Question

To answer this question, suppose that each vehicle is moving at 7.69 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 82.6 kg. The total vehicle masses are 810 kg for the car and 4280 kg for the truck. Note that these values include the masses of the drivers. If the collision time is 0.129 s, (a) what force does the seat belt exert on the truck driver?

Answer 1

Answer:

Force(F) = -80,955.01 N

Explanation:

We need to first determine the impulse that the truck driver received from the car during the collision

So; m₁v₁ - m₂v₂ = (m₁m₂)v₀

where;

m₁ = mass of the truck = 4280 kg

v₁ = v₂ =  speed of the each vehicle = 7.69 m/s

m₂ = mass of the car = 810 kg

Substituting our data; we have:

(4280×7.69) - (810×7.69) = (4280+810)v₀

32913.2 - 6228.9 = (5090)v₀

26684.1 =  (5090)v₀

v₀ = $\frac{26684.1}{5090}$

v₀ = 5.25 m/s

NOW, Impulse on the truck = m (v₀ - v)

= 4280 × (5.25 - 7.69)

= 4280 ×  (-2.44)

= -10,443.2 kg. m/s

Force that the seat belt exert on the truck driver can be calculated as:

Impulse = Force × Time

-10,443.2 kg. m/s = F (0.129)

F = $\frac{-10,443.2}{0.129}$

Force(F) = -80,955.01 N

Thus, the Force that the seat belt exert on the truck driver = -80,955.01 N