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AP Physics C Mechanics
Momentum
2015­12­04
www.njctl.org
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Table of Contents
Click on the topic to go to that section
• Conservation of Linear Momentum
• Impulse ­ Momentum Equation
• Collisions in One Dimension
• Collisions in Two Dimensions
• Center of Mass
• It is Rocket Science
• Ballistic Pendulum
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Conservation of Linear Momentum
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Conservation of Momentum
The most powerful concepts in science are called "conservation principles". Without worrying about the details of a process, conservation principles can be used to solve problems.
If we were to take a snapshot of the initial and final system, comparing the two would provide a lot of information. The last unit presented the Conservation of Energy, which proved helpful in solving problems where the situation was too complex for Newton's Laws to be effectively used. Conservation of Energy is not enough.
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Conservation of Momentum
If two ice skaters are holding hands, but then push away from each other, Conservation of Energy will not be able to determine each skater's velocity after the push. Assume a closed system with no external forces. We need something new. Let's start with Newton's Third Law. Assume the ice is frictionless. Then the force exerted by one skater on the second is equal and opposite to the force exerted by the second skater on the first and there are no external forces.
Stating Newton's Third Law and doing a little bit of substitution
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Conservation of Momentum
Make the assumption that we're not dealing with relativistic conditions (the skaters are moving way slower than the speed of light), and the skaters don't lose any mass in the process. Then, the constant mass terms can be put within the derivative.
This is good, as it looks like we can now say something about the skaters's velocities. The time rate of change of the sum of each skater's mass times velocity is zero ­ it doesn't change. The quantity is defined as linear momentum and will be represented as ­ it is a vector.
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Conservation of Momentum
The initial linear momentum is then equal to the final linear momentum of the system. Thus, linear momentum is conserved in a closed system with no external forces.
For two particles, , and since the initial and final momentum remain constant, we can write:
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Conservation of Momentum
The Energy chapter of this course discussed how potential energy could be calculated for a system that had only conservative forces. The Conservation of Momentum does not put this restriction on forces ­ as long as the forces are internal, momentum is conserved. We will leave the Conservation of Momentum for now, and apply it to problems involving collisions between objects in an upcoming chapter.
Many of the great discoveries in nuclear and particle physics involve smashing atoms and nuclei into each other and seeing what comes out. The Conservation of Momentum is key in these experiments.
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Newton's Second Law restated
Let's take a brief detour and examine Newton's Second Law as it is taught today, and use the definition of momentum.
The sum of forces on a particle is equal to ma. Since we assume the mass is constant, we can bring it inside the derivative.
This is how Newton presented his Second Law in Principia ­ the sum of forces on an object changes its momentum over time.
The more general statement of the Second Law has the benefit that it can be extended to cases where the mass of the objects change ­ not just the velocity. More on this later.
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Momentum is a Vector Quantity
A key difference between momentum and energy is that energy is a scalar, while momentum is a vector.
When there is more than one object in a system, the total momentum of the system is found by the vector addition of the each object's momentum.
Another key difference is that momentum comes in only one flavor (there is no kinetic, potential, elastic, etc.).
The unit for momentum is kg­m/s. There is no special unit for momentum, which presents an excellent opportunity to honor another physicist.
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1 What is the momentum of a 20 kg object with a velocity of +5.0 m/s?
A ­100 kg­m/s
C 0 kg­m/s
D 50 kg­m/s
Answer
B ­50 kg­m/s
E 100 kg­m/s
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2 What is the momentum of a 20 kg object with a velocity of ­5.0 m/s?
B ­50 kg­m/s
C 0 kg­m/s
Answer
A ­100 kg­m/s
D 50 kg­m/s
E 100 kg­m/s
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Momentum of a System of Objects
If a system contains more than one object, its total momentum is the vector sum of each object's individual momentum:
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Momentum of a System of Objects
In order to determine the total momentum of a system :
• Select a direction to be positive for each dimension that's being considered.
• Assign positive values to each momentum in that direction.
• Assign negative values to each momentum in the opposite direction.
• Add the momenta together independently for each dimension.
• Vectorially add each dimension's momentum together to get the total momentum ­ the Pythagoras theorem is used to find the magnitude, and trigonometry is used to find its direction.
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Momentum of a System of Objects
Let's work an example in one dimension. Determine the momentum of a system of two objects: m1, has a mass of 15 kg and a velocity of 16 m/s towards the east and m2,has a mass of 42 kg and a velocity of 6.0 m/s towards the west.
Choose East as positive.
or
to the west
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3 Determine the magnitude of the momentum of a system of two objects: m1, which has a mass of 6.0 kg and a velocity of 13 m/s north and m2, which has a mass of 14 kg and a velocity of 7.0 m/s south. Assume north is positive.
B 15 kg m/s
Answer
A 10 kg m/s
C 20 kg m/s
D ­15 kg m/s
E ­20 kg m/s
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4 Determine the momentum of a system of 3 objects: m1, which has a mass of 7.0 kg and a velocity of 23 m/s north, m2, which has a mass of 9.0 kg and a velocity of 7.0 m/s north and m3, which has a mass of 5.0 kg and a velocity of 42 m/s south. Assume north is positive.
