Impulse Force Model Worksheet
In this model, we learned all about the relationship between momentum, impulse and the different factors that affect them. This unit built on our previous knowledge of Newton’s 2nd and 3rd Laws as well as how to effectively apply formulas to different situations.
Qualitative Impulse-Momentum (Worksheet 1)
This worksheet served to familiarize us with the basic momentum formula and how momentum plays a role in our daily lives. Impulse was also introduced and proved to be synonymous with momentum.
Momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity
Momentum equation: P(symbol for momentum)=m(mass)v(velocity)
The units for momentum are found in kg m/s (kilogram meters per second)
There are several different ways of reaching the same momentum using the equation above.
For example: P=Mv or P=mV- this shows how the same momentum was reached by having either a smaller mass and larger velocity, or larger mass and smaller velocity
Below is an example of a basic problem in which you solve for momentum using the equation.
When dealing with change in momentum, the same equation is used except the formula is change in momentum=(mass)(change in velocity)
The problem below displays change in momentum for two balls using the equation.
The superball had the greatest change in momentum because it ultimately had a larger velocity. Larger velocities automatically mean larger changes in momentum. The key to getting this problem right is remembering to make the velocity of the ball bouncing off the wall negative
Impulse is a measure of change in momentum
Impulse equation: J (symbol for momentum)=(Fnet)(change in time)=(mass)(change in velocity)
The relationship between impulse and momentum is key to proper understanding. Impulse is equal to momentum, therefore, if J=P, then (F)(change in time)=(mass)(change in velocity). Basically, in simpler terms, the equation for impulse can also be used to solve for momentum, it is all based upon what information is given in the problem.
Newton’s 3rd Law- which says that every action has an equal and opposite reaction, applies to momentum as well. This law lets us know that if the same force is applied to 2 objects over the same amount of time, the momentum will be equal— which is also explained in the impulse equation.
EXAMPLE: A hummer and a VW beetle travel at equal velocities towards each other and have a head on collision
which vehicle will experience the greater force of impact: according to Newton’s 3rd Law, the VW beetle hits the hummer with an equal force that the hummer hits the VW beetle
which vehicle will experience the greater change in momentum: the momentum is the same because the forces between the objects are equal across the same period of time. The work that further explains the answer to this question is pictured below.
Basically, what this says is that the impulse of the Hummer is equal and opposite to the negative impulse of the VW beetle because the F and change in t is the same. So, if the impulse is the same, the momentum is the same, but the Hummer has a larger mass and smaller change in velocity whereas the beetle has a smaller mass and larger change in velocity which have been previously stated to both equal the same momentum and proves that the change in momentum is the same
One of the final concepts we learned on the first worksheet dealt with real life situations in which momentum and impulse play a large role. One of the questions we were able to answer is “why are padded dashboards safer than hard dashboards in automobiles?”
ANSWER: When one gets in an automobile accident, the momentum will be the same whether an airbag is present or not because the mass obviously stays the same, but the velocity also remains constant because the car goes from moving to rest regardless. Because the momentum is the same, impulse is also the same, but the difference comes in when the airbag is present. When there is no airbag present, one will hit the dashboard will a harder force in a smaller amount of time when there is a wreck, but when an airbag is present, one will hit the airbag with a smaller force across a greater amount of time. This reduces risk of injury because the force of impact is not nearly as large. The written explanation is below.
Newton’s 2nd Law also plans a role in this unit. Knowing the force F= (mass)(acceleration), this can also be applied to two objects after they collide and it works the exact same way as the momentum equation- when dealing with the same force, the variables are inversely related- as one goes up the other goes down. So if a football player and a ballerina collide with the same force because of Newton’s 3rd Law, the ballerina experiences the greatest acceleration because it has a small mass therefore a large acceleration, and the football player has a larger mass and smaller acceleration
Impulse Forces and Momentum (Worksheet 2)
This worksheet was merely more practice with the formulas assuming we had a deeper understand with how they correlate with one another.
This problem requires you to remember that impulse and momentum are equal, and based on the information given in the question, the impulse equation is used to find the answer that is equal to momentum.
The important element of this problem is remembering that the magnitude of the change in momentum of the ball simply means (mass)(change in velocity)
The following problems are all based off one another.
This problem dates all the way back to when we dealt with acceleration and displacement. The horizontal displacement formula is used to solve for the time because of the values given in the problem
After solving for the time, that can be used in the impulse equation to solve for instantaneous momentum, or simply just the momentum. The force is given in the problem and we just solved for time, so all that needs to be done is a plug and chug
This question integrates a very important theory about objects and the momentum needed to change direction. In this particular problem, the ball is falling with a momentum of .2. When the ball hits the ground it loses that momentum and it becomes 0. In order for the ball to bounce back up, it needs the same momentum but with the opposite sign to be applied to it take it back to its original momentum
Once again, in this problem the impulse and momentum equations are used together, except this time, we are solving for just the force. A particular variable can always be solved for by plugging in the values that were given in the equation and reading the question CAREFULLY.
