Case Studies: Impulse and Force - help3

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 PE student with a wall. In one Case, the wall is padded. In the other Case, the wall is unpadded. 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 baseballs. The PE student has the same mass in each Case. And the question states that the velocity changes from 8 m/s to 0 m/s (stopped) in each Case. And so both Cases have the same velocity change. So by careful reading and the process of elimination, the variable in these collisions is the time. When a PE student hits an unpadded wall, they are stopped immediately. But a padded wall has some "give" to it. And so the collision with the padded wall involves a greater time since the PE student continues moving forward for a few more milliseconds while the padding provides the "give."
 

Momentum Change and Impulse

You will also have to compare the momentum change and the impulse encountered by 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 student collides with a padded wall or an unpadded wall.

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 PE student in the two collisions. So 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 padded wall collision. So colliding with a padded wall leads to a smaller force ... due to the greater collision time. And this is why walls are padded at various locations in gymnasiums. The unpadded wall results in a greater force ... and that hurts.