# Momentum: Collisions

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You're in a car, driving to the gas station. Ahead of you, there is a large truck traveling southbound. Instantly, a small convertible appears to be traveling toward the truck, northbound. The truck driver, who doesn't see the convertible, swerves to the left, hitting the convertible head-on. After the two cars crash, they remain stuck together, and they both travel a few feet southbound, in the original direction of the truck. This act of sticking together after crashing is what we call **collisions**. *But how far to the two cars travel once they crash?* The speed and direction of their travel is determined by laws of **momentum**.

### General Facts About MomentumEdit

The equation for momentum is *p=mv*. It is typically measured in kg•m/s. Momentum refers to the motion of an object. An object in motion has momentum; it has a speed, and it has a mass. The mass and velocity are directly proportional to an object's momentum; if the velocity or the mass of an object increases, its mometum increases. Momentum is a vector quantity, which simply means that it has a direction. Velocity is very similar to speed, but with velocity, the speed must point in a specific direction. Due to this, momentum must has a direction. That direction is the same direction as the velocity of the object.

### Newton's Second Law and MomentumEdit

Newton's Second Law of Motion recognizes that an unbalanced force on an object causes a change in that object's momentum. Basically, the equation for Newton's Second Law can be re-written to identify a relationship containing momentum.

- F
_{net}=ma -> this is the orignal form of Newton's Second Law, which states that the total force on an object is equivalent to the mass of the object multiplied by its acceleration. - F
_{net}=(m∆v)/t -> here, acceleration has been rewritten in terms of velocity, since the acceleration of an object is equivalent to the change in its velocity divided by the time it takes to change velocity. - F
_{net}= ∆(mv)/t -> using mathematical laws of multiplication, the equation can be rewritten to allow for it to be translated to an equation containing momentum. - t•F
_{net}=∆(mv) -> by multiplying the total force by the time, it is easy to isolated ∆(mv), which is the change in momentum. - t•F
_{net}=∆p -> the final equation, which states that the change in momentum is equivalent to the total force on an object multiplied by the time in which the change takes place.

So what does this all mean? Now, it is a little more clear as to how momentum came about. Sir Issaac Newton, an extraordinary man of his time, created three key laws in physics, among other things. By exmaining the second law as performed above, we can see the relationship between momentum and another concept in physics-general motion!

### The Law of Conservation of MomentumEdit

As with all other conservation laws, the Law of Conservation of Momentum states that the total momentum in a system remains constant. This law is derived from Newton's First Law of Motion, which describes inertia: an object is in constant motion, or no motion at all. The equation for momentum conservation is as follows:

P_{initial}=P_{final}

It is important to keep this information in mind while continuing through this website. It's a very crucial piece of information!

### Things to RememberEdit

These fun little images can be found at www.batesville.k12.in.us pencil skirt

### General Sample ProblemsEdit

Based on the information above, let's see if you can answer some general questions on momentum:

a. A 1000-kg car moving 20 m/s northward

b. A 54-kg student moving 2 m/s westward

c. A car has 2000 kg•m/s of momentum. What would happen if the velocity were doubled? If the mass were tripled?

### Collisions and the Conservation of MomentumEdit

In all situations, momentum must be conserved. This is where the **Law of the Conservation of Momentum** comes in. This law states that *the total momentum in the initial situation is exactly equal to that of the final situation.* Simply speaking, when two objects collide, the "before" picture must have the same total momentum as the "final" picture does. We will see how this is preserved in all collisions ecommerce solution uk below.

#### Elastic CollisionsEdit

So what exactly are **elastic collisions**? As we learned before, a collision is simply when to objects hit one another, essentially colliding. An elastic collision, though, relates to the *final picutre* of a collision. What happens after two cars hit each other? In an elastic collision, the two cars, or two object, sort of "bounce off" of one another. It's almost as if someone placed a spring in between the two objects so that they would just bounce right back off of each other.

A perfect example of an elastic collision can be found in a game of pool. When you hit the white ball, it probably hits one of the striped or solid balls, if you have good aim. What happens once the two balls collide? The colored ball and the white ball, if you watch them carefully, do not travel together in the initial direction of the white ball. What tends to happen is that the white ball comes to a halt, following it's inital path, whereas the colored ball continues to travel in the direction of the white ball.

By using the Law of the Conservation of Momentum, we can examine a situation like this one in great detail. Given the image below, we can see that ball 1 and ball 2 have the same mass of 0.5 kg. Ball 1 is travelling at 1.0 m/s eastward, and ball 2 is travelling at 2.0m/s westward. As we determined before, the momentum of every object can be determined by multiplying the objects velocity and mass. The momentums of both balls in the collision are as follows:

- p=mv
- ball 1: p=(0.5 kg)(1.0 m/s East)
- ball 1: p= 0.5 kg•m/s East

- ball 2: p=(0.5 kg)(2.0 m/s West)
- ball 2: p=1.0 kg•m/s West

- ball 1: p=(0.5 kg)(1.0 m/s East)

Consequently, the cumulative momentum for the entire picture is as follows

- p
_{ball 1}+ p_{ball 2}= p_{total} - 0.5 kg•m/s + 1.0 kg•m/s= p
_{total} - 1.5 kg•m/s= p
_{total}

So what happens when they collide in an elastic collision?

