The problem tells us he falls vertically off the ladder (straight down), so we don't need to worry about motion in the horizontal direction.

The equation for momentum is:

We can assume he falls from rest, which allows us to find the initial momentum.

.

From here, we can use the formula for impulse:

We know his initial momentum is zero, so we can remove this variable from the equation.

The problem tells us that his change in time is s seconds, so we can insert this in place of the time.

The only force acting upon man is the force due to gravity, which will always be given by the equation .

In both cases the egg starts with the same velocity, so it has the same initial momentum. In both cases the egg stops moving at the end of its fall, so it has the same final velocity. The only thing that changes is the time and force of the impact. The force is produced by the deceleration resulting from the time that the egg is in contact with its point of impact.

As the time of contact increases, acceleration decreases and force decreases.

As the time of contact decreases, acceleration increases and force increases.

The grass will have less force on the egg, allowing for a lesser acceleration, due to a longer period of impact.

The change in momentum is the final momentum minus the initial momentum, or .

Notice that the problem gives us the final SPEED of the ball but not the final VELOCITY. Since the ball "bounced back," it begins to move in the opposite direction, so its velocity at this point will be negative.

Plug in our values to solve:

If we were to examine the area under the curve of a constant force applied over a certain amount of time we would have a graph with a straight horizontal line.  To find the area of that rectangle we would multiply the base times the height.  The base would be the time (number of seconds the force was applied).  The height would be the amount of force applied during this time.  Force*time is equal to the impulse acting on the object which is equal to the change in momentum of the object.

We know that if the balls fell straight down after the crash, then the total momentum in the horizontal direction is zero. The only motion is due to gravity, rather than any remaining horizontal momentum. Based on conservation of momentum, the initial and final momentum values must be equal. If the final horizontal momentum is zero, then the initial horizontal momentum must also be zero.

In our situation, the final momentum is going to be zero.

Use the given values for the mass of each ball and initial velocity of the first ball to find the initial velocity of the second.

The negative sign tells us the second ball is traveling in the opposite direction as the first, meaning it must be moving east.

The impulse is equal to the change in momentum. The change in momentum is equal to the mass times the change in velocity.  

The yellow ball rebounds higher and therefore has a higher velocity after the rebound.  Since it has a higher velocity after the collision, the overall change in momentum is greater.  Therefore since the change in momentum is greater, the impulse is higher.

This is an example of an elastic collision. We start with two masses and end with two masses with no loss of energy. 

We can use the law of conservation of momentum to equate the initial and final terms.

Plug in the given values and solve for .

The total momentum before the collision is equal to the momentum of each object added together.

Remember that moving to the left means that the object has a negative velocity

Total momentum = 

According to the law of conservation of momentum, the total momentum at the end must equal the total momentum at the beginning.  Since the momentum at the beginning was , the momentum at the end would also be .

Knowns

Unknowns

For elastic collisions we know that the initial and final velocities are related by the equation

We also know that the momentum is conserved meaning that

Since we have two missing variables and two equations, we can now solve for one of the variables using a system of equations

Let’s get the final velocity of car A by itself from the first equation

We can now substitute this equation into our momentum equation.

In our new equation we only have one missing variable, so lets substitute in values and solve.

We can now plug this value back into our equation for  

Knowns

Unknowns

To solve the problem we must consider three different situations.  The first is when the cars are skidding across the ground.  While they are skidding the force of friction is what is resisting their motion and therefore doing work on the cars to slow them down to a stop.

The second situation is the collision itself when the first car has an initial speed and the second car is stopped.  After the collision both cars are traveling with the different speeds as they skid across the ground at different distances.

The third situation is when the first car initially hits the brakes before the collision. While he is skidding to slow down, the force of friction is resisting the motion and therefore doing work to slow the car down some.

To solve this problem we must work from the end of the collision and work backwards to find out what happened before.

To begin, we need to find the speed that the cars are skidding across the ground after the bumpers have been interlocked.

The work-kinetic energy theorem states that the work done is equal to the change in kinetic energy of the object.

Work is directly related to the force times the displacement of the object.

In this case, the force that is doing the work is friction.

Since the cars are on a level surface the normal force is equal to the force of gravity.

The force of gravity is directly related to the mass and the acceleration due to gravity acting on an object.

When we put all these equations together we get

We also know that kinetic energy is related to the mass and velocity squared.

Therefore our final equation should look like

Notice that the mass falls out of the equation since it is in every term.  Also note that the final velocity is 0m/s so this will cancel out as well.

Since both of these terms have a negative we can cancel this out as well.

We can now substitute our variables to determine the velocity of each of the cars just after the crash.

This is the velocity of each of the cars after the collision.  We must now consider our second situation of the collision itself.  During this collision momentum must be conserved.  The law of conservation of momentum states

We can substitute our values for the masses of the cars, the final velocity of the cars after the collision (which we just found) and the initial value of the stopped car.

Now we can solve for our missing variable.

Now we must consider our third situation which is when car A is braking.  We can return to our work-kinetic energy theorem equation again as the force of friction is what is slowing the object down.

This time however, the final velocity is not 0m/s as the car A is still moving when he crashes into car B.  The final velocity is the velocity that we determined the car was moving with before the collision in the second part of the problem.  The masses do still cancel out in this case.