Momentum - HHS Physics · Momentum Impulse.notebook 1 February 06, 2014 Momentum In classical mechanics, linear momentum or translational momentum (plural momenta; SI unit kg m/s,

Momentum Impulse.notebook

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February 06, 2014

Momentum

In classical mechanics, linear momentum or translational momentum (plural momenta; SI unit kg m/s, or, equivalently, N s) is the product of the mass and velocity of an object. For example, a heavy truck moving fast has a large momentum—it takes a large and prolonged force to get the truck up to this speed, and it takes a large and prolonged force to bring it to a stop afterwards. If the truck were lighter, or moving slower, then it would have less momentum.

Newton, first, expressed his Second Law in terms of momentum and not net force. He did not write the equation Fnet = mA, instead, he wrote that motion was proportional to the force impressed. What was meant by motion was the product of multiplying mass and velocity, and impressed force was an unbalanced net force. This quantity is known as linear momentum. Newton stated that a change in linear momentum was proportional to the impressed force. However, a small net force acting over a long period of time could produce the same change in linear momentum as a large force acting over a small time. This means that the relationship is time dependent. Which leads to the equation Fnet = ∆p / ∆t.


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February 06, 2014

Momentum refers to the quantity of motion that an object has. If an object is in motion ("on the move") then it has momentum. Momentum, therefore, can be thought of as "mass in motion." All objects have mass; so if an object is moving, then it has momentum it has its mass in motion. The amount of momentum which an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object.

Momentum = mass x velocity

In physics, the symbol for the quantity momentum is the lower case "p"; thus, the above equation can be rewritten as:

p = m v

where m = mass and v = velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity. The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg m/s.

Why is p used to represent momentum?

Before the word "momentum" was decided upon, they used "impetus". Impetus comes from the latin, "petere," to go towards or rush upon. Petere is where the p comes from.


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February 06, 2014

The Definition of Momentum is:

Momentum: (p) A property of a moving body that the body has by virtue of its mass and motion and that is equal to the product of the body's mass and velocity; broadly : a property of a moving body that determines the length of time required to bring it to rest when under the action of a constant force or moment. A vector quantity that is conserved in collisions between particles and in closed systems; in classical mechanics it is equal to the mass times the velocity of a body, or the vector sum of this product over all the components of a system.

Momentum is a vector quantity. To fully describe the momentum of a 5 kg bowling ball moving westward (180o) at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. The direction of the momentum vector is the same as the direction of the velocity of the ball. If the bowling ball is moving westward at 180o, then its momentum can be fully described by saying that it is 10 kg m/s at 180o. As a vector quantity, the momentum of an object is fully described by magnitude, direction, and unit.


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February 06, 2014

From the definition of momentum, it becomes obvious that an object has a large momentum if either its mass or its velocity is large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum. Yet, if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest; for the momentum of any object which is at rest is 0. Objects at rest do not have momentum they do not have any "mass in motion." Both variables mass and velocity are important in comparing the momentum of two objects.

1. Determine the momentum of the following:

A) 60kg halfback moving eastward at 9 m/s.

B) 1000kg car moving northward at 20 m/s.

C) 40kg student moving southward at 2 m/s.

A) 540 kgm/s eastward

B) 20000 kgm/s northward

C) 80 kgm/s southward


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February 06, 2014

Momentum can only be changed by changing the object's velocity or its mass. Gravity has nothing to do with it

Momentum can't be lost; it can only be transferred. If you catch a football, then the football's momentum goes through you and into the earth (or else you fall down). This will actually change the rotational speed of the earth by a very small amount, but the change is canceled out by the opposite force from when the ball was thrown. This all leads up to conservation of momentum. Remember it has to go somewhere! This means that if you have several objects in a system, perhaps interacting with each other, but not being influenced by forces from outside of the system, then the total momentum of the system does not change over time. However, the separate momenta of each object within the system may change. One object might change momentum, say losing some momentum, as another object changes momentum in an opposite manner, picking up the momentum that was lost by the first.

Law of conservation of linear momentum: The total momentum of the system prior to a collision is equal to the total momentum after the collision. The principle that a system under no external forces will maintain constant linear momentum.


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February 06, 2014

Momentum is how Isaac Newton wrote his laws. Consider a change in momentum with time:

dpdt

since momentum is equal to mV

dpdt =

d(mV)dt = m

dVdt

since is equal to acceleration a.dVdt

dpdt = ma = Fnet

Thus, the derivative of momentum with respect to time is NET FORCE.


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February 06, 2014

Impulse

Impulse ( j ) is defined as the product of the average value of a force and the time during which it acts: the change in momentum produced by the force. A vector quantity given by the integral over time of the force acting on a body, usually in a collision in which the time interval is very brief; it is equal to the change in the momentum of the body. Impulse is calculated by multiplying the force and the time interval over which the force acts.

Impulse ( j ) = F ∆t = ∫ F(t)dt

Impulse is a vector quantity whose direction is the direction of the force. Its SI unit is kg m/s or Ns.

impulse equals the change in momentum.

