Questions1) Explain the physics of a car crash and how the crumple zone, safety belt, and air bag work in concert to increase the drivers survivability. Car crashes do not kill people, after this lab, it was learned that forces is what kill people. People die in car crashes because impact is very fast, and the force of impact too large. If the time of impact can be prolonged, then the acceleration and the force of impact can be reduced. This is because of Newton's 2ndLaw of Motion

We can use the equation

to get

If we increase the time of impact, we can reduce the average force of impact and this will result in injuries that are less severe. There are three main ways to do this:Crumple Zones: These are semi rigid structures at the front and back of the vehicle, which are designed to absorb energy and collapse slowly on impact.Seat Belts are an addition to the defense of the forces. They are designed to stop passengers becoming projectiles inside the car by restraining them in their seats. If rear passengers were not restrained they might kill passengers in the front seat by impacting them from behind, as well as seriously injuring themselves. If front seat passengers were not restrained they could impact the dashboard/steering wheel/windscreen, which are all rigid and will cause serious injury.The final line of defence is air bags. They are designed to inflate if subject to a force above a certain level. They then absorb as much of the kinetic energy of impact as possible when they are hit by a person. Gas is very compressible and very well suited to absorbing the energy of impact. 2) compare the average force experienced by a 60kg person moving at 34 km/h (20mph) being stopped in a distance of 2 cm in one case or 40cm in the other case ( e.g. the effective distance due to the crumple zone and air bag) 34 km/hr = 9.44 m/svf^2 = vi^2 + 2ad0^2 = (9.44m/s)^2 + 2 (a) (.02 m)

a = 2229.94 m/s^2 (for padded dashboard)f = maf = (60kg) (2229.94m/s^2)f = 133796N

0^2 = (9.44m/s)^2 + 2 (a) (.40m)

a = 111.50 m/s^2 (for air bag)

f = maf = (60kg) (111.50m/s^2)f = 6689.8N

A 60kg person experiences an average stopping force about 133796 / 6690 = 20 times larger when stopped in the shorter distance. Therefore, This person is more likely to be injured. 3) If a pressure sensor were incorporated in the experiment, a similar profile to the force impulse should be recorded. Since no crumple zone is included, the air bag must dissipate all the energy in the collision. If p is the overpressure during the collision and V is the volume of vented air, the dissipated energy should be given by the integral: p dVWhich would be the area under the pressure vs volume graph. Choose the air bag collision with highest initial velocity. Assume all kinetic energy is dissipated by the air bag and the average overpressure is 850 pascal. what volume of air (m^3) was pushed out of the bag?K = 1/2 m v^2K = 1/2 (60kg) (9.44m/s)^2K = 2675.93 J

E = p dV2675.93 N*m = (850 N/m^2) (dV)dV =3.148 m^3Volume of air vented out dV =3.148 m^3For the experiment in our lab: K = 1/2 m v^2K = 1/2 (0.6751Kg) (0.67m/s)^2K = 0.1515 J

E = p dV0.1515 N*m = (850 N/m^2) (dV)dV =1.78 cm^3Volume air vented from the zip bag was 1.78 cm^3