B 12 kg m/s
Answer
A ­12 kg m/s
C ­14 kg m/s
D 14 kg m/s
E 15 m/s
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Impulse ­ Momentum Equation
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Impulse­Momentum Theorem
The Conservation of Linear Momentum applies to an isolated system of particles. The overall momentum is conserved, but what about the momentum of each particle?
Start with Newton's Second Law, as expressed in Principia, where we look at all the forces on one of the particles.
Assume the force acts over a time interval t0 to tf, and integrate this expression.
The particle's momentum will change. 20
Impulse­Momentum Theorem
We will examine the specific case where a single, constant, very large force acts on the particle for a very short time, so all other forces need not be considered. The equation simplifies to:
Define Impulse as:
The Impulse­Momentum equation is then:
Impulse is a vector, and it is in the same direction as the change of momentum or velocity of the particle acted on by the force.
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Impulse­Momentum Theorem
The force is not always constant ­ for example when a tennis racquet strikes a tennis ball, the force starts out small, and increases as the ball increases its contact time with the racquet, then decreases as it leaves. The large force at the peak results in a deformation of the ball.
F(N)
t (s)
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Impulse­Momentum Theorem
F(N)
F(N)
The shaded areas are equal in magnitude.
Favg t (s)
t (s)
The force ­ time graph can be used to find the Impulse delivered by the racquet in two ways:
• Find the area underneath the curve, either by integration if the force is specified as a function of time, or by numerical methods.
• Find the average force delivered by the racquet and multiply if by the time interval.
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Impulse­Momentum Theorem
F(N)
F(N)
Favg t (s)
t (s)
• Force known as a function of time: • Average force known:
But what if the force is not easily expressed in terms of time or you don't have a numerical integration capability?
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Impulse­Momentum Theorem
F(N)
F(N)
Favg t (s)
t (s)
The equation also works in reverse (of course). If you have the change in momentum of the object, and the time over which it occurs, the average force can be found.
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Implications of Impulse
Impulse tells us that we can get the same change in momentum with a large average force acting for a short time, or a small average force acting for a longer time.
For a given change of momentum (when a person stops moving because of an impact, like a car accident or falling), the force to the person can be minimized by extending the duration of the velocity reducing event. This is why one should bend their knees during a parachute landing, why airbags are used, and why landing on a pillow hurts less than landing on concrete.
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5 An external force of 25 N acts on a system for 10 s. What is the magnitude of the Impulse delivered to the system?
A 25 N­s
C 150 N­s
D 200 N­s
Answer
B 100 N­s
E 250 N­s
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6 An external force of 25 N acts on a system for 10 s. What is the change in momentum of the system?
A 25 N­s
C 150 N­s
D 200 N­s
Answer
B 100 N­s
E 250 N­s
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7 An average force of 5,000 N acts for 0.03 s on a 2.5 kg object that is initially at rest. What is its velocity after the application of the force?
B 70 m/s
C 60 m/s
Answer
A 80 m/s
D 50 m/s
E 40 m/s
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8 An object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force ­ time graph below to find the net impulse delivered by the force after 6 s.
B 14 N­s
F
(N)
6
4
Answer
A 12 N­s
C 16 N­s
2
D 18 N­s
E 20 N­s
0
2
4
6
t (s)
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9 A 2 kg object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force ­ time graph below to find the net impulse delivered by the force after 6 s.
B 6 N­s
C 8 N­s
F
(N)
6
4
Answer
A 4 N­s
2
D 10 N­s
0
E 12 N­s
2
4
6
t (s)
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10 A 2 kg object at rest experiences a net horizontal force in the +x direction and begins moving. Use the force ­ time graph below to find the object's velocity after 6 s.
B 8 m/s
C 6 m/s
F
(N)
6
4
Answer
A 10 m/s
2
D 5 m/s
0
E 4 m/s
2
4
6
t (s)
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Answer
11 A force described by F(t) = 190t ­ 189t2 is applied by a bat to a 0.145 kg ball. Assume the bat loses contact with the ball when the force decreases to zero. Over what time interval does this force act? The forc
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Answer
12 A force described by F(t) = 190t ­ 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the maximum force delivered to the ball? 34
Answer
13 A force described by F(t) = 190t ­ 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the impulse delivered to the ball? 35
Answer
14 A force described by F(t) = 190t ­ 189t2 is applied by a bat to a 0.145 kg ball. The force acts over a time interval of 1.01s. What is the magnitude of the average force delivered to the ball? 36
Answer
15 A force described by F(t) = 190t ­ 189t2 is applied by a bat to a 0.145 kg ball and delivers an impulse of 32 N/s. What is the velocity of the ball at t = 1.01 s, assuming it started from rest? 37
Collisions in One Dimension
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38
Types of Collisions Objects in an isolated system can interact with each other in two basic ways:
• They can collide.
• If they are stuck together, they can explode (push apart).
In an isolated system both momentum and total energy are conserved. But the energy can change from one form to another.
