Moving objects have momentum Vector quantity

Points in the same direction as the velocity vector Momentum:

Equals the product of an objects mass and velocity Proportional to mass and velocity

p = mv

p = momentum (kg * m/s)m = mass (kg)v = velocity (m/s)

What is the taxi cab’s momentum?* Mass of the taxi = 53 kg* Velocity of the taxi = 1.2 m/s

Answer: p = mv p = (53 kg)(1.2 m/s)

p = 63.6 kg * m/s to the left

v = 1.2 m/s

p = 63.6 kg * m/s

Newton’s 2nd Law ΣF = ma = m(Δv/Δt)ΣF = m(Δv/Δt)

Momentump = mv m = p/v ΣF = (p/v)(Δv/Δt) ΣF = Δp/Δt

If the momentum of an object changes, either mass, velocity, or both change If mass remains the same than velocity

changes acceleration occurs What produces an acceleration?

FORCE Greater the force acting on the object

greater its change in velocity greater its change in momentum

How long the force acts is also important…Stalled car

Apply a force over a brief amount of time produce a change in momentum

Apply the same force over an extended period of time produce a greater change in the car’s momentumA force suspended for a long time produces

more change in momentum than does the same force applied briefly

Both force and time are important in changing momentum

IMPULSE (J) = Δp = pf – pi = mvf – mvi

J = Favg ΔtFavgΔt = mΔv

Favg = mΔv/Δt Impulse (J) = Change in momentum Impulse is also the product of the average

force and the time during which the force is applied.

Vector quantity Units: kg * m/s

A long jumper's speed just before landing is 7.8 m/s. What is the impulse of her landing? (mass = 68 kg) J = Δp J = pf - pi

J = mvf – mvi

J = 0 - (68kg)(7.8m/s) J = -530 kg * m/s

*Negative sign indicates that the direction of the impulse is opposite to her direction of motion

Baseball player swings a bat and hits the ball, the duration of the collision can be as short as 1/1000th of a second and the force averages in the thousands of newtons

The brief but large force the bat exerts on the ball = Impulsive force

View Kinetic books section 8.4- Physics at play: Hitting a baseball

BASEBALL PROBLEMThe ball arrives at 40 m/s and leaves at 49 m/s in the opposite direction. The contact time is 5.0×10−4 s. What is the average force on the ball?

J = Δp = Favg Δt = mΔv

Favg Δt = mΔv

FavgΔt = mΔv

Favg = mΔv/Δt

Favg = (0.14kg)(49 – (-40)m/s)/5.0×10−4 s

Favg = 2.5×104 N

Case 1: Increasing momentumTo increase the momentum of an object

apply the greatest force possible for as long as possible Golfer teeing off and a baseball player trying for

a home runSwing as hard as possible (large force)Follow through with their swing (increase in

time)

Case 2: Decreasing momentumYou are in a car that is out of control Do

you want to hit a cement wall or haystack? In either case, your momentum is decreased by

the same impulse…But, the same impulse does not mean the

same amount of force or the same amount of time rather it means the same PRODUCT of force and time

Case 2 continued: Decreasing momentum

Hit the haystack Extend the impact timeChange in momentum occurs over a long time Small impact force

mv = Ft

Hit the cement wall Change in momentum occurs over a short time Large impact force

mv = Ft

If you want to decrease a large momentum, you can have the force applied for a longer time

** If the change in momentum occurs over a long time Force of impact is small

Examples: Air bags in cars. Crash test video

Ft

If the change in momentum occurs over a short time, the force of impact is large. Karate link Boxing video

Ft

QUESTIONWhen a glass falls, will the impulse be less if

it lands on a carpet than if it lands on a hard floor? NO Impulse is the same for either surface

because the change in momentum is the sameCarpet: More time is available for the change

in momentum smaller force for the impulse Hard floor: Less time is available for the

change in momentum (due to less “give”) larger force for the impulse

Conservation of momentum: Occurs when there are no net external

force(s) acting on the system Result Total momentum of an isolated system

is constantMomentum before = Momentum afterPlaying pool example:

Kinetic books 8.6

Momentump = mv

Conservation of momentumMomentum before = Momentum afterpi1 + pi2 +…+ pin = pf1 + pf2 +…+ pfn

pi1, pi2, …, pin = initial momenta pf1, pf2, …, pfn = final momenta

m1vi1 + m2vi2 = m1vf1 + m2vf2

m1, m2 = masses of objects vi1, vi2 = initial velocities vf1, vf2 = final velocities

A 55.0 kg astronaut is stationary in the spaceship’s reference frame. She wants to move at 0.500 m/s to the left. She is holding a 4.00 kg bag of dehydrated astronaut chow. At what velocity must she throw the bag to achieve her desired velocity? (Assume the positive direction is to the right.)

