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A car with two passengers traveling at 15 m/s collides with a tree. One of the passengers who is not wearing a seat belt strikes the windshield head first and comes to rest in 0.03 s. The area of contact between the head and the windshield is approximately 5x10 m² and the mass of the head is 5.4 kg. The other passenger who is wearing his seat belt comes to rest in 0.50 s. The mass of this passenger is 75 kg. The area of the seat belt in contact with this passenger is about 0.12 m². Find the average force and the force per unit area exerted on the two passengers.
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Respuesta :

In the absence of friction, a force applied to a body for a given amount of time, changes the momentum of the body

The average force per unit area exerted on the passenger with no seatbelt 5,400,000 N/m²

The average force per unit area for the passenger with seat belt 18,750 N/m²

The known parameters are;

The velocity of the car, v₀ = 15 m/s

The time it takes the passenger not wearing seat belt to strike the windshield, [tex]t_{ns}[/tex] = 0.03 s

Area of contact between the head and the windshield, [tex]A_{ns}[/tex] = 5 × 10⁻⁴ m²

The mass of the head, [tex]m_{h}[/tex] = 5.4 kg

Time taken for the passenger wearing seat belt to come to rest, [tex]t_s[/tex] = 0.50 s

The mass of the passenger wearing seatbelt, [tex]m_s[/tex] = 75 kg

Area of the seat belt in contact with the passenger, [tex]A_s[/tex] = 0.12 m²

Required:

To find the average force per unit area exerted on the two passengers

Solution:

According to Newton's second law on motion, force is equal to the product of mass and acceleration

[tex]Acceleration, \overline a = \dfrac{Change \ in \ velocity}{Elapsed \ time} = \dfrac{v - v_0}{t}[/tex]

Therefore, the acceleration, force and average force per unit area of the passenger with no seat belt;

[tex]Acceleration, \overline a_{ns} = \dfrac{0 - 15 \, m/s}{0.03 \, s} = 500 \, m/s^2[/tex]

[tex]F_{ns}[/tex] = [tex]m_{h}[/tex] × [tex]\overline a_{ns}[/tex]

Which gives;

[tex]F_{ns}[/tex] = 5.4 kg × 500 m/s² = 2,700 N

Force per unit area = Force/Area

  • The average force per unit area, for the passenger with no seat belt is given as follows:

[tex]P_{ns} = \dfrac{2,700 \, N}{5 \times 10 ^{-4} \ m^2} = 5,400,000 \ N/m^2[/tex]

The average force per unit area exerted on the passenger with no seatbelt, [tex]P_{ns}[/tex] = 5,400,000 N/m²

  • The acceleration, force and average force per unit area of the passenger with seat belt:

[tex]Acceleration, \overline a_{s} = \dfrac{0 - 15 \, m/s}{0.5 \, s} = 30 \, m/s^2[/tex]

[tex]F_{s}[/tex] = [tex]m_{b}[/tex] × [tex]\overline a_{s}[/tex]

Which gives;

[tex]F_{s}[/tex] = 75 kg × 30 m/s² = 2,250 N

Force per unit area = Force/Area

The average force per unit area, for the passenger with seat belt is therefore;

[tex]P_{s} = \dfrac{2,250\, N}{0.12 \ m^2} = 18, 750\ N/m^2[/tex]

The average force per unit area for the passenger with seat belt, [tex]P_s[/tex] = 18,750 N/m²

Learn more about the concept of forces, change in momentum and pressure here:

https://brainly.com/question/15448893

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