Let's say that ball 1 bounces off of ball 2, and after the collision, it is travelling WESTWARD at 1.5 m/s. That means that the direction as well as the magnitude of the velocity have changed. But remember: the mass of ball 1 is the same. Ball 2 has the same mass as it had before, but now we want to figure out it's final velocity. So we'll use what we know to figure out what we don't know. We can use the mass and final velocity of ball 1 to determine it's momentum:

- p=mv
- p=(0.5 kg)(1.5 m/s West)
- p=0.75 kg•m/s West

According to the Law of Conservation of Momentum, the final momentum of the entire scene is equivalent to the initial momentum of the entire scene. If our inital cumulative momentum is 1.5 kg•m/s, that means that our final momentum has to equal 1.5 kg•m/s as well.

- p
_{initial}=p_{final} - 1.5 kg•m/s=p
_{final}

We know have to use our prior knowledge to figure out the rest of this problem. We'll do it step by step. We know that the final momentum is the sum of an addition problem:

- p
_{final}=p_{ball 1 final}+ p_{ball 2 final}

We know parts of this equation already! Let's fill in what we know:

- 1.5 kg•m/s= 0.75 kg•m/s + p
_{ball 2 final}

Doing an easy form of algebra, we can conclude that the momentum of ball 2 is 0.75 kg•m/s:

- 1.5 kg•m/s= 0.75 kg•m/s + p
_{ball 2 final} - 0.75 kg•m/s= p
_{ball 2 final}

We also know that momentum is equivalent to mass times velocity, so we can rewrite this equation with that information:

- 0.75 kg•m/s= mv

Because we know that the mass of ball 2 remains unchanged, we can plug in 0.5 kg:

- 0.75 kg•m/s= (0.5 kg)v

Again, algebra will help us to isolate the velocity, and conclude the magnitude of it:

- 0.75 kg•m/s= (0.5 kg)v
- 1.5 m/s=v

And there we go! We've found the final speed of ball 2, and in order to make it a velocity, we have to figure out the direction of the ball. Because we know that the direction of the ball reverses after the collision, and this ball was travelling west prior to colliding, we can conclude that the velocity of ball 2 in its final state is 1.5 m/s east.

#### Inelastic CollisionsEdit

What exactly is an **inelastic collision**? Inelastic collisions, similarly to elastic collisions, refer mainly to the final picture of a collision between two objects. The real difference between the two types of collisions is the direction of the objects. In an inelastic collision, the two objects stick together and continue to travel. For example:

A bike rider is cruising along a highway. He stops on the side of the highway and parks his bike on the shoulder. Little does he know that a sedan is coming right for his bike! The sedan crashes into the bike. Don't worry, our biker friend is fine... he was a few meters away from the incident. The sedan, going at a high speed, slows down when it crashes into the bike, and the bike appears to be glued to the front of the car. The two objects sort of become one, and continue to travel in the initial direction of the car. This sticky situation right here is an example of an inelastic collision: the two objects remain together in the final stage of the collision.

All right, let's try and tackle this inelastic collision. When the two objects, the car and the bike, crash, what will their velocity be? This type of question calls for the same methodology that we used for the sample problem about elastic collisions. However, we have to be careful when determining the overall mass this time! Let's start with the initial picture. We have a car, travelling 20. m/s eastward with a mass of 1500 kg. We also have a bicycle, massing 40 kg, at rest, meaning that it has a velocity of 0. m/s. Let's calculate the momentum for each object:

- p
_{car}=mv- p
_{car}=(1500 kg)(20. m/s east) - p
_{car}= 30000 kg•m/s east

- p
- p
_{bike}=mv- p
_{bike}=(40. kg)(0. m/s) - p
_{bike}= 0. kg•m/s

- p

Due to the individual momentums that we calculated, we can use the Law of Conservation of Momentum to determine the total initial momentum in our system:

- p
_{initial}=p_{car}+p_{bike}

- p
_{initial}= 30000 kg•m/s + 0. kg•m/s

- p

Before we continue, we have to remember the other ascpet of the Law of Conservation of Momentum to determine the final total momentum:

- p
_{initial}=p_{final}

- 30000 kg•m/s=p
_{final}

- 30000 kg•m/s=p

Now we know the total momentum in the final stage of the collision! We need to find a total momentum of 30000 kg•m/s. This will be easy-we just have to think for a minute. Now, in the final picture, because the two objects are stuck together, the total mass is the combined total of the bike and the car. Let's try and substitute in this value:

- p
_{final}=p_{car and bike}

- 30000 kg•m/s=(m
_{car}+ m_{bike})(v) - 30000 kg•m/s=(1500 kg + 40. kg)(v)
- 30000 kg•m/s=(1540 kg)(v)
- 19.5 m/s= v east

- 30000 kg•m/s=(m

Now we have answered our question: the total velocity of the final scene is 19.5 m/s east, in the initial direction of the car. That wasn't so bad! There's a trick to all of these collision problems: keep in mind that you have to use all of the information from various laws regarding momentum, and you've got it!

