# This Week: Momentum Monday: What is momentum? p = mv Tuesday:

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This Week: Momentum Monday: What is momentum? p = mv Tuesday:
How does momentum change? (impulse)‏ I = mΔv = fΔt Wednesday: Quiz! Conservation of momentum pinitial=pfinal Thursday: Elastic and inelastic collisions Friday: Quiz! Momentum conservation in 2D Monday: Problem Solving Integrating our knowledge

Today: What is Momentum?

Momentum in the Vernacular
In everyday experience, momentum is the amount “unf” an object has So what factors affect the momentum of an object?

What affects Momentum? Which has more “unf”? A biker going at 20 mph
A car going at 20 mph

What affects Momentum? Which has more “unf”? A biker going at 20 mph
A car going at 20 mph A car will certainly hurt more, why? Because it is more massive (more mass)‏

What affects Momentum? Which has more “unf”? A car going at 10 mph

What affects Momentum? Which has more “unf”? A car going at 10 mph
The faster car will have more “unf”, why? Because faster things are harder to stop

Momentum Defined Momentum is the product of mass and velocity
This is normally written p = m x v Bolded letters denote vectors What are the units of momentum? p = m x v m: kg v: m/s p: kg • m/s :kilogram meters per second

p=mv What is the momentum of a bee that weighs 10 grams and flies at 2 m/s? How does that compare to a tortoise that weighs 1kg and moves at .05m/s?

p=mv What is the momentum of a bee that weighs 10 grams and flies at 2 m/s? 10g=.01kg p=mv=.01x2= .02kgm/s How does that compare to a tortoise that weighs 1kg and moves at .05m/s? p=mv=1x.05= .05kgm/s The tortoise has more momentum.

p=mv Which has more momentum, my car or me?

p=mv Which has more momentum, my car or me? vcar=0
vme= something like 1m/s pcar = mcar * vcar = mcar *0 = 0 pme = mme* vme = mme * something like 1m/s = more than 0

Homework: Review for quiz tomorrow This powerpoint is on the website
Visit the physics classroom for more practice on momentum

Today’s Schedule (Jan 3/4)‏
How does momentum change? (impulse)‏ I = mΔv = fΔt

How Does Momentum of an Object Change?
p=mv Consider Δp=Δmv What does this mean? Why is this not a change in momentum of the object?

How Does Momentum of an Object Change?
p=mv Consider Δp=mΔv What does this mean?

How Does Momentum of an Object Change?
p=mv Consider Δp=mΔv This means that velocity is changing. Unlike Δm, Δv does not imply that the object is falling apart or clumping together

Introducing Impulse Δp is know as impulse (I)‏
Think of impulse as a change from the default path. (Momentum would keep carrying me this direction but I changed it)‏ I had a sudden impulse to “” (I was suddenly did something that was not planned)‏

Δp = I = mΔv A bowling ball (5kg) moving at 5m/s bowls through a set of bowling pins. Right before the ball falls down the shoot it is going 3m/s What is the impulse the bowling pins provide to a bowling ball?

Δp = I = mΔv A bowling ball (5kg) moving at 5m/s bowls through a set of bowling pins. Right before the ball falls down the shoot it is going 3m/s What is the impulse the bowling pins provide to a bowling ball? I = mΔv I = m*(vfinal – vinitial)‏ I = 5*(3-5)=-10kgm/s

Other Ways to Find Impulse
I = mΔv Remember: a = Δv/Δt therefore: Δv = aΔt Substituting in we get: I = maΔt Remember: F=ma Substituting in we get: I=FΔt I = Δp = mΔv = FΔt

I = Δp = mΔv = FΔt In football, a field goal is kicked with the football initially at rest. The football (300g) is kicked at 25m/s.  What was the impulse? 300g=.3kg I = mΔv = m*(Vfinal-Vinitial) = .3*(25-0) = 7.5kgm/s The player's foot was in contact with the ball for .1 seconds.  What is the average force during the time of contact? I = FΔt 7.5 = F * .1 F= 75 N

I = Δp = mΔv = FΔt I want to open a cracked door by throwing a ball at it. If I have two balls of equal mass, one bouncy, and the other clay, which of the two should I throw to achieve my goal? The bouncy one The clay one It doesn’t matter, they are the same Z) I have no clue

Clay ball Bouncy ball

Homework: Period 3: Prepare for the quiz tomorrow
Review the physics classroom “The Impulse- Momentum Change Theorem” (link on website)‏

Today’s Schedule (Jan. 4)‏
Quiz on "What is momentum" Conservation of momentum pinitial=pfinal

Two-Minute Problem A loaded truck collides with a car causing a huge damage to the car. Which of the following is true about the collision? A. The force on the truck is greater than the force on the car B. The force on the car is greater than the force on the truck C. The force on the truck is the same in magnitude as the force on the car D. The car and truck accelerate in the same direction E. During the collision the truck has greater acceleration than the car Z. I have no idea

Conservation of Momentum
Recall Newton's Third law: Every action has an equal and opposite reaction: F1=-F2 When I push on the desk it pushes back on me with equal force in the opposite direction.

