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 the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that experienced by 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 experience 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 moving at 6.30 m/s and that the two vehicles stick together when they crash. 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? (b) What force does the seat belt exert on the car driver? Give the magnitude of the forces.

Answer 1

Here, we will use newton second and third law

we know

F = ma

now if force is equal in magnitude then

a = F/m

then as per the above equation, the object with smaller mass will have greater acceleration.

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Using momentum principle, we have

4280*6.30 + 810 * -6.30 = (4280 + 810)*vf

vf = 4.2948 m/s

so,

now using impulse momentum concept

F * t = change in momentum

For truck driver

F = 82.6 * (4.2948 - 6.3) / 0.129 = 1283.89 N

For car diver

F = 82.6 * (4.2948 - (-6.3)) / 0.129 = 6783.9 N