Impulse = change in momentum

F ∆t = j = ∆p = m ∆v

The phrase "impulse equals change in momentum" is a handy phrase worth memorizing.


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February 06, 2014

Suppose you apply a constant net force, Fnet, to an object of mass m. Newton's Second Law tells you that the object will accelerate, so if it starts with velocity vI, after some time t its velocity will be vF. This situation is diagrammed below.

Start at t=0

VI

Fnet

at time t

VF

Fnet

The acceleration of the object equals its change in velocity divided by the time it takes the velocity to change. In symbols:

A = Δv/Δt = Δv/t

Multiplying both sides of this equation by t gives:

At = Δv = vF vI

The right side of the equation above comes from the fact that the change in velocity equals the final velocity, vF, minus the starting velocity, vI. (Note: We could have just as well started with the kinematics equation vF = vI + at.) This is a valid kinematical statement about the motion.


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February 06, 2014

At = Δv = vF vI

To turn it into a dynamical statement about the motion, multiply both sides of the equation by the object's mass, m:

mAt = mΔv = m(vF vI)

Since Newton's Second Law tells us that the net force on an object equals the product of the object's mass and acceleration, we can replace ma with Fnet in this equation. On the right side, the quantity mass times velocity is called momentum, p.

Fnet t = mΔv = mvF mvI

Fnet t = pF pI = Δp

The quantity on the left, Fnet t, is the impulse exerted on the object by the net force. The quantity on the right of the equation is the object's final momentum minus its starting momentum, which is its change in momentum. Thus:

j = Δp

This is the ImpulseMomentum Equation


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February 06, 2014

Collisions

Collisions between objects are governed by laws of momentum and energy. When a collision occurs in an isolated system, the total momentum of the system of objects is conserved. Provided that there are no net external forces acting upon the objects, the momentum of all objects before the collision equals the momentum of all objects after the collision. If there are only two objects involved in the collision, then the momentum change of the individual objects are equal in magnitude and opposite in direction.

Most text books describe three different types of collisions (perfectly elastic, inelastic, and perfectly inelastic). In reality, it is more true to say there are two ends of a spectrum (range) of collision types. Regardless of what type of collision occurs, the total momentum of the system will always be conserved if there are no external forces present.


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February 06, 2014

Types of Collisions

CollisionType Properties Example

PerfectlyElastic

KE is conservedNo damage to either object • Idealized collisions

Inelastic

MoreElastic

LessElastic

Less damage and heatingLess sound

More damage and heatingMore sound

• Super ball• Golf ball

• Ball bearings

• Basket ball• Tennis Ball

• Under inflated ball

PerfectlyInelastic

• The colliding objects stick together and become one.

• Involves the greatest loss of Kinetic Energy. (keep in mind that if it loses KE then the energy will need to appear as thermal energy, and/or sound)

• The colliding objects usually suffer some kind of permanent damage

• Soft clay• Automobile

accidents• Train coupling

Remind yourself that for any of these collisions, the total momentum of the closed system remains the same and the total energy remains the same (it just may change form from KE to thermal energy, sound, etc.). Both the Momentum and the Energy are always conserved when dealing with a closed system.


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February 06, 2014

Perfectly Inelastic Collision

A perfectly inelastic collision occurs when the maximum amount of kinetic energy of a system is lost. In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system.

In a perfectly inelastic collision there are certain points to be kept in mind:

1. The coefficient of restitution is zero.

2. The colliding objects stick together after the collision.

3. The momentum is conserved.

4. Kinetic energy is not conserved.


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February 06, 2014

Perfectly Inelastic Collision

Next, we will derive an equation for a twobody(Body 1,Body 2) system in a one dimensional collision. In this example, momentum of the system is conserved because there is no friction between the sliding bodies and the surface. But the collision is perfectly inelastic because the two objects stick together.

Before Collision After Collision

frictionless surface

m1 M2

v2 = 0V1

frictionless surface

m1 M2

V1+2

Momentum Before Collisionp1 + p2

m1V1 + m2V2

Momentum after Collisionp'1+2

m1+2V'1+2

===

Note the prime ' denotes post collision conditionsOther notations for post collision velocity is to use the letter u


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February 06, 2014

A 15 g bullet is fired into an 10 kg block of wood and lodges in it. The block, which is free to move on a frictionless surface, has a velocity of 0.65 m/s at 0o after impact. Find the initial velocity of the bullet.

at 0o


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February 06, 2014

A 10,000 Kg railroad car moving at 8.5 m/s at 0o on a horizontal track, collides with a stationary railroad car. After the collision the two cars, which are attached to each other, move off with a velocity of 3.4 m/s at 0o. What is the mass of the second car?