Conservation of momentum and change in kinetic energy can help predict what will happen in these events. 39
Types of Collisions
We differentiate collisions and explosions by the way the energy changes or does not change form. • inelastic collisions: two objects collide, converting some kinetic energy into other forms of energy such as potential energy, heat or sound. • elastic collisions: two objects collide and bounce off each other while conserving kinetic energy ­ energy is not transformed into any other type.
• explosions: an object or objects breaks apart because potential energy stored in one or more of the objects is transformed into kinetic energy.
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Inelastic Collisions
There are two types of Inelastic Collisions. • perfect inelastic collisions: two objects collide, stick together and move as one mass after the collision, transferring kinetic energy into other forms of energy. • general inelastic collisions: two objects collide and bounce off each other, transferring kinetic energy into other forms of energy.
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Elastic Collisions
There is really no such thing as a perfect elastic collision. During all collisions, some kinetic energy is always transformed into other forms of energy.
But some collisions transform so little energy away from kinetic energy that they can be dealt with as perfect elastic collisions. In chemistry, the collisions between molecules and atoms are modeled as perfect elastic collisions to derive the Ideal Gas Law.
Other examples include a steel ball bearing dropping on a steel plate, a rubber "superball" bouncing on the ground, and billiard balls bouncing off each other.
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Explosions
A firecracker is an example of an explosion. The chemical potential energy inside the firecracker is transformed into kinetic energy, light and sound.
A cart with a compressed spring is a good example. When the spring is against a wall, and it is released, the cart starts moving ­ converting elastic potential energy into kinetic energy and sound.
Think for a moment ­ can you see a resemblance between this phenomenon and either an elastic or inelastic collision?
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Explosions
In both an inelastic collision and an explosion, kinetic energy is transformed into other forms of energy ­ such as potential energy. But they are time reversed!
An inelastic collision transforms kinetic energy into other forms of energy, such as potential energy.
An explosion changes potential energy into kinetic energy.
Thus, the equations to predict their motion will be inverted.
The next slide summarizes the four types of collisions and
explosions.
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Collisions and Explosions
Momentum Conserved?
Kinetic Energy Conserved?
Yes
No. Kinetic energy is converted to other forms of energy
Objects stick together
Yes
No. Kinetic energy is converted to other forms of energy
Elastic Collision
Objects bounce off each other
Yes
Yes
Explosion
Object or objects break apart
Yes
No. Release of potential energy increases kinetic energy
Event
Description
General Inelastic Collision
Objects bounce off each other
Perfect Inelastic Collision
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16 Momentum is conserved in which of the following types of collisions?
A Perfect Inelastic
C Elastic
Answer
B Inelastic
D Explosions
E All of the Above
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17 Kinetic energy is conserved in which of the following types of collisions?
B Inelastic
C Elastic
Answer
A Perfect Inelastic
D Explosions
E All of the Above
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Conservation of Momentum
During a collision or an explosion, measurements show that the total momentum of a closed system does not change. The diagram below shows the objects approaching, colliding and then separating.
m Av A
A
mBvB
+x
B
the prime means "after"
A B
mBvB'
mAvA'
A
B
If the measurements don't show that the momentum is conserved, then this would not be a valid law. Fortunately they do, and it is!
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Perfect Inelastic Collisions
In a perfect inelastic collision, two objects collide and stick together, moving afterwards as one object.
Before (moving towards each other) After (moving together)
pA=mAvA
A
pB=mBvB
B
pA'+pB'=(mA+ mB)v'
AB
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Answer
18 A 13,500 kg railroad freight car travels on a level track at a speed of 4.5 m/s. It collides and couples with a 25,000 kg second car, initially at rest and with brakes released. No external force acts on the system. What is the speed of the two cars after colliding?
m1 = 13,50
m2 = 25,00
v1 = 4.5m/
v2 = 0 m/s
v1' = v2' = m1v1+m2v2 =
m1v1+0 = (m
v' = m1v1/(m
= (13,50
= 1.6 m/
in the same 50
Answer
19 A cannon ball with a mass of 100.0 kg flies in horizontal direction with a speed of 250 m/s and strikes a ship initially at rest. The mass of the ship is 15,000 kg. Find the speed of the ship after the ball becomes embedded in it.
m1 = 100kg
m2 = 15,00
v1 = 800m
v2 = 0 m/s
v1' = v2' = m1v1+m2v2 =
m1v1+0 = (m
v' = m1v1/(m1
= (100kg
= 1.7 m/
in the same 51
Answer
20 A 40 kg girl skates at 5.5 m/s on ice toward her 70 kg friend who is standing still, with open arms. As they collide and hold each other, what is their speed after the collision?
m1 = 40kg
m2 = 70kg
v1 = 5.5m/
v2 = 0 m/s
v1' = v2' = 52
Explosions
In an explosion, one object breaks apart into two or more pieces (or coupled objects break apart), moving afterwards as separate objects.
To make the problems solvable at this math level, we will assume:
• the object (or a coupled pair of objects) breaks into two pieces. • the explosion is along the same line as the initial velocity.
