Case Studies: Impulse and Force - Questions 1 Help

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 ball with the floor. In one case, the ball collides and rebounds upward. In the other case, the ball collides and stops ... with no bounce. 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 balls. The question tells you the two balls have the same mass and that the collision time is the same for each. So by careful reading and the process of elimination, the variable in these collisions is the velocity change. In the rebounding collision, the ball encounters a 3 m/s change in velocity (from 2 m/s downward to 1 m/s upward). In the other case, the ball changes its velocity by 2 m/s (from 2 m/s downward to 0 m/s).
 

Momentum Change and Impulse

You will also have to compare the momentum change and the impulse encountered by these two balls. The momentum change is your starting point. Momentum change is the mass multiplied by the velocity change. You have just determined that the rebounding collision has the greater velocity change. And since the two balls have the same mass, the rebounding collision will also have the greater momentum change.

In any collision, the momentum change is equal to the impulse. So if the rebounding collision has the greater momentum change, it will also have the greater impulse.
 

Force

Finally, you will have to use F•∆t = m•∆v to compare the Force experienced by the ball 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 greatest for the rebounding collision. The collision time (∆t) is the same for each Case. So the Case with the greatest momentum change is the Case with the greatest Force. The rebounding collision wins again; it has the greatest Force.