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 7.40 m/s and that they undergo a perfectly inelastic head-on collision. Each driver has mass 76.0 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 0.110 s, what force does the seat belt exert on each driver? (Enter the magnitude of the force.)

force on truck driver   ___________ N

force on car driver     ___________N

Answer 1

As per newton's third law, both car and truck experience same amount of force. but since truck is made of stronger material as compared to car, there is less damage to the truck compared to car. that is why you are better of being in the truck

They do not experiece the same force. the relaxation time in case of truck is large, which reduces the effect of force on the driver.

m = mass of the car = 800 kg

M = mass of truck = 4000 kg

vc = initial velocity of car = - 7.40 m/s

vt = initial velocity of truck = 7.40 m/s

V = final velocity

Using conservation of momentum

m vc + M vt = (m + M) V

(800)(- 7.40) + (4000)(7.40) = (800 + 4000)V

V = 4.93 m/s

for the truck driver :

m = 76

F t = m (V - v)

F (0.110) = (76) (4.93 - 7.40)

F = - 1706.55 N

for car driver :

m = 76

F t = m (V - v)

F (0.110) = (76) (4.93 - 7.40)

F = - 1706.55 N