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 6.50 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 73.0 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? (Enter the magnitude of the force.)

Answer 1

let
m1 = 4000 kg
m2 = 800 kg
u1 = 6.5 m/s
u2 = -6.5 m/s

let v1 and v2 are the velocities after the collision.

we know,

v1 = ( (m1 - m2)*u1 + 2*m2*u2 )/(m1 + m2)

= ( (4000 - 800)*6.5 + 2*800*(-6.5))/(4000 + 800)

= 2.167 m/s

v2 = ( (m2 - m1)*u2 + 2*m1*u1)/(m1 + m2)

= ( (800 - 4000)*(-6.5) + 2*4000*6.5)/(4000 + 800)

= 15.17 m/s

magnitude of acceleration of truck, a_truck = |(v1 - u1)|/t

= |(2.167 - 6.5)|/0.1

= 43.3 m/s^2

magnitude of acceleration of car, a_car = (v2 - u2)/t

= (15.17 - (-6.5))/0.1

= 217 m/s^2

force exerted by seat on the driver in the truck, F_truck = m*a_track

= 73*43.3

= 3161 N <<<<<<<<-----------Answer

force exerted by seat on the driver in the car, F_car = m*a_car

= 73*217

= 15841 N <<<<<<<<-----------Answer

clearly a large amount of force is exerted by the seat belt on the car driver.