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Explosions
Here's a schematic of an explosion. Can you see how it's the time reversed picture of a perfectly inelastic collision? Before (moving together)
pA+pB=(mA+ mB)v
A B
After (moving apart)
pB'=mBvB'
pA'=mAvA'
A
B
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Answer
21 A 5 kg cannon ball is loaded into a 300 kg cannon. When the cannon is fired, it recoils at 5 m/s. What is the cannon ball's velocity after the explosion?
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Answer
22 Two railcars, one with a mass of 4000 kg and the other with a mass of 6000 kg, are at rest and stuck together. To separate them a small explosive is set off between them. The 4000 kg car is measured travelling at 6 m/s. How fast is the 6000 kg car going?
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Elastic Collisions
In an elastic collision, two objects collide and bounce off each other, as shown below, and both momentum and kinetic energy are conserved.
This will give us two simultaneous equations to solve to predict their motion after the collision.
Before (moving towards)
pA=mAvA
A
pB=mBvB
B
After (moving apart)
pA'=mAvA'
A
pB'=mBvB'
B
57
Elastic Collisions
Before (moving towards)
pA=mAvA
A
pB=mBvB
B
After (moving apart)
pA'=mAvA'
A
pB'=mBvB'
B
58
Elastic Collisions
Since both of these equations describe the motion of the same system, they must both be true ­ hence they can be solved as simultaneous equations. That will be done on the next slide, and an interesting result will appear. 59
Elastic Collision Simultaneous Equations Conservation of Momentum
Conservation of Kinetic Energy
m1v1 + m2v2 = m1v1' +m2v2'
½m1v12 + ½m2v22 = ½m1v1'2 +½m2v2'2
m1v1 ­ m1v1' = m2v2' ­ m2v2 m1v12 + m2v22 = m1v1'2 +m2v2'2
m1(v1 ­ v1') = m2(v2' ­ v2) m1v12 ­ m1v1'2 = m2v2'2 ­ m2v22 m1(v12 ­ v1'2) = m2(v2'2 ­ v22)
m1(v1 + v1')(v1 ­ v1') = m2(v2' + v2)(v2' ­ v2)
m1(v1 + v1')(v1 ­ v1') = m2(v2' + v2)(v2' ­ v2)
m1(v1 ­ v1') = m2(v2' ­ v2) v1 + v1' = v2' + v2
v1 ­ v2 = ­(v1' ­ v'2)
60
Properties of Elastic Collisions
By solving the conservation of momentum and constant kinetic energy equations simultaneously, the following result appeared:
v1 ­ v2 = ­(v1' ­ v'2)
Do you recognize the terms on the left and right of the above equation? And, what does it mean?
The terms are the relative velocities of the two objects before and after the collision. It means that for all elastic collisions ­ regardless of mass ­ the relative velocity of the objects is the same before and after the collision.
61
Answer
23 Two objects have an elastic collision. Before they collide, they are approaching with a velocity of 4 m/s relative to each other. With what velocity do they move apart from one another after the collision?
62
Answer
24 Two objects have an elastic collision. Object m1, has an initial velocity of +4.0 m/s and m2 has a velocity of ­3.0 m/s. After the collision, m1 has a velocity of 1.0 m/s. What is the velocity of m2?
63
Answer
25 Two objects have an elastic collision. Object m1, has an initial velocity of +6.0 m/s and m2 has a velocity of 2.0 m/s. After the collision, m1 has a velocity of 1.0 m/s. What is the velocity of m2?
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Properties of Elastic Collisions
Using the conservation of kinetic energy and the conservation of momentum equations, and solving them in a slightly different manner (the algebra is a little more intense), the following equations result: .
You should derive these two equations ­ .
.
the intense algebra is fun.
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Properties of Elastic Collisions
Using these equations, we will predict the motion of three specific cases, and relate them to observation:
• Two same mass particles collide.
.
• A very heavy particle collides with a light particle at rest.
• A light particle collides head on with a heavy particle at rest.
.
.
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Collision of same mass particles
For a collision of two particles of the same mass, m = m1 = m2. When this substitution is made into the two equations on the left (try it!), they simplify to:
.
v1' = v2 and v2' = v1
m
m
m
m
v'2 = v1
v'1 = v2
v1
v2
.
67
Collision of same mass particles
v1' = v2 and v2' = v1
.
The particles exchange velocities. A good example of this is playing billiards. The cue ball and the striped and solid balls are all the same mass. When the cue ball hits a solid ball, the cue ball stops, and the solid ball takes off with the cue ball's velocity. m
m
m
m
v'2 = v1
v'1 = v2
v1
v2
.
68
Collision of same mass particles
v1' = v2 and v2' = v1
.
The particles exchange velocities. Another good example is found in nuclear reactor engineering. When Uranium 235 fissions, it releases 3 high speed neutrons. These neutrons are moving too fast to cause another Uranium 235 nucleus to fission, so they have to be slowed down. What type of atom would be of most use here? m
m
m
m
v'2 = v1
v'1 = v2
v1
v2
.
69
Collision of same mass particles
v1' = v2 and v2' = v1
.