VARIABLES: Mass of astronaut ma = 55 kg

Mass of bag mb = 4 kg

Initial velocity of astronaut via = 0 m/s

Initial velocity of bag vib =0 m/s

Final velocity of astronaut vfa = -0.5 m/s

Final velocity of bag vfb = ? EQUATION:

m1vi1 + m2vi2 = m1vf1 + m2vf2

mavia + mbvib = mavfa + mbvfb

0 = mavfa + mbvfb

Vfb = - (mavfa / mb)

Vfb = - ((55kg)(-0.5m/s))/(4kg) = 6.875 m/s

Collision of objects Demonstrates the conservation of momentumWhenever objects collide in the absence of

external forces net momentumbefore collision = net momentumafter collision

Momentum is conserved in ALL TYPES of collisionsElastic Collisions

Objects collide without being permanently deformed and without generating heat

Inelastic Collisions Colliding objects become distorted (tangled or

coupled together) and generate heat

ProblemConsider a 6-kg fish that swims toward and

swallows a 2-kg fish that is at rest. If the larger fish swims at 1 m/s, what is its velocity immediately after lunch?

net momentumbefore collision = net momentumafter collision

(net mv)before = (net mv)after

(6kg)(1m/s) +(2kg)(0) = (6kg + 2kg)(vafter)

vafter = ¾ m/s

ProblemConsider a 6-kg fish that swims toward and

swallows a 2-kg fish that is moving towards the larger fish at 2 m/s. If the larger fish swims at 1 m/s, what is its velocity immediately after lunch?

net momentumbefore collision = net momentumafter collision

(net mv)before = (net mv)after

(6kg)(1m/s) +(2kg)(-2m/s) = (6kg + 2kg)(vafter)

vafter = 1/4 m/s

Perfectly Elastic collisionsNot common in the everyday world

Some heat is generated during collisionsDrop a ball and after it bounces from the floor,

both the ball and the floor are a bit warmer

At the microscopic level perfectly elastic collisions are common Electrically charged particles bounce off one

another without generating heat

Examples of Examples of Perfectly Perfectly

ELASTIC CollisionsELASTIC Collisions

• Electron scattering• Hard spheres (Pool balls)

Elastic collisionKinetic energy is conserved

KE before = KE afterKE = 1/2mv2

Momentum is conserved in any collision Elastic or inelastic

ELASTIC collisions in 1-dimension1. Conservation of Kinetic Energy:

2. Conservation of Momentum:

• Rearrange both equations and divide:€

12m1v1i

2 + 12m2v2i

2 = 12m1v1 f

2 + 12m2v2 f

2 (1)

m1v1i m2v2i m1v1 f m2v2 f (2)

m1 v1i2 v1 f

2 m2 v2 f2 v2i

2 (1)

m1 v1i v1 f v1i v1 f m2 v2 f v2i v2 f v2i m1 v1i v1 f m2 v2 f v2i (2)

v1i v1 f v2 f v2i

v1i v2i v1 f v2 f

Final velocities in Head-On Two-Body Elastic Collisions (v2i = 0 m/s)

€

v1 f =m1 −m2

m1 + m2

⎛

⎝ ⎜

⎞

⎠ ⎟v1i

€

v2 f =2m1

m1 + m2

⎛

⎝ ⎜

⎞

⎠ ⎟v1i

• Catching a baseball: Video• Football tackle• Cars colliding and sticking• Bat eating an insect

Examples of PerfectlyExamples of PerfectlyINELASTIC CollisionsINELASTIC Collisions

