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 an egg with the ground. In one case, the egg collides is dropped from a 5-meter height. In the other case, the egg is dropped from a 1-meter height. 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 eggs. The question tells you the two eggs 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. When an egg is dropped from a higher height, it reaches a greater speed before hitting the ground. So the egg dropped from 5 meters has a greater velocity when it strikes the ground. It thus experiences the greater velocity change during the collision.
 

Momentum Change and Impulse

You will also have to compare the momentum change and the impulse encountered by these two eggs. The momentum change is your starting point. Momentum change is the mass multiplied by the velocity change. You have just determined that the egg dropped from 5 meters has the greater velocity change. And since the two eggs have the same mass, it is the egg dropped from 5 meters that will have the greater momentum change.

In any collision, the momentum change is equal to the impulse. So if the egg dropped from 5 meters 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 egg 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 egg dropped from 5 meters. 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 egg dropped from 5 meters wins again​; it experiences the greatest Force.