The particles exchange velocities. An atom with a similar mass to neutrons. A Hydrogen atom moving slower than the neutrons would be great ­ the neutron would exchange velocities, slowing down, while speeding up the hydrogen atom. Operational reactors use water for this purpose (heavier than hydrogen, but has many other benefits ­ nuclear engineering is not as simple as elastic collisions!).
m
m
m
m
v'2 = v1
v'1 = v2
v1
v2
.
70
Large mass striking lighter mass at rest
For this case, m1 >> m2 (>> is the symbol for much greater), which means that m1 ± m2 ≈ m1. The equations simplify to:
.
v1' = v1 and v2' = 2v1
m1
m1
m2
m2
v1
v2 = 0
v'1 ≈ v1
v'2 = 2v1
71
Large mass striking lighter mass at rest
v1' = v1 and v2' = 2v1
.
The large mass particle continues on its way with a slightly smaller velocity. The small particle moves at twice the initial speed of the large mass particle. Picture a bowled bowling ball striking a beach ball at rest.
m1
m1
m2
m2
.
v1
v2 = 0
.
v'1 ≈ v1
v'2 = 2v1
72
Small mass striking heavier mass at rest
For this case, m1 << m2 (<< is the symbol for much less than), which means that m1 ± m2 ≈ m2. The equations simplify to:
.
v1' = ­v1 and v2' ≈ 0
m2
m2
m1 v
1
m1
v'1
v2 = 0
v'2 = 0 73
Small mass striking heavier mass at rest
v1' = ­v1 and v2' ≈ 0
.
The large mass particle remains mostly at rest. The small particle reverses direction, with the same magnitude as its initial velocity. Picture a moving golf ball hitting a stationary bowling ball. m2
m2
m1 v
1
m1
v'1
v2 = 0
v'2 = 0 74
Answer
26 A bowling ball has a velocity of +v when it collides with a ping pong ball that is at rest. The velocity of the bowling ball is virtually unaffected by the collision. What will be the speed of the ping pong ball?
v1­v2 = 75
Answer
27 A baseball bat has a velocity of +v when it collides with a baseball that has a velocity of ­2v. The bat barely changes velocity during the collision. How fast is the baseball going after it's hit?
76
28 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +6.0 m/s and m 2 is ­3.0 m/s. What is the velocity of m1 after the collision?
Answer
v1 = +6m/s
v2 = ­3m/s
v1' = ?
v2' = ?
When iden
elastic col
v1' = v2 = v2' = v1 = So the vel
77
29 Two objects with identical masses have an elastic collision: the initial velocity of m1 is +6.0 m/s and m 2 is ­3.0 m/s. What is the velocity of m2 after the collision?
Answer
v1 = +6m/s
v2 = ­3m/s
v1' = ?
v2' = ?
When iden
elastic coll
v1' = v2 = ­
v2' = v1 = 6
So the velo
78
Answer
30 A golf ball is hit against a solid cement wall, and experiences an elastic collsion. The golf ball strikes the wall with a velocity of +35 m/s. What velocity does it rebound with?
79
Collisions in Two Dimensions
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80
Conservation of Momentum in Two Dimensions
Momentum vectors (like all vectors) can be expressed in terms of component vectors relative to a reference frame This, of course also applies to three dimensions, but we'll stick with two for this chapter!
This means that the momentum conservation equation p = p' can be solved independently for each component:
81
Example: Collision with a Wall
Consider the case of a golf ball colliding elastically with a hard wall, rebounding with the same velocity, where its angle of incidence equals its angle of reflection.
Is momentum conserved in this problem?
m
p'
py'
py
px' θ
px θ
p
m
The solid lines represent the momentum of the ball (blue ­ prior to collision, red ­ after the collision). The dashed lines are the x and y components of the momentum vectors.
82
Example: Collision with a Wall
Momentum is not conserved!
An external force from the wall is being applied to the ball in order to reverse its direction in the x axis.
However, since we have an elastic collision, the ball bounces off the wall with the same speed that it struck the wall. Hence, the magnitude of the initial momentum and the final momentum is equal: m
p'
py'
py
px' θ
px θ
p
m
Now it's time to resolve momentum into components along the x and y axis.
83
Example: Collision with a Wall
m
p'
py'
py
Apply the Impulse Momentum Theorem in two dimensions:
px' θ
px θ
p
m
What does this tell us?
84
Example: Collision with a Wall
m
p'
py'
py
Momentum is conserved in the y direction,
but it is not conserved in the x direction.
px' θ
px θ
p
m
The initial and final momentum in the y direction
are the same. The initial and final momentum
in the x direction are equal in magnitude, but
opposite in direction ­ they reverse.
85
31 A tennis ball of mass m strikes a wall at an angle θ relative to normal then bounces off with the same speed as it had initially. What was the change in momentum of the ball?
A ­mv
B ­2mv
C ­mv cosθ
D ­2mv cosθ
86
B ­mv
m
p'
C ­2mv
D ­mv cosθ E ­2mv cosθ py'
py'
px' θ
Answer
32 A tennis ball of mass m strikes a wall an an angle θ relative to normal and then bounces off with the same speed as it had initially. What is the change in momentum of the ball?