M1 = 10,000 Kg

V1 = 8.5 m/s at 0oM2

V2 = 0M1 + M2

V'1+2 = 3.4 m/s at 0o


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February 06, 2014

Two blocks are moving towards each other on a frictionless surface. Block 1 has a mass of M and a velocity of 4 m/s right; and block 2 has a mass of 2M and a velocity of 1 m/s left. The 2 blocks collide in a perfectly inelastic collision, what is the velocity of the 2 stuck together blocks?

1 2

V1 = 4 m/s V2 = 1 m/s V'1+2 = ?

M 2M 3M

Before Collision After Collision

1 2


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February 06, 2014

Two Dimensional Perfectly Inelastic Collision Problems

Two 1000 kg cars approach an intersection at a 90o angle and collide inelastically, sticking together after the collision. What is the velocity (speed and direction) of the twocar clump of twisted metal immediately after the collision?

Velocity of car 1 = 20 m/s east. The

momentum of car 1 before the collision is 20,000 kg m/s East Velocity of car 2 = 10

m/s north.The momentum of car 2 before the collision is 10,000 kg m/s North

In the collision between the two cars, total system momentum is conserved. Yet this might not be apparent without an understanding of the vector nature of momentum.


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February 06, 2014

Momentum, like all vector quantities, has a magnitude (size), a direction, and unit. When considering the total momentum of the system before the collision, the individual momentum of the two cars must be added as vectors. That is, 20,000 kg m/s East must be added to 10,000 kg m/s North. The sum of these two vectors is not 30,000 kg m/s; this would only be the case if the two momentum vectors had the same direction. Instead, the sum of 20,000 kg m/s East and 10,000 kg m/s North is 22,360.7 kg m/s at an angle of 26.6o North of East.

Since the two momentum vectors are at right angles, their sum can be found using vector addition. Draw the 2 vectors so they have a common tail point.

p2 =10,000 kg m/s

p1 =20,000 kg m/s


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February 06, 2014

p2 =10,000 kg m/s

p1 =20,000 kg m/s

We know if using the parallelogram method of addition of vectors the resultant would be the diagonal of the parallelogram formed be the 2 vectors as seen below.

R

Using numerical vector addition find the X and Y components for p1 and p2.

p2 =10,000 kg m/s

p1 =20,000 kg m/s

X = 0Y = 10,000 kg m/s

X = 20,000 kg m/sY = 0


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February 06, 2014

p2 =10,000 kg m/s

p1 =20,000 kg m/s

X = 0Y = 10,000 kg m/s

X = 20,000 kg m/sY = 0

Add all the X components to find Rx

Rx = 20,000 kg m/s

Add all the Y components to find Ry

Ry = 10,000 kg m/s

Find the magnitude of R by using the Pythagorean theorem.

R = √[(Rx)2 + (Ry)2] = 22,360.7 kg m/s

Find θ using the arc tan function

θ = / Tan1(Ry/Rx)/ = 26.6o

Find φ using the proper φ formula. φ = 26.6o


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February 06, 2014

The value 22,360.7 kg m/s at 26.6o is the total momentum of the system before the collision; and since momentum is conserved, it is also the total momentum of the system after the collision. Since the cars have equal mass (and ONLY because they have equal mass), the total system momentum is shared equally by each individual car. In order to determine the momentum of either individual car, this total system momentum must be divided by two (approx. 11,200 kg m/s). Once the momentum of the individual cars are known, the aftercollision velocity is determined by simply dividing momentum by mass (v' = p/m)

p'1+2 = m1+2 V'1+2

V'1+2 = p'1+2 / m1+2

V'1+2 = (22,360.7 kg m/s at 26.6o) / 2000kg

V'1+2 = 11.2 m/s at 26.6o


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February 06, 2014

Two cars approach an intersection at a 90o angle and collide inelastically, sticking together after the collision. What is the velocity (speed and direction) of the twocar clump of twisted metal immediately after the collision?

M1 = 1200 kgV1 = 15 m/s east

M2 = 1400 kgV2 = 25 m/s north


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February 06, 2014

Two cars approach an intersection at a 90o angle and collide inelastically, sticking together after the collision. What is the velocity of car 1, if the wreckage has a velocity of 18.72 m/s at 29.1o immediately after the collision?

M1 = 1200 kgV1 = ? east

M2 = 1000 kgV2 = 20 m/s north

V'1+2 = 18.72 m/s at 29.1o


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February 06, 2014

A 7,000 Kg truck, traveling with a velocity 5 m/s at 0o, collides with a 1,200 Kg car moving 18 m/s at 220o as in the figure below. After the collision, the two vehicles remain tangled together. With what velocity will the wreckage begin to move immediately after the crash?

CarTruck

220o


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February 06, 2014

Momentum Conservation in Explosions

As discussed in a previously, total system momentum is conserved for collisions between objects in an isolated system. For collisions occurring in isolated systems, there are no exceptions to this law. This same principle of momentum conservation can be applied to explosions. In an explosion, an internal impulse acts in order to propel the parts of a system (often a single object) into a variety of directions. After the explosion, the individual parts of the system (that is often a collection of fragments from the original object) have momentum. If the vector sum of all individual parts of the system could be added together to determine the total momentum after the explosion, then it should be the same as the total momentum before the explosion. Just like in collisions, total system momentum is conserved.


