Most of us know intuitively that in a head-on collision between a large dump tru

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? 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 in 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. What about the two drivers? Do they experience the same forces? To answer this question, suppose each vehicle is initially moving at 10.0 m/s and they undergo a perfectly inelastic head-on collision. Each driver has mass 90.0 kg. Including the drivers, the total vehicle masses are 810 kg for the car and 4010 kg for the truck. 

If the collision time is 0.140 s, what force does the seatbelt exert on the truck driver?
 _______N
 
What average force does the seatbelt exert on the car driver?
 _______N

Explanation / Answer

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 ---------------------------------------------------------------------- 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 (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 (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

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