A 0
px' θ
p
m
87
33 A tennis ball of mass m strikes a wall an an angle θ relative to normal and then bounces off with the same speed as it had initially. What is the change in momentum of the ball in the y direction?
A 0
p'
py'
C ­2mv
D mv E 2mv py'
px' θ
Answer
B ­mv
m
px' θ
p
m
88
General Two Dimensional Collisions
m1
m2
We'll now consider the more general case of two objects moving in random directions in the x­y plane and colliding. Since there is no absolute reference frame, we'll line up the x­axis with the velocity of one of the objects.
89
General Two Dimensional Collisions
Before
m1
After
m2
p2 = 0
m1
m2
p2 = ?
This is not a head on collision ­ note how m1 heads off with a y component of velocity after it strikes m2. Also, did you see how we rotated the coordinate system so the x axis is horizontal?
To simplify the problem, we will specify that mass 2 is at rest.
The problem now is to find the momentum of m2 after the collision.
90
General Two Dimensional Collisions
Before
m1
After
m2
p2 = 0
m1
m2
p2 = ?
This will be done by looking at the vectors first ­ momentum must be conserved in both the x and the y directions. Since the momentum in the y direction is zero before the collision, it must be zero after the collision.
And, the value that m1 has for momentum in the x direction must be shared between both objects after the collision ­ and not equally ­ it will depend on the masses and the separation angle.
91
General Two Dimensional Collisions
Here is the momentum vector breakdown of mass 1 after the collision:
m1
m2 ?
m2 needs to have a component in the y direction to sum to zero with m1's final y momentum. And it needs a component in the x direction to add to m1's final x momentum to equal the initial x momentum of m1:
m2
and this is the final momentum for mass 2 by vectorially adding the final px and py.
92
34 After the collision shown below, which of the following is the most likely momentum vector for the blue ball?
A
B
before
m2
after
m2 ?
m1
Answer
m1
C
D
E
93
General Two Dimensional Collisions
Now that we've seen the vector analysis, let's run through the algebra to find the exact velocity (magnitude and direction) that m2 leaves with after the collision.
12.0 kg­m/s
after
before
m1 20.0 kg­m/s
60.0°
m1
m2
m2
θ
There is a bowling ball with momentum 20.0 kg­m/s that strikes a stationary bowling pin and then the bowling ball and pin take off as shown above. What is the final velocity of the pin?
94
General Two Dimensional Collisions
12.0 kg­m/s
after
before
m1 20.0 kg­m/s
60.0°
m1
m2
m2
θ
Given:
Find:
95
General Two Dimensional Collisions
12.0 kg­m/s
after
before
m1 20.0 kg­m/s
60.0°
m1
m2
m2
θ
Use Conservation of Momentum in the x and y directions.
x direction y­direction
96
General Two Dimensional Collisions
12.0 kg­m/s
after
before
m1 20.0 kg­m/s
60.0°
m1
m2
m2
θ
Now that the x and y components of the momentum of mass 2 have been found, the total final momentum is calculated.
97
35 A 5.0 kg bowling ball strikes a stationary bowling pin. After the collision, the ball and the pin move in directions as shown and the magnitude of the pin's momentum is 18.0 kg­m/s. What was the velocity of the ball before the collision?
?
ball
pin
53.1°
18 kg­m/s
30°
after
Answer
before
y­direction
p1y + p2y = p'1y +
0 + 0 = p'1y +18
0 = p'1y ­ 9
p'1y = 9 kg­m/s
98
Perfect Inelastic Collisions in Two Dimensions
One common kind of inelastic collision is where two cars collide and stick at an intersection.
In this situation the two objects are traveling along paths that are perpendicular just prior to the collision.
Before
m2
After
mm
1 2
p2
θ
p'
m1
p1
99
Perfect Inelastic Collisions in Two Dimensions
Before
m2
mm
1 2
After
p'x
p2
θ
p'
m1
p'y
p1
p­conservation in x: in y: final momentum: final velocity: final direction:
100
Answer
36 Object A with mass 20.0 kg travels to the east at 10.0 m/s and object B with mass 5.00 kg travels south at 20.0 m/s. They collide and stick together. What is the velocity (magnitude and direction) of the objects after the collision?
101
Center of Mass
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102
Center of Mass
All the problems we've solved have assumed that we're dealing with point particles ­ this is shown most clearly when free body diagrams were used to find the acceleration of objects due to multiple forces. The object was actually represented by a point!
Other than making the problem simpler (which is good), why did the computed answers actually match what happened to the larger, more extended object? Simple solutions are no good unless they represent reality.
To answer this question, the center of mass will be defined, and its relationship to the conservation of momentum will be shown.
103
Center of Mass
There is a collection of particles, with different masses, in space, each with an x and y component (x, y) specifying their location. Define the center of mass as having coordinates (xcm, ycm), where:
104
Center of Mass
A particle's position is described by a position vector, r:
Therefore rcm is expressed as:
This is a "weighted average" of each particle ­ more massive particles contribute more to the coordinates of the center of mass
What's next? Think Kinematics.
105
Center of Mass
Let's derive the velocity and the acceleration of the center of mass by taking the time derivatives position. First the velocity.