### Sample Regents ProblemsEdit

These questions are from Barron's Regents Review Book.

- A 0.50 kg object traveling at 2.0 m/s east collides with a 0.30 kg object traveling at 4.0 m/s west. After the collision, the 0.30 kg object is traveling at 2.0 m/s east. What are the magnitude and direction of the velocity of the first object?
- A 2.0 kg cart traveling east with a speed of 6 m/s collides with a 3.0 kg cart traveling west. If both carts come to a rest immediately after the collision, what was the speed of the westbound car before the collision?
- 6 m/s
- 2 m/s
- 3 m/s
- 4 m/s

- A 4.0 kg mass is moving at 3.0 m/s toward the right, and a 6.0 kg mass is moving at 2.0 m/s toward the left on a horizontal, frictionless table. If the two masses collide and remain together after the collision, their final momentum is
- 1.0 kg•m/s
- 24 kg•m/s
- 12 kg•m/s
- 0 kg•m/s

- When two objects collide, there will be no net change in the
- velocity of each object
- displacement of each object
- kinetic energy of each object
- total momentum of the objects

- A 1.0 kg object moving east with a velocity of 10 m/s collides with a 0.50 kg object that is at rest. Neglecting friction, what is the momentum of the system after the collision?
- 15 kg•m/s E
- 15 kg•m/s W
- 10 kg•m/s E
- kg•m/s W

- A 20. kg cart traveling east with a speed of 6.0 m/s collides witha 30. kg cart traveling west. If both carts come to rest after the collision, what was the speed of the westbound cart before the collision?
- 0 m/s
- 9.0 m/s
- 3.0 m/s
- 4.0 m/s

Solutions to NYS Regents Exam Problems

### Sample National Exam ProblemsEdit

To make things seem more nationally crucial, here are sample problems found on both the SAT II Physics exam and the AP Physics exams.

On the SAT II Physics, there are very few problems invloving collisions. However, below is a sample problem that can be solved using the knowledge of collision from Barron's SAT II Physics Review Book moving company.

1. Block *B* is at rest on a horizontal, frictionless surface. It has a mass of 20 kg. Block *A* slides toward it with a velocity of 6.0 m/s. It has a mass of 10 kg. After block *A* hits block *B* they remain together. What is the final velocity of the combined blocks?

Below are a few sample problems from the Princeton Review AP Physics B&C Book.

2. Two balls roll toward each other. The red ball has a mass of 0.5 kg and a speed of 4 m/s just before impact. The green ball has a mass of 0.3 kg and a speed of 2 m/s. After the head-on collision, the red ball continues forward with a speed of 2 m/s. Find the speed of the green ball after the collision.

3. Two balls roll toward each other. The red ball has a mass of 0.5 kg and a speed of 4 m/s just before impact. The green ball has a mass of 0.3 kg and a speed of 2 m/s. If the collision is completely inelastic, determine the velocity of the composite object after the collision.

Solutions to National Exam Problems

**References**Edit

The following websites are external links posted throughout this page:

- General Facts About Momentum
- Newton's Second Law of Motion
- Law of Conservation of Momentum
- Newton's First Law of Motion
- Things to Remember
- Collisions and the Conservation of Momentum

**Resources**Edit

For more comprehensive information on momentum and collisions, please feel free to browse the following websites:

- www.sparknotes.com This website describes collisions in one-dimension, and takes a slightly different approach to describing the concept of collisions in momentum.
- regentsprep.org This website has seven subtopics, all of which cover different aspects of momentum itself. One of the subtopics explains different types of collisions.
- www.glenbrook.k12.il.us This website, beyond the referenced link, gives a number of animations of collision forms.
- www.phys.hawaii.edu This site has an interactive applet that allows its user to chnage various conditions in a collision setting.
- www.colorado.edu If you would like to examine an experiment performed in a school setting, this website describes lab procedure and contains follow-up questions.
- lectureonline.cl.msu.edu This website offers a virtual air track, and you can plug in your own values to create elastic and inelastic collisions on the virtual, frictionless air track.
- Barron's Regents Review Book The Barron's Regents Review book is a perfect study aid for the Physical Setting/Physics New York State Regents Exam.
- Barron's SAT II Physics Review Book Another Barron's Review book, but this one is geard toward a National Exam: the SAT II. This book gives expert information about momentum in general, but it does not focus on collisions themselves.
- Princeton Review AP Physics B&C Book The Princeton Review AP book gives intensive information and step-by-step problems involving collisions and momentum in general. The book's format is also very easy to follow.