Conservation of Momentum Worksheet 3/4
These worksheets introduced us to the conservation of momentum and the equations that go along with it. We were able to learn about how to calculate unknown variables in each equation and then use that information to solve for other values.
The Conservation of Momentum says that the momentum before the collision will always be equal to the momentum after the collision (p total before=p total after)
When dealing with 2 objects that are unstuck or not together the equation: mava+mbvb is used. This is simply multiplying the mass and velocities of each object together which gives us the total momentum of the 2 objects
When dealing with 2 objects that are stuck together, the equation is: (ma+mb)vab. Because the objects are stuck together, you simply add the masses of the 2 and then multiply it by whatever velocity the whole system is moving with
Working with these problems, it is essential to remember, once again, to make the velocities negative if objects are traveling in different directions
When setting up problems that deal with the conservation of momentum, a few preliminary steps are required.
1. A picture of the situation must be drawn for visualization. This picture should include all the values that are given in the problem so the exact variables that we need to solve for should be more clear
2. Before the equals sign represents before the collision and after the equals sign is after the collision, so after carefully reading the question, decide which formulas are needed for the situation
3. A graph must also be drawn to represent the momentum before and after the collision. Is is basically a written form of the picture of the situation that was drawn. The objects masses and velocities are listed, and multiplied together to get the momentum. Remembering that certain velocities are negative, the momentum is shaded in on either side (positive or negative momentum) based off a scale that is personally choses and the total momentum from the collision before can be used to make sure the momentumthat was solved for the collision after is the same
This is a classic unstuck, stuck problem. Two items were separate before the collision and end up being together after the collision. A picture to illustrate the situation is drawn.
The graph contains each individual object and its velocity listed and then multiplied across and represented on the graph.
Because the objects go from unstuck to stuck, the formula used is mava+mbvb=(ma+mb)vab and the values are simply plugged in. The only variable there is no number for is the vab, so it becomes very easy to plug and chug. After getting the vab, it can be added to the graph (already shown in shading) which represents after the collision and multiplied with the mass to make sure the momentum is the same as it was before.
42. This problem is an unstuck unstuck situation. This problem is a little different because you have to remember that just because two objects collide, doesn't always mean they will stick together.
This picture represents what happened in the situation
The graph shows the masses and velocities of the 2 objects before they collided, as well as the masses and a missing velocity (although already shaded representing the answer after it was reached) for one of the objets after the collision
With the unstuck-unstuck momentum conservation equation, the values are plugged in. None of the velocities have to be made negative because both objects are traveling in the same direction. After the velocity of the second object is found, it is plugged back in to the graph and checked to make sure the momentums are equal
All in all, these problems test out understanding of the situation as well as how to apply the formulas. When all the details are accounted for such as making the velocities are positive or negative, making sure the picture is drawn accurately, and making sure you understand what happened, it is quite easy to solve for the variables. JUST REMEMBER THE CONSERVATION OF MOMENTUM: the momentum before is always equal to the momentum after
IS MOMENTUM STILL CONSERVED AT AN ANGLE?
The final type of problem we did in this unit is figuring out if momentum is still conserved when objects collide at angles.
The problem below illustrates a ball flying at 20 m/s before colliding with another ball at rest and bouncing off at a 45 degree angle. Both balls are 1 kg. Our job was to compare the momentum of both balls before and after using the unstuck unstuck conservation of momentum formula. Ignore the velocities written on the x and y axes as how to get there will be explained shortly.
Before using the equation, we needed to solve for the missing velocity of one of the balls. There are several different ways that this can be done:
1. pythagorean theorem: a^s+b^2=c^2
2. sin cos tan- which is very familiar at his point
3. 45, 45, 90 triangle rules for the value of the sides
Which method is used depends on the information given and whichever is easiest
In this particular problem cosine was used because we had an angle and a hypotenuse. Using that, we solved for the velocity that the ball went after the collision and got 9.97 which is approximately 10.
It was quite easy to fill in values for the balls before the collision because the masses were both 1, and one ball was going 20 m/s and the other was at rest. After the collision, the value for the final velocity of one of the balls was missing and required us to solve. Using the diagram, we were solved for the final velocity of “ball A” to use in the equationThe goal is to see if momentum is conserved, so the 10 was only plugged into the final velocity for one of the ball which then leaves us to solve for the velocity of the other ball which is shown below. In this situation, the velocity of the second ball ended up being the same as the first ball, but this is a rare occasion and will not always happen.
