9 Chapter 9: Linear Momentum

Textbook Chapter 9: Linear Momentum and Collisions

Section 9.1: Momentum and Impulse

Textbook Section 9.1: Linear Momentum

Textbook Section 9.2: Impulse and Collisions

The of an object is calculated as its velocity times its mass, and given the symbol p. As mass is a scalar and velocity is a vector, momentum is also a vector quantity.

    \[\vec{p} = m \vec{v}\]

The concept of momentum comes from the force from Newton’s Second Law.

    \[\vec{F} = m \vec{a} = \frac{d}{dt} (m \vec{v}) = \frac{d}{dt}(\vec{p})\]

Momentum has units of kg m/s.

Impulse (\vec{J}, units of N s) is calculated as force times time, and is the same thing as a change in momentum. A change in momentum usually is caused by a change in velocity, but can be due to a change in mass, but either way, it’s an impulse. We will only consider cases where the force on the object is constant during the impulse.

    \[\vec{J} = \vec{F} \times \Delta t = \vec{p}_2 - \vec{p}_1\]

Section 9.2: One-Dimensional Momentum Conservation

Textbook Section 9.1: Linear Momentum

Textbook Section 9.3: Conservation of Linear Momentum

Textbook Section 9.4: Types of Collisions

Momentum is a conserved quantity in the absence of external forces (for example, when we’re not worrying about friction). Because friction is ubiquitous in the real world, we will be considering momentum conserved right before a collision and right after a collision. Momentum problems are collision problems. The longer we go back before or after the collision, the more likely it is that momentum is not conserved due to external friction. Internal forces are forces that the objects in question (colliding objects) exert on each other; external forces are forces exerted on the object by other objects not part of our system.

A collision is a short interaction between two objects. Momentum will be conserved in collisions (same momentum just before the collision as just after the collision) in the case that internal forces are much greater than external forces. Momentum can be transferred between objects in a collision, but will not be lost.

Because momentum is conserved in our collisions, we can write the following equations:

    \[p_{A,i} + p_{B,i} = p_{A,f} + p_{B,f}\]

    \[m_A v_{A,i} + m_B v_{B,i} = m_A v_{A,f} + m_B v_{B,f}\]

Here are the steps for solving a conservation of momentum problem:

Draw a picture for before the collision and after the collision.

Pick a coordinate system that simplifies the problem.

Break momentum into x and y coordinates if necessary (see Two Dimensional Momentum Conservation).

Apply the conservation of momentum equation(s) to solve for the requested variable.

There are two types of we will consider in this class. The first is an , where the objects bounce off each other (for example, colliding billiard balls or a soccer ball hitting a wall and rebounding). In an elastic collision, both momentum and kinetic energy are conserved (\vec{p}_i = \vec{p}_f, KE_i = KE_f).

The second kind of collision is an , where the objects stick together after colliding. An example of this might be a person running and jumping into a car, or two snowballs colliding in midair. Momentum is still conserved in inelastic collisions, but kinetic energy is {\bf not} conserved — some of the kinetic energy turns into heat and chemical bonds, so KE_f < KE_i.

Section 9.3: Two Dimensional Momentum Conservation

Textbook Section 9.5: Collisions in Multiple Dimensions

Momentum is a vector, so any momentum problem can be considered a two-dimensional problem, and must be treated that way if all the velocity vectors do not lie in a straight line.

We can now write down the equations for momentum conservation in two dimensions.

    \[m_{A} v_{A,x,i} + m_B v_{B,x,i} = m_A v_{A,x,f} + m_B v_{B,x,f}\]

    \[m_{A} v_{A,y,i} + m_B v_{B,y,i} = m_A v_{A,y,f} + m_B v_{B,y,f}\]

As per normal when working in two dimensions, we keep the x and y components separate, and they do not influence each other. To find the overall momentum (magnitude and direction), follow the same rules as vector magnitudes and directions from Chapter 2.

In Class Group Problem 9.1:

Calculate the momentum of a 3,500 kg car moving at 30 m/s.

How fast does a 0.45 kg baseball have to be moving to have the same momentum as the car?

In Class Group Problem 9.2:

You are pushing your friend on a sled (55 kg total). If you push with a constant force for 5 seconds and your friend goes from 1.5 m/s to 4.5 m/s, what impulse did you deliver?

In Class Group Problem 9.3:

A 3,500 kg car traveling west at 20 m/s runs into a 4,500 kg car traveling east at 15 m/s. If the cars stick together after the collision, what is the final velocity of the wreck?

Calculate the kinetic energy of the cars before and after the collision in the previous question. Was kinetic energy conserved? What kind of collision was this?

In Class Group Problem 9.4:

A 60 kg person is standing on an icy road with a \mu_K of 0.150. She shoots a bullet of mass 0.050 kg at 1200 m/s towards the north. How far will she slide across the ice before coming to a stop?

