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 7.20 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 70.0 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 0.140 s, what force does the seat belt exert on each driver? (Enter the magnitude of the force.)

Answer 1

first, we find final velocity of car/truck after the collision

let care be denoted as 1 and truck be denoted as 2

so,

Let car is moving in positive direction and truck is moving in negative direction.

m1v1i + m2v2i = ( m1 + m2)v

800 * 7.20 - 4000 * 7.20 = ( 800 + 4000)v

solve for v, we get

v = - 4.8 m/s ( don't worry about the sign, it just tells direction)

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So, Now we use impulse momentum

for car

Force * time = mass ( final velocity - initial velocity)

Force * 0.140 = 70 ( - 4.8 - 7.2)

Force = 6000 N

for truck

Force * 0.140 = 70 ( - 4.8 - (-7.2))

Force = 1200 N

_________

Therefore, car driver feels much higher force !!!

Therefore, its better to be in a truck