Case Studies: Impulse and Force - help4

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 collisions of two water balloons with the catcher's hands. In one case, the balloon contains 150 mL of water. In the other case, the balloon contains 600 mL of water. 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 water balloons. The question tells you the two water balloons change their velocity from 8 m/s to 0 m/s (brought to a stop); they have the same velocity change. And the collision time is stated to be the same for each. So by careful reading and the process of elimination, the variable in these collisions is the mass of the water balloons. A 600-mL water balloon will have more mass than a 150-mL water balloon.
 

Momentum Change and Impulse

You will also have to compare the momentum change and the impulse encountered by these two water balloons. The momentum change is your starting point. Momentum change is the mass multiplied by the velocity change. You have just determined that the mass of the 600-mL balloon is greater. And since the two balloons have the same velocity change, it is the 600-mL water balloon that will have the greater momentum change.

In any collision, the momentum change is equal to the impulse. So if the 600-mL water balloon 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 water balloon 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 600-mL water balloon. 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 600-mL water balloon wins again; it experiences the greatest Force.