Physics Chapter 6Physics Chapter 6

Momentum and Its Conservation

Linear MomentumLinear Momentum

The velocity and mass of an object determine what is needed to change its motion.

Linear Momentum (ρ) is the product of mass and velocityρ =mvUnit is kgm/s

Example 1. Find the magnitude of the momentum of each cart shown below, before the collision.

Example 1Example 1

Example 1. Find the magnitude of the momentum of each cart shown below, before the collision.

Given:

Solution:

Find:

mA = 450 kg mB = 550 kg A = 4.50 m/s B = 3.70 m/s

ρ A and ρ B

ρ A = mAA = (450 kg)(4.50 m/s) = 2025 = 2.0 x 103 kg . m/s

ρ B = mBB = (550 kg)(3.70 m/s) = 2035 = 2.0 x 103 kg . m/s

Example 2. The man in this picture has a mass of 85.0 kg. If he strikes the ground at 7.7 m/s, what is his change in momentum (ρ) Assume his mass remains the same after the collision.

ρ = m(f - i) = (85.0 kg)((0.0 m/s – (-7.7 m/s)) = 650 kg . m/s

Given:

Find:

Solution:

m = 85.0 kg f = 0.0 m/si = -7.7 m/s (negative for down)

ρ

If the velocity is changed by an outside force, then the momentum is also changed.F=ma = mv/t = ( ρ /t)Ft = m v = mvf-mvi= ρ Ft = ρ = “impulse” in N st is the time during which the force is applied.

This is called the “Impulse-Momentum Theorem”.

Example 3. A player exerts a force (assume constant) of 12.5 N on a ball over a period of 0.35 s. Find the total impulse .

Given:

Find:

Solution:

F = 12.5 N t = 0.35 s

Impulse

Impulse = Ft = (12.5 N)(0.35 s) = 4.4 N . s

Example 4:Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car, which stops it without splashing it back (kind of a fake problem, but that’s OK). What is the force exerted by the water on the car each second.m = 1.5 kg vi=20m/s vf=0m/s t=1s

FC t = ρ

FC means force on water by car (Fw is force on car by water which is equal and opposite)

FC = (ρ f – ρi) / t

FC = (mvf – mvi) / t

FC = m (vf – vi) / t

FC = 1.5 kg (0 m/s – 20 m/s) / 1 s

FWC,x = -30 N FCW,x = 30 NFC = -30 N FW = 30 N

FC= ρ /t

Stopping DistanceStopping DistanceStopping distance depends on

the Impulse-Momentum Theorem.A heavy object takes longer to

stop…Also, A change in momentum

over a longer time requires less force.Ex: nets, air mattresses, padding

(see fig. 6-5)

Calculating time and Calculating time and distancedistance

Stopping time can be calculated using Ft = p, and solving for t.

Stopping distance can be calculated using x = ½(vi+vf)t

ExampleExample

A 720.0 kg beryllium sphere traveling to the west slows down uniformly from 230.0 m/s to 12.00 m/s. How long does it take the sphere to decelerate if the force on the sphere is 840 N to the east? How far does the sphere travel during the deceleration?

First find change in First find change in timetime

m=720.0kgvi=-230.0 m/s vf=-12.00m/sF= +840N Ft=pt=p/F=(mvf-mvi)/F=[(720kgX-12.00m/s)-(720.0kgX-230.0m/s)]/840N=186.9s

Then find change in xThen find change in x

x=1/2 (vi+vf)t=

1/2(-230.0m/s-12.00m/s)(186.9s)

=-22614.9m = 22610 m to the West

Newton’s 3Newton’s 3rdrd Law LawAs two objects collide, the force that

one exerts on the other multiplied by the time (Ft) is the impulse for each.

Newton’s 3rd law tells us that the force on mass1 is equal and opposite to the force on mass2 when they collide, which leads to conservation of momentum.Forces in real collisions are not constant

(see fig. 6-9, pg. 220)

Conservation of Conservation of Momentum in Momentum in

CollisionsCollisionsWhen two objects collide, the momentum of the individual objects change, but the total momentum remains the same.p1,i + p2,i = p1,f + p2,f

m1v1,i + m2v2,i = m1v1,f + m2v2,f

Friction is disregarded

Example 1Example 1

A 76 kg person, initially at rest in a stationary 45 kg boat, steps out of the boat and onto the dock. If the person moves out of the boat with a velocity of 2.5 m/s to the right, what is the final velocity of the boat?

List Knowns and List Knowns and Unknowns and Unknowns and

formula…formula…vi,boat=0 vi,person=0vf,boat=? vf,person=2.5m/smperson=76kg mboat=45kg

mpvp,i + mbvb,i = mpvp,f + mbvb,f

0 + 0 = (76kg)(2.5m/s) + (45kg)(vb,f)

Vbf= - (76X2.5)/45 = -4.2m/s(negative means the boat is moving

in the opposite direction)

Example 2. A proton with an initial velocity of 235 m/s strikes a stationary alpha particle(He) and rebounds off with a velocity of –188 m/s. Find the final velocity of the alpha particle.

Mass of proton= 1.67 x 10 -27 kg Mass of alpha particle (He)= 6.64 x 10 -27 kg

Given:

Find:

Original Formula:

mp = 1.67 x 10-27 kg mHe = 6.64 x 10-27 kgp,i = 235 m/s He,i = 0.0 m/s p,f = -188 m/s

He,f

fHe,Hefp,piHe,Heip,p νm νm νm νm Now, the initial velocity of He (alpha particle) is zero, so cross it out!