Steel ball demo I = Δp = mΔv = FΔt
Compare the force of Ball1 on Ball 2 to Ball 2 on Ball 1 Compare the time of contact for Ball 1 with Ball 2 to the time of contact for Ball 2 with Ball 1

Steel ball demo Compare the force of Ball1 on Ball 2 and
F12 = -F21 Compare the time of contact for Ball 1 with Ball 2 to the time of contact for Ball 2 with Ball 1 Δt12 = Δt21

Derivation of Conservation of Momentum
Recall from last class that I = Δp = mΔv = FΔt F1 = -F2 Δt1 = Δt2= Δt I1= F1 Δt I2 = F2 Δt = - F1 Δt I1=-I2 In any interaction, momentum gain of one object is equal to the loss of momentum from another

Quick check An astronaut (80kg) in space kicks off of his space shuttle at 1.5m/s. What is the impulse that he provides to the space shuttle? (Assume that away from the space shuttle is the positive direction)‏ What is the impulse that the space shuttle provides to him? What is the total change in momentum of the space shuttle and the astronaut together? 120 kgm/s 240 kgm/s 0 kgm/s -120 kgm/s Z) I have no idea

Total Momentum of a System is conserved
If there are no outside forces acting on an system, the momentum of that system remains constant; it is conserved This is the property momentum conservation. ptotal = Σp = p1 + p2 + p3 + … ptotalinitial = ptotalfinal

Momentum Conservation
ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal A 10 kg object moves at a constant velocity 2 m/s to the right and collides with a 4 kg object moving at a velocity 5 m/s to the left. Which of the following statements is correct? A. The total momentum before and after the collision is 20 kg·m/s B. The total momentum before and after the collision is 40 kg·m/s C. The total momentum before and after the collision is 10 kg·m/s D. The total momentum before and after the collision is 30 kg·m/s E. The total momentum before and after the collision is zero

Momentum Conservation
ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal A freight car A with a mass of 24,000 kg moves at a constant velocity of 8 m/s on a horizontal railroad track and collides with an empty stationary car B with a mass of 24,000 kg. After the collision the car A is moving at 3 m/s in the same direction. What is the velocity of car B after the collision? A. 1 m/s B. 3 m/s C. 5 m/s D. 7 m/s D. 11 m/s

Momentum Conservation
ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal The same situation as before: A freight car A with a mass of 24,000 kg moves at a constant velocity of 8 m/s on a horizontal railroad track and collides with an empty stationary car B with a mass of 24,000 kg. However, after the collision the car A is moving at 3 m/s in the opposite direction. What is the velocity of car B after the collision? A. 1 m/s B. 3 m/s C. 5 m/s D. 7 m/s D. 11 m/s

ptotailnitial = ptotalfinal
ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal Time to get tricky, you may need a calculator: A loaded freight car A with a mass of 24,000 kg moves at a constant velocity of 8 m/s on a horizontal railroad track and collides with an empty stationary car B with a mass of 8,000 kg. After the collision the cars stick to each other and moves like one object. What is the velocity of two cars after the collision? A. 2 m/s B. 4 m/s C. 6 m/s D. 8 m/s D. 12 m/s

Homework: Periods 2 and 3:
We do not have a quiz tomorrow, however I expect you to be able to do the practice quiz on the website. If you don’t feel comfortable with it take extra care reviewing the physics classroom Review the Physics Classroom “Momentum Conservation Principle” (link on website)‏

Today’s Schedule (Jan. 5)‏
Summary of Equations Schedule for the next week and a half Test on Jan. 13th (Friday next week)‏ Today’s material: Review momentum equation uses

Momentum in a Nutshell p = mv Δp = I = mΔv = FΔt
ptotal = Σp = p1 + p2 + p3 + … ptotailnitial = ptotalfinal

p=mv What is the momentum of a bee that weighs 10 grams and flies at 2 m/s? 10g=.01kg p=mv=.01x2= .02kgm/s How does that compare to a tortoise that weighs 1kg and moves at .05m/s? p=mv=1x.05= .05kgm/s The tortoise has more momentum.

I = Δp = mΔv = FΔt In football, a field goal is kicked with the football initially at rest. The football (300g) is kicked at 25m/s.  What was the impulse? 300g=.3kg I = mΔv = m*(Vfinal-Vinitial) = .3*(25-0) = 7.5kgm/s The player's foot was in contact with the ball for .1 seconds.  What is the average force during the time of contact? I = FΔt 7.5 = F * .1 F= 75 N

ptotal = Σp = p1 + p2 + p3 + … An astronaut (80kg) in space kicks off of his space shuttle at 1.5m/s. What is the impulse that he provides to the space shuttle? (Assume that away from the space shuttle is the positive direction)‏ What is the impulse that the space shuttle provides to him? What is the total change in momentum of the space shuttle and the astronaut together? 120 kgm/s 240 kgm/s 0 kgm/s -120 kgm/s Z) I have no idea

ptotailnitial = ptotalfinal
ptotal final = m₁f x v ₁f + m ₂f x v₂f ptotal initial = m₁i x v ₁i + m ₂i x v₂i ptotailnitial = ptotalfinal

No Official Homework We will be working on more difficult problems from here on out If you are not comfortable with the concepts covered today you will get lost and not benefit from our practice problems Review kinematics if the projectile motion was difficult

Egg Throw! (bring a jacket)‏

Other Similar Impulse Applications
Climbing ropes Airbags Circus Nets Diving Golf/Baseball: following through on your swings

Sample Test Question: Ball A of mass 0.10 kg is sliding at 1.4 m/s on the horizontal tabletop of negligible friction shown above. It makes a head-on collision with a target ball B of mass 0.50 kg at rest at the edge of the table. As a result of the collision, the Ball A rebounds, sliding backwards at 0.70 m/s immediately after the collision. (a) Calculate the speed of Ball B immediately after the collision. (b) Calculate the horizontal displacement d.

