1 Unit Six: Linear Momentum and Collisions John Elberfeld [email protected] 518 872 2082 GE253 Physics.

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Unit Six:Linear Momentum

and Collisions

John Elberfeld

[email protected]

518 872 2082

GE253 Physics

2

Schedule

• Unit 1 – Measurements and Problem Solving• Unit 2 – Kinematics• Unit 3 – Motion in Two Dimensions• Unit 4 – Force and Motion• Unit 5 – Work and Energy• Unit 6 – Linear Momentum and Collisions• Unit 7 – Solids and Fluids• Unit 8 – Temperature and Kinetic Theory• Unit 9 – Sound• Unit 10 – Reflection and Refraction of Light• Unit 11 – Final

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Chapter 6 Objectives• Compute linear momentum and the components of

momentum.• Relate impulse and momentum, and kinetic energy and

momentum.• Explain the condition for the conservation of linear

momentum and apply it to physical situations.• Describe the conditions on kinetic energy and

momentum in elastic and inelastic collisions.• Explain the concept of the center of mass and compute

its location for simple systems, and describe how the center of mass and center of gravity are related.

• Apply the conservation of momentum in the explanation of jet propulsion and the operation of rockets.

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Reading Assignment

• Read and study College Physics, by Wilson and Buffa, Chapter 6, pages 179 to 210

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Written Assignments

• PREPARE FOR A UNIT EXAM II on all material covered in the course

• Do all the assignments on the handouts as partial preparation

• Review all your quizzes

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Introduction

• We know F = m a, Newton’s Second Law

• Now we will manipulate that equation to learn more about impulse and momentum

• Masses with large momentums are hard to stop

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Introduction

• Newton’s First Law describes momentum by explaining that an object at rest remains at rest, and an object in motion continues in motion at a constant speed in a straight line unless acted upon by an outside force

• The formula, p = mv, give momentum (p) a value

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Linear Momentum

• The momentum of an object is the product of its mass and velocity. It is a vector quantity because velocity is a vector quantity.

• The unit of momentum is kilogram times meter divided by second :

• p = m v (kg m/s)• Both velocity and momentum are vectors

and both use signs to indicate directions

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Practice

• A 100-kg football player runs with a velocity of 4.0 m/s straight down the field. A 1.0-kg artillery shell leaves the barrel of a gun with a muzzle velocity of 500 m/s.

• Which has a greater momentum (magnitude), the football player or the shell?

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Calculations

• P = mv• P = 100kg 4 m/s = 400 kg m/s (player)• P = mv• P = 1 kg 500 m/s = 500 kg m/s (shell)• The artillery shell has more momentum,

but not that much more

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Conservation of Momentum

• Momentum Before = Momentum After• The total momentum of a system of

objects (or group under consideration) is constant - ALWAYS

• For example, the momentum of two cars before they collide is equal to the momentum of the two cars are the collision

• Momentum Before = Momentum After

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Total Momentum

• The total momentum of a system (P) is the sum of the momentum of all its particles (p).

• You can write this as:

• P = P1 + P2 + P3 + …

• Remember to use vector addition and not just the sum of the magnitudes

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Vector Addition

• Add horizontal and vertical components

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When Angles are Involved

• Use vector addition

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Impulse and Momentum

• You can express Newton’s second law in terms of momentum as:

• F = m a = m (v-v0)/ Δt = m Δv / Δt• F Δt = m Δv • F Δt = Impulse• m Δv = change in momentum• Impulse = change in momentum• Remember to use vector addition

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Momentum in a System

• Part (a) shows a particle hitting a wall and bouncing straight back.

– The change in momentum of the ball is twice the initial momentum and points away from the wall.

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Force and Momentum

• A projectile is a good example of changing momentum.

• In the figure shown below, the momentum of the projectile changes constantly in both direction and magnitude.– Force of gravity is

constantly changing the momentum

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Impulse

• Thus impulse is defined as Fave Δt and: • Fave Δt = m Δv = ΔP• Impulse = change in momentum• The unit of impulse is Newton times

second, which is the same as that of momentum.

• Impulse is used for collisions and other phenomena where a force is applied for a short period of time.