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February 06, 2014

A 32 kg bomb which is initially at rest explodes into 2 pieces as shown below. Piece #2 has a mass of 12 kg and piece #1 has a speed of 57.8 m/s at 0o. What is the speed of piece #2?


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February 06, 2014

Two girls (masses m1 and m2) are on skates and at rest. The girls are close to each other, and are facing each other. Girl 1 pushes against girl 2 causing girl 2 to move backwards.

A) Assuming the girls move freely on skates (no friction), find a formula to calculate the speed of girl 1.

B) Calculate the speed of girl 1, If the mass of girl 1 is 54 kg, the mass of girl 2 is 42 kg, and girl 2’s speed is 2.4 m/s.

C) Calculate the impulse felt by girl 1 and girl 2. (Use the values in B)

D) Calculate the force felt by both girl 1 and girl 2, if the duration of the push was 0.8 s.


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February 06, 2014

A 20.0 kg cannonball is fired from a 2.40 × 103 kg cannon. If the cannon recoils with a velocity of 3.5 m/s backwards,what is the velocity of the cannonball?


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February 06, 2014

A coal barge with a mass of 1.36 × 104 kg drifts along a river. When it passes under a coal hopper, it is loaded with 8.4 × 103 kg of coal. What is the speed of the unloaded barge if the barge after loading has a speed of 1.3 m/s?

VV' = 1.3 m/s


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February 06, 2014

A child jumps from a moving sled with a speed of 2.2 m/s and in the direction opposite the sled’s motion. The sled continues to move in the forward direction, but with a new speed of 5.5 m/s. If the child has a mass of 38 kg and the sled has a mass 68 kg, what is the initial velocity of the sled?

Vs+c = ? V'sled = 5.5 m/s at 0o

Vc = 2.2 m/s at 180o


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February 06, 2014

An 80 kg stunt man jumps out of a window and falls 45 m to an airbag on the ground.

A) How fast is he falling when he reaches the airbag? (Initial velocity is zero)

B) He lands on a large airfilled target, coming to rest in 1.5 s. What average force does he feel while coming to rest?

C) What average force does he feel if he had fallen 45 m and landed on the ground (impact time = 10 ms = .001 s))?

Air bag

45 m


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February 06, 2014


A) How fast is he falling when he reaches the airbag? (Initial velocity is zero, ignore air resistance)


C) What average force does he feel if he had fallen 35 m and landed on the ground (impact time = 10 ms = 0.001 s))?

Air bag

35 m

(SHOW ALL WORK)

Round ALL answers to the TENTHS place.


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February 06, 2014

A railroad car of mass 2.50 x 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.00 m/s.

(a) What is the speed of the four cars after the collision?

(b) How much energy is lost in the collision?


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February 06, 2014

A railroad car of mass 30000 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the oppsite direction with an initial speed of 2.00 m/s.


(b) How much energy is lost in the collision?


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February 06, 2014

Perfectly Elastic Collision

A perfectly elastic collision is defined as one in which there is no loss of kinetic energy in the collision. An inelastic collision is one in which part of the kinetic energy is changed to some other form of energy in the collision. Any macroscopic collision between objects will convert some of the kinetic energy into internal energy and other forms of energy, so no large scale impacts are perfectly elastic. Momentum is conserved in inelastic collisions, but one cannot track the kinetic energy through the collision since some of it is converted to other forms of energy. Collisions in ideal gases approach perfectly elastic collisions, as do scattering interactions of subatomic particles which are deflected by the electromagnetic force. Some largescale interactions like the slingshot type gravitational interactions between satellites and planets are perfectly elastic.


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February 06, 2014

Collisions between hard spheres may be nearly elastic, so it is useful to calculate the limiting case of an elastic collision. The assumption of conservation of momentum as well as the conservation of kinetic energy makes possible the calculation of the final velocities in twobody collisions.

Momentum is Conserved

p1 + p2 = p'1 + p'2

m1V1 + m2V2 = m1V'1 + m2V'2


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February 06, 2014

Kinetic Energy is Conserved

KE1 + KE2 = KE'1 + KE'2

1/2m1V12 + 1/2m2V22 = 1/2m1V'12 + 1/2m2V'22

Multiplying all terms by 2

m1V12 + m2V22 = m1V'12 + m2V'22


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February 06, 2014

Consider two particles, m1 and m2, with velocities v1 and v2 respectively. They hit in a perfectly elastic collision at an angle, and both particles travel off at an angle to their original displacement, as shown below:

Y Axis

X Axis

M1

V1 at θ1

M2

V2 = 0 θ'1θ'2

V'1

V'2 = 0


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February 06, 2014

To solve this problem we again use our conservation laws to come up with equations that we hope to be able to solve. In terms of kinetic energy, since energy is a scalar quantity, we need not take direction into account, and may simply state:

1/2m1V12 + 1/2m2V22 = 1/2m1V'12 + 1/2m2V'22

m1V12 + m2V22 = m1V'12 + m2V'22

Whereas in the one dimensional problem we could only generate one equation for the conservation of linear momentum, in two dimensional problems we can generate two equations: one for the xcomponent and one for the ycomponent.


