**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:

*m _{t}a_{t} = m_{c}a_{c}*

It is given that the mass of the truck is twice the mass of the car, so we can write the equation as such:

**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:

*
P _{c} + P_{t} = P_{f}*And since

m

**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 = ½ m _{c}v_{c}^{2} + ½ m_{t}v_{t}^{2} = ½ (1c)(10^{2} m/s) + ½ (2c)(-10^{2} m/s) = 150m J*
So, a great deal of the kinetic energy is lost in the collision (as is expected from an inelastic collision by definition).

KE after = ½ (m_{c} + m_{t}) V_{f}^{2} = ½ (3c)(3.3^{2} 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%