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 10.0 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 100.0 kg. Including the drivers, the total vehicle masses are 900 kg for the car and 4100 kg for the truck. The collision time is 0.080 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!)

(l) What is the average force on the car driver from the seatbelt? (Sign matters!)



Final thoughts on this problem: Note that Newton's 3rd Law is valid: the force the truck exerts on the car is equal and opposite to the force the car exerts on the truck. In terms of momentum, the impulse the truck gives the car is equal and opposite to the impulse the car gives the truck. But, this does NOT mean the forces exerted on the individual drivers is equal and opposite! The force the seatbelt exerts on the truck driver is much less than the force the seatbelt exerts on the car driver. But that is okay; these two forces are NOT part of a Newton's 3rd Law force pair.

Answer 1

let m1 = 4100 kg, v1 = 10 m/s

m2 = 900 kg, v2 = -10 m/s

a) Px_Before = m1*v1 + m2*v2

= 4100*10 + 900*(-10)

= 32000 kg.m/s

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

= (4100*10 + 900*(-10))/(4100 + 900)

= 6.40 m/s

c) Px_after = Px_Before

= 32000 kg.m/s

d) vx_f = 6.40 m/s

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

= m1*(vf - v1)

= 4100*(6.4 - 10)

= -14760 N.s

f) impulse on the car = m2*(vf - v2)

= 900*(6.4 - (-10))

= 14760 N.s

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

= -14760/0.08

= -184500 N

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

= 14760/0.08

= 184500 N

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

= m*(vf - v1)

= 100*(6.4 - 10)

= -360 N.s

j) impulse on the car driver = m*(vf - v2)

= 100*(6.4 - (-10))

= 1640 N.s

k) on truck driver, Favg = impulse/time

= -360/0.08

= -4500 N

l) on car driver, Favg = impulse/time

= 1640/0.08

= 20500 N