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February 06, 2014

X Direction

Let's start with the X component. Our initial momentum in the X direction is given by: m1 V1x + m2 V2x . (Note V1x is the X component of V1, and V2x is the X component of V2). After the collision, each particle maintains a component of their velocity in the X direction, which can be calculated using trigonometry. Our final momentum in the X direction is given by: m1 V'1x + m2 V'2x (Note V'1x is the X component of V'1, and V'2x is the X component of V'2). Thus our equation for the conservation of linear momentum in the X direction is:

m1V1x + m2V2x = m1V'1x + m2V'2x

m1V1cosθ1 + m2V2cosθ2 = m1V'1cosθ'1 + m2V'2cosθ'2


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February 06, 2014

Y Direction

Let's now look at the Y component. Our initial momentum in the Y direction is given by: m1 V1y + m2 V2y . (Note V1y is the Y component of V1, and V2y is the Y component of V2). After the collision, each particle maintains a component of their velocity in the Y direction, which can be calculated using trigonometry. Our final momentum in the Y direction is given by: m1 V'1y + m2 V'2y (Note V'1y is the Y component of V'1, and V'2y is the Y component of V'2). Thus our equation for the conservation of linear momentum in the Y direction is:

m1V1y + m2V2y = m1V'1y + m2V'2y

m1V1sinθ1 + m2V2sinθ2 = m1V'1sinθ'1 + m2V'2sinθ'2


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February 06, 2014

We now have three equations: conservation of kinetic energy, and conservation of momentum in both the X and Y directions. With this information, is this problem solvable? Recall that if we are given only the initial masses and velocities we are working with four unknowns: V'1, V'2, θ'1 and θ'2. We cannot solve for four unknowns with three equations, and must specify an additional variable. Perhaps we are trying to make a pool shot, and can tell the angle of the ball being hit by where the hole is, but would like to know where the cue ball will end up. This equation would be solvable, since with the angle the ball will take to hit the pocket we have specified another variable.

m1V12 + m2V22 = m1V'12 + m2V'22

m1V1x + m2V2x = m1V'1x + m2V'2x

m1V1y + m2V2y = m1V'1y + m2V'2y

V1x = V1 cos θ1 V1y = V1 sin θ1 V'1x = V'1 cos θ'1 V'1y = V'1 sin θ'1V2x = V2 cos θ2 V2y = V2 sin θ2 V'2x = V'2 cos θ'2 V'2y = V'2 sin θ'2


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February 06, 2014

Consider two billiard balls, M1 and M2, with velocities V1 and V2 respectively. They hit in a perfectly elastic off center collision, and both balls travel off at an angle to their original displacement. Determine V'1, V'2, and θ'1.

Y Axis

X Axis

M1 = 1.5 kg

V1 = 2 m/s at 0o

M2 = .5 kg

V2 = 0 θ'1 θ'2 = 40o

V'1

V'2


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February 06, 2014

The solution to this problem is extremely difficult involving trigonometry and extensive algebra. The resulting equations for its solution are below courtesy of wikipedia.

In a center of momentum frame at any time the velocities of the two bodies are in opposite directions, with magnitudes inversely proportional to the masses. In an elastic collision these magnitudes do not change. The directions may change depending on the shapes of the bodies and the point of impact. For example, in the case of spheres the angle depends on the distance between the (parallel) paths of the centers of the two bodies. Any nonzero change of direction is possible: if this distance is zero the velocities are reversed in the collision; if it is close to the sum of the radii of the spheres the two bodies are only slightly deflected.

Assuming that the second particle is at rest before the collision, the angles of deflection of the two particles, θ1 and θ2, are related to the angle of deflection θ in the system of the center of mass by:

The velocities of the particles after the collision are:


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February 06, 2014

Consider two billiard balls, denoted by subscripts 1 and 2. Let m1 and m2 be the masses, V1 and V2 the velocities before collision, and V'1 and V'2 the velocities after collision. If the collision is perfectly elastic, determine the final velocities of the two balls.

1 12 2

m1 = 2 kg

V1 = 2 m/s

m2 = 4 kg

V2 = 0

m1 = 2 kg

V'1 = ?

m2 = 4 kg

V'2 = ?

+


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February 06, 2014

Consider two billiard balls, M1 and M2, with velocities V1 and V2 respectively. They hit in an elastic off center collision, and both balls travel off at an angle to their original displacement. Determine V'1, V'2, and θ'1.

