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 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 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. But what about the two drivers? Do they expernence the same forces? To answer this question, suppose that each vehicle is inibially moving at 9.0 m/s and that they undergo a perfectly inelastic head-on collision. (In an inelastic collision, the two objects move together as one object after the collision.)Each driver has a mass of 70.0 kg. Including the drivers, the total vehicle masses are 870 kg for the car and 4070 kg for the truck. The collision time is 0.140 s. Choose coordinates such that the truck is initially moving in the positive x direction, and the car is initially moving in the negative x direction. (a) What is the total x-component of momentum BEFORE the collision? Enter a number What is the x-component of the CENTER-OF-MASS velocity BEFORE the collision? (c) What is the total x-component of momentum AFTER the collision? Enter a number Dimensionally incorrect. Please check the type or dimension of your unit. (d) What is the x-component of the final velocity of the combined truck-car wreck? Enter a number ) What impulse did the truck receive from the car during the collision? (Sign matters) Enter a number What impuise did the car receive from the truck during the collision? (Sign matters) Enter a number what is the average force on the truck from the cor during the collision? (Sign mattersi) Enter a number b) What is the averape force on the car from the truck during the colision? (Sign matters!) Enter a number truck driver experience from his seatbelt? (Sign matters) Enter a number What impuise did the car driver experience from his seatbelt? (Sign matters) Enter e number k) What is the average force on the truck driver from the seatbelt? (Sign Enter a number U) What is the average force on the car driver from the seatbelt? (Sign matters!) Enter a number

Explanation / Answer

let m1 = 4070 kg, v1 = +9 m/s

v2 = 870 kg, v2 = -9 m/s

a) Px_Before = m1*v1 + m2*v2

= 4070*9 + 870*(-9)

= 28800 kg.m/s

b) vcm_before = (m1*v1 + m2*v2)/(m1+m2)

= (4070*9 + 870*(-9))/(4070 + 870)

= 5.83 m/s

c) Px_after = Px_Before

= 28800 kg.m/s

d) vx_f = 5.83 m/s

e) impulse on the truck = change in momentum of the truck

= m2*(vf - v2)

= 4070*(5.83 - 9)

= -12902 N.s

f) impulse on the car = m1*(vf - v1)

= 870*(5.83 - (-9))

= 12902 N.s

g) on truck, Favg = impulse on truck/time

= -12902/0.14

= -92157 N

h) on car, Favg = impulse on car/time

= 12902/0.14

= 92157 N

i) impulse on the truck driver = change in momentum of the truck driver

= m*(vf - v2)

= 70*(5.83 - 9)

= -222 N.s

j) impulse on the car driver = m*(vf - v1)

= 70*(5.83 - (-9))

= 1038 N.s

k) on truck driver, Favg = impulse/time

= -222/0.14

= -1586 N

l) on car driver, Favg = impulse/time

= 1038/0.14

= 7414 N

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