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 experience the same forces? To answer this question, suppose that each vehicle is initially 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 80.0 kg. Including the drivers, the total vehicle masses are 880 kg for the car and 4080 kg for the truck. The collision time is 0.150 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? (b) 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? (d) What is the x-component of the final velocity of the combined truck-car wreck? (e) What impulse did the truck receive from the car during the collision? (Sign matters!) (f) What impulse did the car receive from the truck during the collision? (Sign matters!) (g) What is the average force on the truck from the car during the collision? (Sign matters!) (h) What is the average force on the car from the truck during the collision? (Sign matters!) (i) What impulse did the truck driver experience from his seatbelt? (Sign matters!) (j) What impulse did the car driver experience from his seatbelt? (Sign matters!) (k) What is the average force on the truck driver from the seatbelt? (Sign matters!) -1701 N ------ correct (l) What is the average force on the car driver from the seatbelt? (Sign matters!) 7899 N ------ correct please show the steps how you got the parts from a to j

Explanation / Answer

part a:

total x component of momentum=mass of car*veloicty of car+mass of truck*veloicty of truck

=880*(-10)+4080*10=32000 kg.m/s

b)x component of center of mass veloicty before collision=total momentum/total mass

=32000/(4080+880)=6.4516 m/s

c)as there is no external force, total momentum will remain conserved.

hence total x component of momentum after collision=total x component of momentum before collision=32000 kg.m/s

d)final veloicty=total momentum/total mass=32000/(4080+880)=6.4516 m/s

e)impulse received by the truck=change in momentum=mass*(final velocity-initial veloicty)

=4080*(6.4516-10)=-1.4477*10^4 N.s

f)impulse received by the car=mass*(final speed-initial speed)

=880*(6.4561-(-10))

=1.4481*10^4 N.s

g)average force on the truck=impulse/time=-1.4477*10^4/0.13=1.1136*10^5 N

h)average force on the car=impulse/time=1.4481*10^4/0.13=1.1139*10^5 N

i)impulse received by the truck driver=mass*(final speed-initial speed)

=80*(6.4516-10)

=-283.86 N.s

l)impulse received by the car driver=mass*(final speed-initial speed)

=80*(6.4516-(-10))

=1316.1 Ns

k)force received by the truck driver=mass*(final speed-initial speed)/time

=80*(6.4516-10)/0.13

=-2183.6 N

l)force received by the car driver=mass*(final speed-initial speed)/time

=80*(6.4516-(-10))/0.13

=1.0124*10^4 N

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.