Y Axis

X Axis

M1 = 1.5 kg

V1 = 2 m/s at 0o

M2 = .5 kg

V2 = 0 θ'1 = 15.6o

θ'2 = 40o

V'1

V'2 = 1.96 m/s


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February 06, 2014

Two billiard balls collide as shown below: What is the velocity of ball #2 after the collision?

m1 = 0.17 kgV1 = 2.5 m/s at 0o

m3 = 0.16 kgV3 = 1.0 m/s at 180o

30o

m3 = 0.16 kgV'3 = 0.62 m/s at 210o

m1 = 0.17 kgV'1 = ?

Elastic Collision


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February 06, 2014

Ballistic Pendulum

In a ballistic pendulum an object of mass m is fired with an initial speed v0 at a pendulum bob. The bob has a mass M, which is suspended by a rod of length L and negligible mass. After the collision, the pendulum and object stick together and swing to a maximum angular displacement θ as shown.

The first part of the problemdeals with the conservation of

momentum.

The second part of the problemdeals with the conservation of

mechanical energy.


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February 06, 2014

A 0.05 kg bullet with velocity 150 m/s is shot into a 3 kg ballistic pendulum. Find how high the pendulum rises after the bullet gets stuck inside

V1 = 150 m/sm1 = 0.05 kg

V2 = 0M2 = 3 kg

V'1+2 = ?m1 + M2 = 3.05 kg

Start the problem by finding V'1+2 using the conservation of momentum.

p1 + p2 = p'1+2

Since the pendulum is initially at rest p2 equals zero. Also, V1+2 acts in the positive X direction so the problem reduces to a one dimensional perfectly inelastic collision problem.


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February 06, 2014

p1 + p2 = p'1+2

p1 = p'1+2

m1V1 = m1+2V'1+2

(0.05 kg)(150 m/s) = (3.05 kg)V'1+2

7.5 kgm/s = (3.05 kg)V'1+2

2.46 m/s = V'1+2

0

Next we apply our knowledge of energy. The pendulum is frictionless, therefore, this is a conservative system.

Solving for V'1+2


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February 06, 2014

At B all the mechanical energy of the pendulum is in KE since hB = 0. The velocity at b is the velocity we calculated earlier V1+2= VB.

hB = 0

BhC = ?

VC = 0

MEB = KEB = 1/2 M1+2VB2

At C all the mechanical energy of the pedulum is in GPE since VC = 0.

C

MEC = GPEC = M1+2ghC

MEB = MEC

1/2 M1+2VB2 = M1+2ghC1/2 VB2 = ghC

= hCVB2

2gVB = √(2ghC)


52

February 06, 2014

= hCVB2

2g

= hC(2.46 m/s)2

2(10 m/s2)

0.303 m = hC

Solving for hC:


53

February 06, 2014

A 0.06 kg bullet with velocity 250 m/s is shot into a 3.5 kg ballistic pendulum. Find how high the pendulum rises after the bullet gets stuck inside

V1 = 250 m/sm1 = 0.06 kg

V2 = 0M2 = 3.5 kg V'1+2 = ?

m1 + M2 = ?

Solve the following:


54

February 06, 2014

A pendulum consisting of a 0.5 kg ball is released from the position as shown below and strikes a block of mass 5 kg. The block slides a distance S before stopping due to a frictional force (The coefficient of friction is 0.2). Find the distance S if the ball rebounds to an angle of 15 degrees.

L = 1 m

37o

S


55

February 06, 2014

Solution Strategies1) Break down the problem to smaller parts

1. Pendulum2. Collision3. Sliding

2) Ask what the key concept of each part might be1. Pendulum Oscillations, Conservation of Mechanical Energy, Simple

Harmonic Motion.2. Collision Momentum and Impulse3. Sliding Kinematics (the Big 3), Dynamics, Nonconservative Energy,

Work.3) Isolate each part and determine the given values. Organize these values with a picture. Figure out what the unknowns of interest are.4) Determine a conceptual path to the solution, and devise a solution method.5) Solve.


56

February 06, 2014

PendulumThe first part of the problem deals with the swinging of a pendulum. It strikes an object and rebounds to a different height. The collision between the pendulum bob and the block is obviously a momentum problem so we must know the velocity of the bob just before and just after the collision. We can do this by using the concept of conservation of mechanical energy.

Before Collision

L = 1 m

37o

A

B

MEA = MEB

MEA = KEA + GPEA + EPEA

MEA = mghA

MEB = KEB + GPEB + EPEB

MEB = 1/2 mVB2

mghA = 1/2 mVB2

VB = √2ghA

VA = 0hA = ? VB = ?

hB = 0

0 0

hA

H

00

We are missing hA, but this can be found using trigonometry.


57

February 06, 2014

Finding hA

L = 1 m

37o

A

B

hA

H L θH H = L cosθ

hA = L H

hA = L L cosθ

hA = L(1 cosθ)

hA = L(1 cosθ)

hA = 1m(1 cos37o)

hA = 0.201 m

VB = √2ghA

VB = √2(10m/s2)(0.201 m)VB = 2.0 m/s

L


58

February 06, 2014

L = 1 m

15o

D

C

MED = MEC

MED = KED + GPED + EPED

MED = mghD

MEC = KEC + GPEC + EPEC

MEC = 1/2 mVC2

mghD = 1/2 mVC2

VC = √2ghD

VD = 0hD = ? VC = ?

hC = 0

0 0

hD

H

00

We are missing hD, but this can be found using trigonometry.