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Design for High Impulse

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Practice

• A shuffleboard player pushes a 0.25-kg puck, which is initially at rest, so that a constant horizontal force of 6.0 N acts on it through a distance of 0.50 m. (Neglect friction

• (a) What is the kinetic energy and the speed of the puck when the force is removed?

• (b) How long does the force act?• (c) What is the change in momentum?• (d) What is the impulse?• (Notice this builds on previous chapters!)

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Calculations

• Conservation of energy and velocity• W = F d = mv2/2• W = 6.0 N .5 m = .25kg v2/2• KE = 3.0 Nm = 3.0 J• v2 = 3.0 Nm·2 / .25kg • v = 4.90 m/s

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Find the Time

• F = ma

• 6.0N = .25kg a

• a = 24m/s2

• v = v0 + a Δt

• 4.90 m/s = 0 + 24m/s2 Δt

• Δt = .204 s

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Momentum and Impulse

• v = 4.90 m/s

• Momentum = P = m v

• m v = .25 kg 4.90 m/s = 1.23 kg m/s

• Impulse = F Δt

• F Δt = 6 N .204 s

• Impulse = 1.23 Ns

• Impulse = Change in Momentum

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Practice

• A golfer drives a 0.10-kg ball from an elevated tee, giving it an initial horizontal speed of 40 m/s (about 90 mi/h).

• The club and the ball are in contact for 1.0 ms (millisecond).

• What is the average force exerted by the club on the ball during this time?

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Calculation

• FΔt = m Δv

• F 1 x 10-3s = .10 kg 40m/s

• F = 4,000 N in the direction of the ball’s motion

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Practice

• The golfer in the previous example drives the ball with the same average force, but “follows through” on the swing so as to increase the contact time to 1.5 ms.

• What effect will this change have on the initial horizontal speed of the drive?

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Time Affects Results

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Calculations

• FΔt = m Δv

• 4000N 1.5 x10-3s = .10 kg v

• v = 60 m/s

• Applying the same force for a longer period of time creates a greater change in momentum

• The ball has a higher initial velocity (60 m/s –not 40m/s) and travels farther

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Practice

• A tennis ball with a mass of .1 kg moving at 25 m/s is smashed back with a velocity of 35 m/s?

• What is the change in velocity?

• What is the impulse?

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• Before

• After

Sketch a diagram and label

.1 kg

25 m/s

.1 kg

35 m/s

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Solve

• Pay close attention to signs– Set motion to the right as positive

• m1 = .1 kg

• v1i = +25 m/s

• v1f = -35 m/s

• m1 v1f - m1v1i = (.1kg)(-35m/s)-(.1kg)(25m/s)

• Impulse = change in momentum =• -6.0 kg m/s (to the left)• Impulse = -6.0 N s

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Find the force

• F Δt = m (vf – vi)• Impulse = -6.0 N s • If the applied force happens over 0.1

seconds, what is the average force?• F Δt = - 6.0 Ns = F (0.1s)• F = -6.0 N s/0.1s = - 60.0 N• A force of 60.0 N to the left was applied for

.1 s to make the tennis ball go back with a velocity of 35 m/s.

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Conservation of Linear Momentum

• Momentum BEFORE = Momentum AFTER• Using Impulse = Change in Momentum:

• F Δt = m (vf – vi)• For every action, there is an equal and opposite

reaction, so the force on one object in a collision is equal but opposite to the force on the other object, and times are equal

• In a collision, the positive increase in momentum is balanced by a negative decrease in momentum, so the total momentum is not changed

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Practice

• Two masses, m1 = 1.0 kg and m2 = 2.0 kg, are held on either side of a lightly compressed spring.

• They are joined together by a light string.

• When you burn the string (negligible external force), the masses move apart on the frictionless surface, with m1 having a velocity of 1.8 m/s to the left.

• What is the velocity of m2? • This apparatus illustrates

how you can have a system

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Calculations

• First, decided motion to the right is positive:• Since there are no outside force affecting

material within the system, the total change in momentum = 0

• Momentum before the collision = momentum after the collision

• m01v01 + m02v02 = m1v1 + m2v2

• 0 + 0 = 1.0kg(-1.8m/s) + 2 kg v2

• 1.8 kg m/s = 2 kg v2

• v2 = .9 m/s to the right because it is positive

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Collisions• When you study the

collision of two objects, you need to know only their masses and velocities before and/or after the collision.

