Case Studies: Impulse and Force - help5

During a collision, an object experiences an impulse that changes its momentum. The impulse is equal to the momentum change. Knowing that impulse is the product of Force•∆Time and that momentum change is the product of Mass•∆Velocity, one can use the Force•∆Time = Mass•∆Velocity relationship as a guide to thinking about how alterations in m, ∆t, and ∆v affect the force in a collision.

In this question, you will have to compare two collisions of a driver. In one Case, the driver is stopped by an airbag. In the other Case, the driver is stopped by a hard steering wheel. Here's how to think about the physics of these collisions:
 

The Variable

First you must determine what the variable is. It is either the velocity change (Delta V), the collision or contact time, or the mass of the driver. The driver has the same mass in each Case - 50 kg. And the question states that the velocity changes from 12 m/s to 0 m/s (stopped) in each Case. And so both Cases have the same velocity change. By careful reading and the process of elimination, the variable in these collisions is the time. When a driver hits a hard steering wheel, he is stopped immediately. But an air bag has some "give" to it. And so the collision with the air bag involves a greater time since the driver continues moving forward for a few more milliseconds since the air bag provides some "give."
 

Momentum Change and Impulse

You will also have to compare the momentum change and the impulse for these two Cases. The momentum change is your starting point. Momentum change is the mass multiplied by the velocity change. You have just determined that the mass and the velocity change is the same for both Cases. And so there is no difference in momentum change for these two cases. The momentum change is the same whether the 50-kg driver hits an air bag or a hard steering wheel.

In any collision, the momentum change is equal to the impulse. So if the two Cases have the same momentum change, they will also have the same impulse.
 

Force

Finally, you will have to use F•∆t = m•∆v to compare the Force experienced by the driver in the two collisions. The force is the momentum change divided by the collision time ... that is, m•∆v/∆t. The numerator in this expression is the momentum change (m•∆v). You have just determined that it is the same for both Cases. You also have determined that the collision times (∆t) is greater for the air bag collision. So colliding with an air bag leads to a smaller force ... due to the greater collision time. And this is why we place air bags in cars. The hard steering wheel results in a greater force ... and that hurts.