Question

Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that exerted on the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false; Newton's third law tells us that both objects are acted upon by forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at 6.40 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 72.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4,000 kg for the truck. If the collision time is 0.100 s, what force does the seat belt exert on each driver? force on truck driver N force on car driver N

Answer 1

Mass of the car with driver, m1 = 810 kg

Mass of the truck with driver, m2 = 4010 kg

Initial velocity of the car, v1 = 10 m/s

Initial velocity of the truck, v2 = - 10 m/s

In the perfectly inelastic collision,

Common velocity of the system, v = ( m1 v1 + m2 v2 ) / (m1+m2)

                                                    = (8100 - 40100)/(810+4010)

                                                    = - 6.64 m/s

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Mass of each driver, M = 90 kg

Collision time, t = 0.140 s

(a)

Force exerted by belt on the truck driver:

Ftruck= M (v-v2)/t

        = 90 * (-6.64+10)/0.140

        =2.16 x 10^3 N

(b)

Force exerted by belt on the car driver:

Fcar = M (v-v1)/t

        = 90 * (-6.64-10)/0.140

        =1.07 x 10^4 N