• You may be interested to know that in physics, three-body problems (three objects colliding with each other at once) are impossible to solve to exact precision

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Elastic Collisions

• In elastic collisions, total kinetic energy is conserved.

• In other words, the total kinetic energy of all the particles before and after the collision remains the same.

• Kf = Ki • Final KE = Initial KE• When a Super Ball hits a hard surface, it

bounces off with nearly the same speed as before the collision, therefore (ideally) its kinetic energy is the same after the collision and the collision is elastic.

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Inelastic Collision

• In contrast, an inelastic collision is one in which some (or all) of the total kinetic energy of the particles is lost.

• Kf < Ki • Since a soft rubber ball does not bounce up

from the ground back to the same height it is dropped from, we know that some of its kinetic energy was lost in the collision, so its collision with the ground is inelastic.

• The lost kinetic energy turns into heat

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Collisions and Impulse

• There are two types of collisions– Perfectly elastic collision

• When kinetic energy is conserved in a collision, it is called a perfectly elastic collision.

– Perfectly Inelastic collisions• If kinetic energy is not conserved, the

collision is called inelastic.

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Elastic Collisions

• In an elastic collision between two particles, the conservation of momentum and kinetic energy is given by:

• m1v012/2 + m2v02

2/2 = m1v12/2 + m2v2

2/2• m1v01 + m2v02 = m1v1 + m2v2

• Kinetic energy before the collision = Kinetic energy after the collision

• Momentum before the collision = momentum after the collision

• When one of these masses is initially at rest, these equations become simpler.

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Elastic Collisions

• When one of the particles is stationary, the equations for conservation of kinetic energy and momentum become:

• m1v012/2 = m1v1

2/2 + m2v22/2

• m1v01 = m1v1 + m2v2

• By manipulating these equations:

• v1 = v01(m1-m2)/(m1+m2)

• v2 = v01(2m1)/(m1+m2)

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Practice

• A 0.30-kg object with a speed of 2.0 m/s in the positive x-direction has a head-on elastic collision with a stationary 0.70-kg object located at x = 0 m.

• What is the distance separating the objects 2.5 s after the collision?

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Calculations

• m1v012/2 = m1v1

2/2 + m2v22/2

• m1v01 = m1v1 + m2v2

• These equations lead to:• v1 = v01(m1-m2)/(m1+m2) • v2 = v01(2m1)/(m1+m2) • v1 = 2m/s(.3kg -.7kg )/(.3kg +.7kg) = -.8m/s• v2 = 2m/s(2 x .3kg )/(.3kg +.7kg) = +1.2 m/s• Distance = (1.2m/s)2.5s – (-.8m/s) 2.5s• Distance = 5 m

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Extreme Elastic Collisions

• A small ball bounces off a large ball• A .5 kg ball moves at 8 m/s to the right and

collides with a stationary ball with a mass of 9 kg• Obviously, the small ball is going to bounce back

off the big ball• The big ball will move very slowly to the right• To conserve kinetic energy, since the bowling

ball is barely moving, the small ball must be moving almost as fast away from the big ball as it was just before it hit the ball

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Calculations

• m1v012/2 = m1v1

2/2 + m2v22/2

• m1v01 = m1v1 + m2v2

• These equations lead to:• v1 = v01(m1-m2)/(m1+m2) • v2 = v01(2m1)/(m1+m2) • v1 = 8m/s(.5kg - 9kg )/(.5kg + 9kg) =-7.15m/s• v2 = 8m/s(2x.5kg )/(. 5kg + 9kg) = +.84 m/s• Velocity of the big ball is small, and the

velocity of the small ball is slightly less, but in the opposite direction (just like common sense!)