After Collision

hD = L(1 cosθ)

hD = 1m(1 cos15o)

hD = 0.034 m

VD = √2ghA

VB = √2(10m/s2)(0.034 m)VB = 0.825 m/s

VB is negative because of the direction


59

February 06, 2014

The CollisionBefore After

mA = 0.5 kgVA = 2.0 m/s

A B A B

mB = 5.0 kgVB = 0

mA = 0.5 kgV'A = 0.825 m/s

mB = 5.0 kgV'B = ?

pA + pB = p'A + p'BmAVA = mAV'A + mBV'B

(0.5 kg)(2.0 m/s) = (0.5 kg)( 0.825 m/s) + (5.0 kg) V'B1 kgm/s = 0.4125 kgm/s + (5.0 kg) V'B

1.4125 kgm/s = (5.0 kg) V'B

V'B = 0.2825 m/s

0


60

February 06, 2014

The Slide

Sm = 5.0 kgVA = 0.2825 m/s

m = 5.0 kgVB = 0

Find the frictional force.FN

Fg = mg

FN = mgf = μK FN = μK mgf = 0.2 (5.0 kg)(10 m/s2)f = 10 N

fFNet = f

FNet = 10 N

Rx = FNet Ry = 0


61

February 06, 2014

The Slide

Sm = 5.0 kgVA = 0.2825 m/s

m = 5.0 kgVB = 0

FNet = 10 N at 180o

WNet = ΔKEFNet S cos θ = 1/2 m (VF2 VI2)

10 N S cos 180o = 1/2 (5 kg) (02 (0.2825 m/s)2)( 10 N) S = 0.1995 J

S = 0.01995 mS = 20 cm


62

February 06, 2014

Two discs of masses m1 = 2 kg and m2 = 8 kg are placed on a horizontalfrictionless surface. Disc m1 moves at a constant speed of 8 m/s in +xdirection and disc m2 is initially at rest. The collision of two discs is elastic and the directions of two velocities are presented by the diagram.

A) What is the xcomponent of the initial momentum of disc m1?B) What is the ycomponent of the initial momentum of disc m1?C) What is the xcomponent of the initial momentum of disc m2?D) What is the ycomponent of the initial momentum of disc m2?E) What is the initial momentum of the system?F) What is the xcomponent of the final momentum of disc m1?G) What is the xcomponent of the final momentum of disc m2?H) What is the ycomponent of the final momentum of disc m2?I) What is the ycomponent of the final momentum of disc m1?J) What is the final vector velocity of disc m1?K) What is the final vector velocity of m2?L) Determine if the collision was perfectly elastic.

m1

v1

m2

m1

v1'

m2 30o

v2'Before Collision After Collision


63

February 06, 2014

A) What is the xcomponent of the initial momentum (p1x) of disc m1?

p1 = mV1

p1x = mV1x

p1x = (2 kg)(8 m/s)p1x = 16 kg m/s

B) What is the ycomponent of the initial momentum (p1y) of disc m1?

p1 = mV1

p1y = mV1y

p1y = (8 kg)(0 m/s)p1y = 0 kg m/s

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

positive because its in the postive X direction


64

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

C) What is the xcomponent of the initial momentum (p2x) of disc m2?p2 = mV2

p2x = mV2x

p2x = (2 kg)(0 m/s)p2x = 0

D) What is the ycomponent of the initial momentum (p2y) of disc m2?p2 = mV2

p2y = mV2y

p2y = (8 kg)(0 m/s)p2y = 0 kg m/s


65

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

E) What is the initial momentum (pinitial) of the system?

ptotalx = Σpix = p1x + p2x

ptotalx = 16 kg m/s + 0 = 16 kg m/s

ptotaly = Σpiy = p1y + p2yptotaly = 0 + 0 = 0

pinitial = ptotalx + ptotaly

Using VECTOR ADDITIONpinitial = 16 kg m/s at 0o


66

February 06, 2014

Since momentum is conservedtotal momentum before the collision = total momentum after the collision

p = p'

16 kg m/s at 0o = p'

The components of momentum before and after the collision must also be equal

ptotalx = p'totalx16 kg m/s = p'totalx

ptotaly = p'totaly0 = p'totaly

positive because its in the postive X direction


67

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270 = 0

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

F) What is the xcomponent of the final momentum (p1x')of disc m1?Because the direction of m1 after the collision is 270o which causes V1x' to equal 0, (since p = mV) then momentum also MUST equal zero!Thus

p1x' = 0


68

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270 = 0

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

G) What is the xcomponent of the final momentum (p'2x) of disc m2?