106
Center of Mass
Take the derivative of the position vector of the center of mass with respect to time.
where
Multiply through by M, and we get the total momentum of the system.
What is the significance of this last equation?
107
Center of Mass
It shows that the total momentum of a system of objects is equal to the total mass of the system times the velocity of its center of mass. The system acts the same as if all of its mass was concentrated at the center of mass.
Rewriting:
If the momentum of the system is constant (no net external forces acting), then no matter what the individual objects are doing, the velocity of their center of mass remains constant.
108
Center of Mass
One more step to finalize the description of the motion of the center of mass. Take the derivative of the momentum vector of the center of mass with respect to time.
Newton's Second Law
This shows that conservation of momentum applies equally to a solid object as well as a system of particles.
109
Center of Mass
In a system of objects, both internal and external forces are present. Internal, from the interaction of the objects, and external coming from outside the system, so we write:
Newton's Third Law ­ the internal forces are action reaction forces and their net effect on the system is zero, so ΣFint = 0.
This enables us to work with solid, macroscopic bodies ­ the body accelerates as if all of the external forces are acting on the center of mass of all its component parts.
110
Separating Masses
The below photograph shows 3 different fireworks rockets that were sent up into the air and then exploded. The center of mass of each rocket kept moving up, and the particles symmetrically spread out from each one. For each rocket, the velocity of the center of mass is equal to the weighted average of the velocities of each particle, resulting in the starburst pattern.
The acceleration of each particle (after the explosion) and the center of mass all equal ­g.
http://www.flickr.com/photos/stewart/126122/
111
Answer
37 A wire is bent into the below shape. What are the coordinates for the center of mass of the wire?
112
38 A missile is launched with velocity v0, and explodes mid flight into over 1000 fragments. What is the velocity of the center of mass of the system after the explosion?
B v0 C 2v0 Answer
A 0
D ­v0 E ­2v0 113
Explosions in Two Dimensions
The Black object explodes into 3 pieces (blue, red and green). We want to determine the momentum of the third piece. p'2
p'1
p'3
During an explosion, the total momentum is unchanged, since no EXTERNAL force acts on the system.
• By Newton's Third Law, the forces that occur between the particles within the object will add up to zero, so they don't affect the momentum.
• If the initial momentum is zero, the final momentum is zero. • The third piece must have equal and opposite momentum to the sum of the other two. Move the dashed box to see the third piece's momentum.
114
Explosions in Two Dimensions
before: px = py = 0
p'2
p'1
p'3
θ
The Black object explodes into 3 pieces (blue, red and green). We want to determine the momentum of the third piece. In this case the blue and red pieces are moving perpendicularly to each other, so: after: p'1x + p'2x + p'3x = 0
p'1y + p'2y + p'3y = 0
115
39 A stationary cannon ball explodes into three pieces. The momenta of two of the pieces is shown below. What is the direction of the momentum of the third piece?
A C Answer
B D E 116
Answer
40 A stationary 10.0 kg bomb explodes into three pieces. A 2.00 kg piece moves west at 200.0 m/s. Another piece with a mass of 3.00 kg moves north with a velocity of 100.0 m/s. What is the velocity (speed and direction) of the third piece?
117
It is Rocket Science
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118
It is Rocket Science
On July 20, 1969, the American Apollo 11 spacecraft landed on the moon, and six hours later, Neil Armstrong became the first human to walk on the moon.
While the spacecraft was enroute to the moon, the New York Times published the following article on the next slide.
119
It is Rocket Science
JULY 17, 1969: On Jan. 13, 1920, Topics of The Times, an editorial­
page feature of The New York Times, dismissed the notion that a rocket could function in a vacuum and commented on the ideas of Robert H. Goddard, the rocket pioneer, as follows: ''That Professor Goddard, with his 'chair' in Clark College and the countenancing of the Smithsonian Institution, does not know the relation of action to reaction, and of the need to have something better than a vacuum against which to react ­­ to say that would be absurd. Of course he only seems to lack the knowledge ladled out daily in high schools.''
Further investigation and experimentation have confirmed the findings of Isaac Newton in the 17th century and it is now definitely established that a rocket can function in a vacuum as well as in an atmosphere. The Times regrets the error.
http://www.nytimes.com/2001/11/14/news/150th­
anniversary­1851­2001­the­facts­that­got­away.html
120
It is Rocket Science
The 1920 article referenced Newton's Third Law, but how did they misinterpret it?
And how is Newton's Third Law related to the Conservation of Momentum?
121
It is Rocket Science
A rocket operates by expelling burned fuel through its tail. The fuel has both mass and velocity, so it has momentum. Before launch, the rocket was at rest, so its momentum was zero.
Define the system as the rocket and its fuel (the onboard and the expelled fuel), and since there are no external forces acting on it, momentum must be conserved.
The rocket is launched by burning the fuel and expelling it. It then has a momentum opposite the expelled fuel and moves forward to keep the total momentum of the system equal to zero.
122
It is Rocket Science
The action ­ reaction forces described by Newton's Third Law and the free body diagrams from his Second Law are:
• The rocket exerts a force on the burned fuel as it expels it out the tail.