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Another Extreme Elastic

• A bowling ball (9kg) moves down the gutter (4m/s) and collides with a identical, stationary ball

• What happens defies common sense

• The moving ball stops completely, and the ball it hits moves off with the same velocity as the incoming ball in this ELASTIC collision

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Calculations

• m1v012/2 = m1v1

2/2 + m2v22/2

• m1v01 = m1v1 + m2v2

• These equations lead to:• v1 = v01(m1-m2)/(m1+m2) • v2 = v01(2m1)/(m1+m2) • v1 = 4m/s(9kg - 9kg )/(9kg + 9kg) = 0 m/s• v2 = 4m/s(2x9kg )/(9kg + 9kg) = 4 m/s• The first ball stops, and the second moves

off with the same velocity as the incoming ball in the ELASTIC collision.

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Identifying Collisions

• Usually collision between bowling balls, pool balls, superballs… are elastic and kinetic energy is conserved

• Collisions between cars and trucks or velcro covered balls are inelastic and kinetic energy is NOT conserved

• Momentum is ALWAYS conserved for all types of collisions

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Inelastic Collisions

• In inelastic collisions, the total kinetic energy after and before the collisions does not remain the same

• Final momentum ALWAYS equals initial momentum

• In the first diagram, two velcro-covered balls with equal and opposite momentum collide head-on; as a result of the collision, they stick together and stop.

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Inelastic Collisions

• In the second diagram, one velcro-covered ball is moving and the other is at rest.

• As a result of the collision, they stick and move together at a speed less than that of the initially moving ball.

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Inelastic Collisions

• In the previous screens, the first collision is clearly inelastic because the total kinetic energy of the two balls after the collision is zero.

• In the second collision, however, you need to calculate total kinetic energy of the two balls to confirm KE was lost.

• Since there are no external forces acting on the two balls, momentum is conserved:

• m1v01 + m2v02 = m1v1 + m2v2

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Practice

• In the example of inelastic collision on the previous screens, a 1.0-kg ball with a speed of 4.5 m/s strikes a 2.0-kg stationary ball and sticks to it.

• (a) What is the speed of the balls after the collision?

• (b) What percentage of the initial kinetic energy do the balls have after the collision?

• (c) What is their total momentum after the collision?

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Calculations

• m1v01 + m2v02 = m1v1 + m2v2

• 1kg 4.5m/s + 0 = (1kg + 2kg) v• v = 1.5 m/s• KB = m1v01

2/2 + m2v022/2

• KA= m1v12/2 + m2v2

2/2• KB = 1kg (4.5m/s)2/2 = 10.125 J• KA = (3kg)(1.5m/s)2/2 = 3.375 J• KA/KB =3.375 J / 10.125 J = 33%• P = m1v01 = 1kg 4.5m/s = 4.5 kg m/s – same

before and after

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Results

• The end velocity of the joined objects is in the same direction as the object that had the greatest magnitude of momentum

• Because Kinetic energy turned to heat, the collision was inelastic

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Center of Mass

• The center of mass of a solid object or a collection of particles is a point at which the entire mass of that solid object or collection of particles is considered to be concentrated.

• The location of the center of mass for a system with n particles is defined as:

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Practice

• Three masses, 7.0 kg, 5.0 kg, and 4.0 kg, are located at positions (-4.0, 0), (3.0, 0), and ( 6.0, 0), respectively, in meters from the origin.

• Where is the center of mass of this system?

-4 630

?

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Calculations

• xcm = 7kg(-4m) +5kg3m+4kg6m 7kg + 5kg + 4kg

• The center of mass is .688 m to the right of the origin. A single force at that point will hold up all the masses

-4 630

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Center of mass

• There does not have to be any actual material at the location of the center of mass

• Where is the center of mass of a doughnut? Is it inside the material part of the doughnut?

• Where is the center of mass for the letter “L”

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Practice

Find the center of mass if 4kg is at .4m and 7kg is a .9m?

4kg .4m 7kg .9m0m

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Calculations

• Xcm = (m1X1+m2x2)/(m1+m2)

• Xcm = (4kg .4m+7kg .9m)/(4kg+7kg)

• Xcm = .718 m

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Summary

• What is momentum?• What is the impulse-momentum theorem?• Why is linear momentum conserved?• What is the difference between elastic and

inelastic collisions?• What is the center of mass of an object?• How does the conservation of momentum

explain jet propulsion?