Inelastic collisionKinetic energy is NOT conserved

KE before ≠ KE after

Momentum is conserved in any collision Elastic or inelastic

Perfectly INELASTIC collisionsin 1-dimension

• Final velocities are the same

€

m1v1i + m2v2i = m1 + m2( )v f

Problem

A 5879-lb (2665 kg) Cadillac Escalade going 35 mph smashes into a 2342-lb (1061 kg) Honda Civic also moving at 35 mph (15.64 m/s) in the opposite direction. The cars collide and stick.

a) What is the final velocity of the two vehicles?

a) m1v1i + m2v2i = (m1 +m2)vf

(2665kg)(15.64m/s) + (1061kg)(-15.64m/s) = (2665 + 1061kg)vf

vf = 6.73 m/s = 15.1 mph

Collisions

Momentum is always conserved in a collision Collision video Classification of collisions:

ELASTIC Both energy & momentum are conserved

INELASTIC Momentum conserved, not energy Perfectly inelastic -> objects stick Lost energy goes to heat

“Average” location of mass An object can be treated as though all

its mass were located at this pointFor a symmetric object made from a

uniformly distributed material, the center of mass is the same as its geometric center

Equation:

xcm = m1x1 + m2x2 + …mnxn / m1 + m2 + …mn

xCM = x position of center of mass

mi = mass of object i

xi = x position of object i

View section 8.20 in Kinetic books* Specifically example 1- Center of mass

problem

Center of Mass Video Balancing Activity: video demo

Key Facts:Newton’s 2nd Law (F = ma)

To accelerate an object Net force must be applied

To change the momentum of an object exert an impulse on it

The momentum of a system cannot change unless it is acted on by external forces

Law of Conservation of Momentum In the absence of an external force, the

momentum of a system remains unchanged Examples in which the net momentum is the

same before and after the event:Radioactive decayCars collidingStars exploding

Conservation of Momentum

mv(initial) = mv(final)

An astronaut of mass 80 kg pushes away from a space station by throwing a 0.75-kg wrench which moves with a velocity of 24 m/s relative to the original frame of the astronaut. What is the astronaut’s recoil speed? (0.75kg)(24m/s) = 80kg(v)v = 0.225 m/s

QuestionNewton’s 2nd law states that if no net force

is exerted on a system, no acceleration occurs. Does it follow that no change in momentum occurs? Yes, because no acceleration (a = Δv/t)

means no change in velocity and no change in momentum (p = mΔv)

Also, no net force means no net impulse (J = Ft) J = Δp no change in momentum

QuestionNewton’s 3rd law states that the force a

rifle exerts on a bullet is equal and opposite to the force the bullet exerts on the rifle. Does is follow that the impulse the rifle exerts on the bullet is equal and opposite to the impulse the bullet exerts on the rifle? Yes, because the rifle acts on the bullet and bullet

reacts on the rifle during the same time intervalSince time is equal and force is equal and

opposite for both Impulse, Ft, is also equal and opposite for both (Impulse – vector quantity and can be canceled)

• The law of conservation of momentum can be derived from Newton’s 2nd and 3rd laws• Newton’s 2nd law F = ma• Newton’s 3rd law Forces are equal but

opposite* Refer to Kinetic Books- 8.7 For step-by-step derivation

Collisions:Momentum- Useful concept when applied to

collisions

In a collision, two or more objects exert forces on each other for a brief instant of time, and these forces are significantly greater than any other forces they may experience during the collision

ProblemA proton (mp=1.67x10-27 kg) elastically collides with a target proton which then moves straight forward. If the initial velocity of the projectile proton is 3.0x106 m/s, and the target proton bounces forward, what are

a) The final velocity of the projectile proton?b) The final velocity of the target proton?0.0 m/s3.0 x 106 m/s

Elastic collision in 1-dimensionFinal equations for head-on elastic

collision:

• Relative velocity changes sign

• Equivalent to Conservation of Energy

m1v1i m2v2i m1v1 f m2v2 f

v1i v2i v1 f v2 f

Problem

An proton (mp=1.67x10-27 kg) elastically collides with a target deuteron (mD=2mp) which then moves straight forward. If the initial velocity of the projectile proton is 3.0x106 m/s, and the target deuteron bounces forward, what are

a) The final velocity of the projectile proton?b) The final velocity of the target deuteron?

vp = -1.0 x 106 m/svd = 2.0 x 106 m/s

Head-on collisions with heavier objects always lead to reflections