• The burned fuel exerts a force on the rocket ship as it leaves.
Burned fuel
Rocket
And that's how Newton's Third Law relates to the Conservation of Momentum!
123
It is Rocket Science
This is a slightly more complex problem than the skaters pushing off of each other or an archer shooting an arrow. In those cases, the masses of each object remains the same (the arrow's mass is negligible compared to the archer) .
124
It is Rocket Science
For the more general case, assume the rocket is already moving at a constant veloicty and has a certain momentum. It then lights its engines and starts burning fuel. There is no external force so the momentum right before the burn must equal the momentum at any time.
The rocket and the burned fuel both increase momentum but the momentum at any time during the fuel burn is constant.
125
It is Rocket Science
Conservation of Momentum gives:
Initial velocity of rocket relative to earth
Mass of rocket
Change in velocity of rocket
Mass of fuel expelled in Δt
Velocity of burned fuel relative to rocket
(v ­ vef) is the velocity of the expelled burned fuel relative to the earth. We need to have the same reference frame for all
velocities.
126
It is Rocket Science
Some algebraic manipulation:
Take the limit of Δv and Δmef as Δt approaches zero:
As the mass of the expelled fuel increases, the mass of the rocket plus unburned fuel decreases by the same amount . 127
It is Rocket Science
Integrate from v 0 to vf and from M r0 (initial mass of rocket with fuel) to Mrf (final mass of rocket with remaining, unburned fuel). v ef is the velocity of the expelled fuel which is kept constant by the pilot:
Rocket equation
128
41 A missile is launched from rest. The initial mass of the missile and its fuel is 125 kg. The fuel's exhaust velocity is 2500 m/s. How much fuel is used to accelerate the rocket to a speed of 926 m/s?
B 20 kg
C 39 kg
Answer
A 7.8 kg
D 50 kg
E 86 kg
129
Ballistic Pendulum
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130
Ballistic Pendulum
Before they were rendered obsolete by modern sensors, Ballistic Pendulums were used to find the velocities of projectiles (bullets) .
But, they still are used in Physics Labs, where a combination of Conservation of Momentum, and Conservation of Total Mechanical Energy can be used to find the approximate velocity of a projectile.
l
l
m
h
vM = 0
v
M
M+m
v'
A more exact solution requires the use of rotational dynamics to account for the moment of inertia of the block.
131
Ballistic Pendulum
A bullet of mass m is fired into a block of mass M and remains embedded in the block. The block is attached to a stand with a fixed length, l, of string/wire and moves as shown below. What type of collision is this? What is conserved?
l
l
m
h
vM = 0
v
M
M+m
v'
132
Ballistic Pendulum
This is a perfectly inelastic collision, so linear momentum is conserved, but kinetic energy is not conserved ­ some of the bullet's energy goes into non conservative forces, such as friction and material deformation of the block and bullet.
After the collision, assume that there are zero non conservative external forces acting on the system. What conservation law can now be used?
l
l
m
h
vM = 0
v
M
M+m
v'
133
Ballistic Pendulum
Conservation of Total Mechanical Energy. After the collision, the bullet­block system rises a height h, above its initial position, and the string makes an angle θ with the vertical. That is shown in the picture on the right where one of the strings has been removed to see the angle θ more clearly.
θ
l
l
m
h
vM = 0
v
M
M+m
v'
134
Ballistic Pendulum
A little trigonometry to find the change in gravitational potential energy due to the block rising after the bullet's impact. l
lcosθ
θ
l ­ lcosθ
h
v'
M+m
135
Ballistic Pendulum
Using Conservation of Momentum at the impact point:
l
m
vM = 0
v
M
Right after the bullet hits, all the energy is kinetic. At the peak of the swing, all the energy is potential:
l
lcosθ
θ
l ­ lcosθ
h
v'
M+m
136
Ballistic Pendulum
Substitute the first into the second equation:
l
m
vM = 0
v
l
lcosθ
M
θ
l ­ lcosθ
h
v'
M+m
137
Ballistic Pendulum
By measuring the angle subtended by the block as it rises, and knowing the length of the string and the masses of the block and the bullet, the initial velocity of the bullet is determined.
l
m
vM = 0
v
l
lcosθ
M
θ
l ­ lcosθ
h
v'
M+m
138
42 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block and embeds within the block (completely inelastic collision). Find the velocity of the bullet ­ block system.
A 3.2 m/s
C 5.0 m/s
v
Answer
B 4.7 m/s
D 5.9 m/s
E 6.3 m/s
139
43 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block attached to a 2.0 m long string and embeds within the block (completely inelastic collision). Find the maximum height that the bullet ­ block system moves to.
A 1.0 m
Answer
Before
B 1.5 m
v
C 1.8 m
D 2.2 m
After
E 2.3 m
2 m
h
140
44 A 13 g bullet is fired at 460 m/s into a stationary 1.0 kg block attached to a 2.0 m long string and embeds within the block (completely inelastic collision). Find the angle θ that is subtended by the string at the maximum height of the bullet ­ block system.
Answer
A 570
B 620
C 65
D 77
0
0
2 m
h
E 840
141