In Class Group Problem 9.5:

Alice throws a 150 g lump of clay at a 750 g box sitting at rest on a horizontal surface. After the clay hits the box and sticks to it, the clay and box slide across the table 1.9 m before coming to a stop. If the \mu_K = 0.520, how fast was the clay moving just before it hit the box?

In Class Group Problem 9.6:

Ball A of mass 2.0 kg, moving at 1.5 m/s, hits ball B of mass 3.5 kg, originally at rest. What are the final velocities of the balls assuming a completely elastic collision?

In Class Group Problem 9.7:

Car A of mass 1500 kg is initially moving at 32 m/s 40^{\circ} north of east. It collides and sticks to car B of mass 1800 kg, which was originally moving at 27 m/s 32^{\circ} south of west. What is the final velocity (magnitude and direction) of the wreck?

In Class Group Problem 9.8:

You have been hired as a technical consultant for an early-morning cartoon series for children to make sure that the science is correct. In the script, a wagon containing two boxes of gold (total mass of 150 kg) has been cut loose from the horses by an outlaw. The wagon starts from rest 50 meters up a hill with a 6^{\circ} slope. The outlaw plans to have the wagon roll down the hill and across the level ground and then crash into a canyon where his confederates wait. But in a tree 40 meters from the edge of the canyon wait the Lone Ranger (mass 80 kg) and Tonto (mass 70 kg). They drop vertically into the wagon as it passes beneath them. The script states that it takes the Lone Ranger and Tonto 5 seconds to grab the gold and jump out of the wagon, but is this enough time for them to escape the wagon before it goes over the cliff? You assume that the wagon rolls with negligible friction.

In Class Group Problem 9.9:

You are looking forward to the end of the semester to start your great new summer job: treasure-hunting in the Caribbean! Your employer has discovered the captain’s logbook of the 40,000 ton luxury liner, the Hedonist, which left Miami in 1925 and never returned. In addition to the log, there is a long list of jewelry and other valuables in the ship’s safe. The ship sank when it collided with a freighter and the wreckage was never found. The log tells that the Hedonist was going due south at a speed of 20 knots in the calm seas through a rare fog just before the collision. While in the fog, it was struck broadside by the 60,000 ton freighter, the Ironhorse, which was traveling west at 10 knots. The log tells the exact location (latitude and longitude) of the liner just before the collision. Of course, your employer is keeping that information secret for now. The log also notes that the freighter’s bow pierced the hull of the liner, so the two ships were stuck together when they sank. To prepare for your summer job, you are given the assignment of determining the search area by calculating the velocity (magnitude and direction) of the ships just after collision.

In Class Group Problem 9.10:

Your friend has just been in a traffic accident and is trying to negotiate with the insurance company of the other driver to pay for fixing her car. She believes that the other car was speeding and therefore the accident was the other driver’s fault. She knows that you have a knowledge of physics and hopes that you can prove her conjecture. She takes you out to the scene of the crash and describes what happened. She was traveling North when she entered the fateful intersection. There was no stop sign, so she looked in both directions and did not see another car approaching. It was a bright, sunny, clear day. When she reached the center of the intersection, her car was struck by the other car which was traveling East. The two cars remained joined together after the collision and skidded to a stop. The speed limit on both roads entering the intersection is 50 mph. From the skid marks still visible on the street, you determine that after the collision the cars skidded 63 feet at an angle of 50^{\circ} north of east before stopping. She has a copy of the police report which gives the make and year of each car. At the library you determine that the weight of her car was 2800 lbs. and that of the other car was 2000 lbs., where you included the driver’s weight in each case. The coefficient of kinetic friction for a rubber tire skidding on dry pavement is 0.82. It is not enough to prove that the other driver was speeding to convince the insurance company. She must also show that she was under the speed limit.

Practice Exam Question 9.1:

These questions are provided for practice purposes. There is no guarantee a problem similar to this one will be on your exam.

A wooden block of mass 5.50 kg rests on a at surface with \mu_K = 0.260. A bullet (mass 0.080 kg) is shot into the block, and exits the other side of the block having lost half its velocity in the process. If the block slides 2.40 m before coming to a stop after being hit by the bullet, how fast was the bullet traveling just before it hit the block?

Practice Exam Question 9.2:

These questions are provided for practice purposes. There is no guarantee a problem similar to this one will be on your exam.

You (67 kg) are sliding on your sled (5.0 kg) down a snowy hill on a 12^{\circ} incline with a coefficient of kinetic friction of \mu_K = 0.180. When you left the top of the hill (6.0 m above ground level) at rest, your cousin then pushed a 10.0 kg snowball down the hill after you so it would travel faster than you and hit you on your way down the hill. The snowball, moving at 18 m/s, hits you when you are halfway down the hill and sticks to you.

How fast are you moving halfway down the hill, just before you get hit with the snowball?

How fast are you moving just after colliding with the snowball?

How fast are you moving at the bottom of the hill?


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Introductory Physics Resources by Adria C Updike is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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