text{ kg}
    Initial velocity of car (vciv_{ci}) = +40 m/s+40 \text{ m/s}
    Mass of truck (mtm_t) = 5000 kg5000 \text{ kg}
    Initial velocity of truck (vtiv_{ti}) = −20 m/s-20 \text{ m/s} (since it's moving head-on, in the opposite direction)
    After the collision, the car and truck move together with a common final velocity (vfv_f).

  2. Apply the law of conservation of linear momentum.

    The total momentum before the collision equals the total momentum after the collision:

    mcvci+mtvti=(mc+mt)vfm_c v_{ci} + m_t v_{ti} = (m_c + m_t) v_f
  3. Substitute the values into the equation.

    (1000 kg)(+40 m/s)+(5000 kg)(−20 m/s)=(1000 kg+5000 kg)vf(1000 \text{ kg})(+40 \text{ m/s}) + (5000 \text{ kg})(-20 \text{ m/s}) = (1000 \text{ kg} + 5000 \text{ kg}) v_f 40000 kg⋅m/s−100000 kg⋅m/s=(6000 kg)vf40000 \text{ kg} \cdot \text{m/s} - 100000 \text{ kg} \cdot \text{m/s} = (6000 \text{ kg}) v_f
  4. Solve for the final velocity (vfv_f).

    −60000 kg⋅m/s=(6000 kg)vf-60000 \text{ kg} \cdot \text{m/s} = (6000 \text{ kg}) v_f vf=−60000 kg⋅m/s6000 kgv_f = \frac{-60000 \text{ kg} \cdot \text{m/s}}{6000 \text{ kg}} vf=−10 m/sv_f = -10 \text{ m/s}

    The negative sign indicates that the final velocity is in the opposite direction to the car's initial motion (i.e., in the direction the truck was initially moving).

    The velocity of the thief's car immediately after the collision is 10 m/s in the direction of the truck’s initial motion\boxed{\text{10 m/s in the direction of the truck's initial motion}}.

    4.5:

  5. Calculate the impulse experienced by the car.

    Impulse (JJ) is the change in momentum (Δp\Delta p). We will calculate the impulse on the car.
    Jc=Δpc=mcvf−mcvciJ_c = \Delta p_c = m_c v_f - m_c v_{ci}
    Using the values from 4.4:
    mc=1000 kgm_c = 1000 \text{ kg}
    vci=+40 m/sv_{ci} = +40 \text{ m/s}
    vf=−10 m/sv_f = -10 \text{ m/s}

    Jc=(1000 kg)(−10 m/s)−(1000 kg)(+40 m/s)J_c = (1000 \text{ kg})(-10 \text{ m/s}) - (1000 \text{ kg})(+40 \text{ m/s}) Jc=−10000 kg⋅m/s−40000 kg⋅m/sJ_c = -10000 \text{ kg} \cdot \text{m/s} - 40000 \text{ kg} \cdot \text{m/s} Jc=−50000 kg⋅m/sJ_c = -50000 \text{ kg} \cdot \text{m/s}
  6. Calculate the average force exerted during the collision.

    The average force (FavgF_{avg}) is equal to the impulse divided by the time interval (Δt\Delta t).
    The collision lasts Δt=0.5 s\Delta t = 0.5 \text{ s}.

    Favg=JcΔtF_{avg} = \frac{J_c}{\Delta t} Favg=−50000 kg⋅m/s0.5 sF_{avg} = \frac{-50000 \text{ kg} \cdot \text{m/s}}{0.5 \text{ s}} Favg=−100000 NF_{avg} = -100000 \text{ N}

    The magnitude of the average force is 100000 N100000 \text{ N}.

  7. Compare the calculated average force with the fatal injury threshold.

    The research states that forces greater than 85000 N85000 \text{ N} may cause fatal injuries.
    Our calculated average force magnitude is 100000 N100000 \text{ N}.
    Since 100000 N>85000 N100000 \text{ N} > 85000 \text{ N}, the collision could result in a fatal injury.

    The collision could result in a fatal injury because the calculated average force of 100000 N\boxed{\text{100000 N}} is greater than the threshold of 85000 N85000 \text{ N}.

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