p'totalx = 16 kg m/s

p'totalx = p'1x + p'2xSince p1x' = 0

Then

p'totalx = 0 + p'2xp'totalx = p'2x

16 kg m/s = p'2x positive because its in the postive X direction


69

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270 = 0

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

H) What is the ycomponent of the final momentum (p2y') of disc m2?

p2'

p2x' = 16 kg m/s

p2y'30o

Solve using trigonometry.

tan 30o = p2y'/p2x'

p2y' = p2x' tan 30o

p2y' = 16 kg m/s tan 30o

p2y' = 9.2 kg m/s


70

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270 = 0

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

I) What is the ycomponent of the final momentum (p'1y) of disc m1?

p'totaly = 0

p'totaly = p'1y + p'2yThen

0 = p'1y + p'2yp'1y = p'2y

p'1y = 9.2 kg m/s


71

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270 = 0

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

J) What is the final vector velocity (V1') of disc m1?Find p1'

p1' = p1x' + p1y'p1' = 0 + (9.2 kg m/s)

Using VECTOR ADDITIONp1' = 9.2 kg m/s at 270o


72

February 06, 2014

p1' = m1 V1'

V1' =p1'm1

V1' =9.2 kg m/s at 270o

2 kg

V1' = 4.6 m/s at 270o


73

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270 = 0

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

J) What is the final vector velocity (V2') of disc m1?Find p2'

p2' = p2x' + p2y'p2' = 16 kg m/s + 9.2 kg m/s

Using VECTOR ADDITIONp2' = 18.5 kg m/s at 30o


74

February 06, 2014

p2' = m2 V2'

V2' =p2'm2

V2' =18.5 kg m/s at 30o

8 kg

V2' = 2.3 m/s at 270o


75

February 06, 2014

m1

v1

m2

m1

v1'

m2 30o


V1x = 8 m/sV1y = 0

V2x = 0V2y = 0 V1x' = V1x' cos 270 = 0

V1y' = V1y' sin 270

V2x' = V2x' cos 30V2y' = V2y' sin 30

L) Determine if the collision was perfectly elastic.If a collision is perfectly elastic then kinetic energy is conserved.

KE1 + KE2 = KE1' + KE2'½m1V12 + ½m2V22 = ½m1V12' + ½m2V22'

½(2 kg)(8 m/s)2 + ½(8 kg)(0)2 = ½(2 kg)(4.6 m/s)2 + ½(8 kg)(2.3 m/s)2

64 J + 0 = 21.16 J + 21.16 J64 J ≠ 42.32 J

Kinetic Energy is not conserved.


76

February 06, 2014

3040

15 kg

Ft1Ft2


77

February 06, 2014

Draw an FBD

Fg = 150N

30

Ft1

40

Ft2

Fg = 150N

Ft1

70

Ft2

60

X = FT2 cos 70

Y = FT2 sin 70

X = FT1

Y = 0

X = 75.0 NY = 129.9 N

Rx = 0

FT1 FT2 cos 70 75.0 N = 0

FT1 = FT2 cos 70 + 75.0 N

FT1 = 122.3 N

Ry = 0

FT2 sin 70 129.9 N = 0

FT2 sin 70 = 129.9 N

FT2 = 129.9 Nsin 70

FT2 = 138.2 N


78

February 06, 2014

5520

15 kg

Ft1Ft2


79

February 06, 2014

An 8,000 Kg truck, traveling with a velocity 12 m/s at 0o, collides with a 1,400 Kg car moving 22 m/s at 220o as in the figure below. After the collision, the two vehicles remain tangled together. With what velocity will the wreckage begin to move immediately after the crash?

CarTruck

220o


80

February 06, 2014

Two girls (masses m1 and m2) are on skates and at rest. The girls are close to each other, and are facing each other. Girl 1 pushes against girl 2 causing girl 2 to move backwards.

A) Assuming the girls move freely on skates (no friction), find a formula to calculate the speed of girl 1.

B) Calculate the speed of girl 1, If the mass of girl 1 is 48 kg, the mass of girl 2 is 62 kg, and girl 2’s speed is 1.8 m/s.

C) Calculate the impulse felt by girl 1 and girl 2. (Use the values in B)

D) Calculate the force felt by both girl 1 and girl 2, if the duration of the push was 0.7 s.


81

February 06, 2014


A) How fast is he falling when he reaches the airbag? (Initial velocity is zero)


C) What average force does he feel if he had fallen 65 m and landed on the ground (impact time = 10 ms = .001 s))?

Air bag

65 m


82

February 06, 2014

A railroad car of mass 30,000 kg is moving with a speed of 9.00 m/s to the right. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the opposite direction with an initial speed of 3.00 m/s.


(b) How much energy is lost in the collision?9 m/s 3 m/s

Momentum - HHS Physics · Momentum Impulse.notebook 1 February 06, 2014 Momentum In classical mechanics, linear momentum or translational momentum (plural momenta; SI unit kg m/s,

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