 a) Newton's Third Law states that "whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body." Consequently, it can be said that the magnitude, Ft, of the net force on the truck is equal to the magnitude, Fc, of the net force on the car.

b) Since we know that F = ma and we know that the magnitudes of the forces are equal (part A), it can be said that:
mtat = mcac
It is given that the mass of the truck is twice the mass of the car, so we can write the equation as such:
(2x)at = (1x)ac
which gives the proportion that the acceleration of the truck is equal to half the acceleration of the car.

c/d) Car crashes are often completely inelastic, with much of the kinetic energy going into deforming the cars. In this case momentum is always conserved, meaning that the total momentum of the system of the truck plus the car before the collision is equal to the total momentum after the collision. Using c for car, t for truck, and f for final, the conservation of momentum equation is:
Pc + Pt = Pf
And since P = mv,
mcvc + mtvt = (mc + mt) Vf
So, Vf = [mcvc + mtvt] / (mc + mt)
If we take east as the positive direction, then the car’s velocity goes into the equation with a negative sign, so:
Vf = [ (1x)(-10 m/s) + (2x)(10 m/s) ] / (3x) = 3.3 m/s, which is 3.3 m/s east, and
Pf = (3.3 m/s)(3x) = 10 kg m/s
Thus, the truck’s momentum in the positive direction decreases by 10 kg m/s, while the car’s momentum in the positive direction increases by 20 kg m/s.

e) The change in kinetic energy can be found by adding up the kinetic energy before and after the collision (where c represents the mass of the car):
KE before = ˝ mcvc2 + ˝ mtvt2 = ˝ (1c)(102 m/s) + ˝ (2c)(-102 m/s) = 150m J
KE after = ˝ (mc + mt) Vf2 = ˝ (3c)(3.32 m/s) = 16.6c J
KE lost = 150c J – 16.6c J = 133.4c J
Percentage of KE lost = 100% x 133.4c / 150c = 88.9%
So, a great deal of the kinetic energy is lost in the collision (as is expected from an inelastic collision by definition).