(a) Calculate the speed of the 0
(a) Calculate the speed of the 0.50 kg target ball immediately after the collision. We use conservation of momentum to solve this one. The tabletop is 1.20 m above a level, horizontal floor. The target ball is projected horizontally and initially strikes the floor at a horizontal displacement d from the point of collision.

(b) Calculate the horizontal displacement d.
This is a projectile motion problem. We know the ball’s horizontal velocity and the height of the table, so we can easily find the horizontal distance it travels as it falls. Time to fall: The ball has a horizontal velocity of 0.42 m/s (which we just figured out), so the distance d is simply:

2 Dimensional Momentum A B C D E F
Which of the following are possible final momenta for the case below? A B C D E F

Full 2D Problem on Board Ball 1 (1kg) travelling at 5m/s collides with Ball 2 (4kg) travelling at 3m/s After the collision Ball 1 is travelling 1m/s at 60 degrees above the x-axis. What is the velocity (direction and speed) of Ball 2?

Tricky Extension In another experiment on the same table, the target ball B is replaced by target ball C of mass 0.10 kg. The incident ball A again slides at 1.4 m/s, as shown above left, but this time makes a glancing collision with the target ball C that is at rest at the edge of the table. The target ball C strikes the floor at point P, which is at a horizontal displacement of m from the point of the collision, and at a horizontal angle of 30 from the + x-axis, as shown above right. (c) Calculate the speed v of the target ball C immediately after the collision. (d) Calculate the y-component of incident ball A's momentum immediately after the collision.

Tricky Extension (c) Calculate the speed v of the target ball C immediately after the collision. This sounds very hard, but it’s actually quite simple. We know the distance C traveled (1.5 m) and we know how long it is in the air before it hits – same as the previous problem time. Using the horizontal distance and the time to fall we can find the horizontal velocity, which is the velocity it began with, which is the velocity right after the collision. Find v:

Tricky Extension (d) Calculate the y-component of incident ball A's momentum immediately after the collision. We know that momentum is conserved in the x and y directions. So we can sum momentum in the y direction. We know that this momentum has to add up to be zero because ball A had no initial momentum in the y direction.

There will be a lab tomorrow. Decide on the details of your experiment: How much mass will you have on Cart B? What will the initial velocities be? Do you want both carts to be moving, or just have one hit the other? Tomorrow you will come in and perform your experiment, be ready! Review the calculations and how you will take the data. Do any one of the three worksheet problems posted on the website

Practice Test Questions:
Rocket Turbine Try out the problems in the worksheet section

Rockets! A rocket engine works by shooting particles out at incredibly high speeds.  How many grams of fuel would a large model rocket (5kg) need to eject to speed up from 0 to 50 m/s?  Exhaust is ejected at a speed of 1,333 m/s.  (Assume the mass of the rocket does not change significantly, the speed of the exhaust is constant and that gravity is negligable.)

Rockets! Can we use momentum conservation?
Yes, there are no outside forces on our system since we are ignoring gravity Lets setup our momentum conservation: Initial momentum rocket = pinitial = 0 Final momentum rocket = m rocket x v rocketfinal = 5 * 50 = 250 kgm/s Total final momentum = p rocket + p exhaust = 0

Rockets! Final momentum rocket = m rocket x v rocketfinal = 5 * 50 = 250 kgm/s Total final momentum = p rocket + p exhaust = 0 Because momentum is conserved, the exhaust must have a momentum equal and opposite to the rocket: p exhaust = -250 = mexhaust x vexhaust note that the v in this equation is negative because the exhaust is ejected downward.  This cancels the negative momentum resulting in a positive mass Vexhaust=1333m/s X = 250/1333 = kg = 188 grams

Turbine Problem A stream of water strikes a stationary turbine blade, as the drawing illustrates. The incident water stream has a velocity of m/s, while the exiting water stream has a velocity of m/s. The mass of water per second that strikes the blade is 25.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.

Turbine Problem Velocityi = +18.0 m/s, Velocityf = -18.0 m/s.
Water enters at 25.0 kg/s. Find average force on the water Which sort of problem is this? A Total momentum B Impulse C Conservation of momentum

Turbine Problem Velocityi = +18.0 m/s, Velocityf = -18.0 m/s.
Water enters at 25.0 kg/s. Find average force on the water B Impulse We are dealing with change in momentum and forces over a period of time Try solving now on your own.

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