Given:

Find:

Original Formula:

mp = 1.67 x 10-27 kg mHe = 6.64 x 10-27 kgp,i = 235 m/s He,i = 0.0 m/s p,f = -188 m/s

He,f

fHe,Hefp,pip,p νm νm νm

Solve for He,f

Working Formula:

He

fp,pip,pfHe, m

νm - νm ν

Given:

Find:

Working Formula:

mp = 1.67 x 10-27 kg mHe = 6.64 x 10-27 kgp = 235 m/s He = 0.0 m/s p,f = -188 m/s

He,f

He

fp,ip,pfHe, m

)ν - (νm ν

The mass of the proton doesn’t change, so

sm /106

gk 10 x 6.64

m/s)) (-188 - m/s (235gk1.67x10

ν

27-

27

fHe,

Perfectly Inelastic Perfectly Inelastic CollisionsCollisions

A collision is perfectly inelastic when they collide and move together as one mass with a common velocity.m1v1,i + m2v2,i = (m1 + m2)vf

Example 1. A moving railroad car, mass=M, speed=3m/s, collides with an identical car at rest. The cars lock together as a result of the collision. What is their common speed afterward?

Vi1 Vi2=0

MM

before

Vf?

MM

after

M(+Vi1) + 0= (M+M)(Vf)

Vf = MVi1 / 2M = Vi1 / 2

Vf = (3m/s) / 2 = 1.5m/s

x x

Vi1 Vi2=0

MM

before

Vf?

MM

after

p1i = p2f

Kinetic Energy and Kinetic Energy and Perfectly Inelastic Perfectly Inelastic

CollisionsCollisions

During perfectly inelastic

collisions, some kinetic energy is lost because the objects are deformed.Once mass and velocities are

known (using conservation of momentum), solve for initial and final kinetic energies (1/2mv2) using the formula from Ch. 5 to determine energy lost.

KEi = KE1,i + KE2,i KEf = KE1,f + KE2,f

KE = KEf - KEi

Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.

Given:

Find:

Original Formula:

mr = 1.5 kg mb = 2.8 kg r,i = 4.5 m/s b,i = 0 m/s

f

fν)m (m νm νm brib,bir,r

Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.

Given:

Find:

Original Formula:

mr = 1.5 kg mb = 2.8 kg r,i = 4.5 m/s b,i = 0 m/s

fν)m (m νm νm brib,bir,r

The black ball starts at rest, so

f

Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.

Given:

Find:

Original Formula:

mr = 1.5 kg mb = 2.8 kg r,i = 4.5 m/s b,i = 0 m/s

fν)m (m νm brir,r

The black ball starts at rest, so

f

Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.

Given:

Find:

Working Formula:

mr = 1.5 kg mb = 2.8 kg r = 4.5 m/s b = 0 m/s

)m (m

νm ν

br

rrf

f

Example 1. A red clay ball with a mass of 1.5 kg and an initial velocity of 4.5 m/s strikes and sticks to a 2.8 kg black clay ball at rest. Find the final velocity of the combined mass.

Given:

Find:

Solution:

mr = 1.5 kg mb = 2.8 kg r,i = 4.5 m/s b,i = 0 m/s

m/s 1.6 1.569 g)k 2.8 gk (1.5

m/s) g)(4.5k (1.5

)m (m

νm ν

br

ir,rf

f

Recall the formula for kinetic energy.

22

1 mv KE

Was the kinetic energy conserved in example 1?

J 15.1875 m/s) kg)(4.5 (1.5 vm KE 22

12rr2

1red

J 0 m/s) kg)(0.0 (2.8 vm KE 22

12rr2

1black

KEbefore = 15 J

J 5.3 m/s) kg)(1.6 (4.3 vm KE 22

12fbr2

1after

KEbefore = 15 J KEafter = 5.3 J

KE = 5.3 J-15 J = -9.7 J

Where does the missing kinetic energy go?

Some energy is converted to internal energy, some is converted into sound.

Elastic CollisionsElastic CollisionsTwo objects collide and return

to their original shape with no loss of Kinetic Energy.

The two objects move separately after the collision.

Total momentum and KE remain constant. See page 230.Most collisions lose energy as

sound, internal elastic potential energy and friction…we assume perfect collisions.

Formula for Formula for elastic elastic collisionscollisions

m1v1,i + m2v2,i = m1v1,f + m2v2,f ½ m1v1,i

2 + ½ m2v2,i2 = ½ m1v1,f

2 + ½

m2v2,f2

Example 1. A 0.025 kg blue marble sliding to the right on a frictionless surface with a velocity of 17 m/s makes an elastic head-on collision with 0.025 kg red marble moving to the left with a velocity of 15 m/s. After the collision, theblue marble moves to the left with a velocity of 15.0 m/s. Find the velocity of the red marble.

Example 1.

Given:

Find:

Original Formula:

mb = 0.025 kg mr = 0.025 kg b,i = 17 m/sr,i, = -15 m/s b,f = -15 m/s

r,f

fr,rfb,bir,rib,b νm νm νm νm

r

fb,bir,rib,bfr, m

νm - νm νm ν

=17m/s=17m/s

kg 0.025

m/s) kg)(-15 (0.025 - m/s) kg)(-15 (0.025m/s) kg)(17 (0.025

m

νm - νm νm ν

r

fb,bir,rib,bfr,

Now…you can plug the numbers into the conservation of kinetic energy formula to check your work which they will ask you to do!

(Remember – there is NO LOSS OF KINETIC ENERGY with elastic collisions!!)

½ m1v1,i2 + ½ m2v2,i

2 = ½ m1v1,f2 + ½ m2v2,f

2

½ m1v1,i2 + ½ m2v2,i

2 = ½ m1v1,f2 + ½ m2v2,f

2

½ (.025)(17)2+ ½ (.025)(-15)2 = ½ (.025)(-15)2+ ½ (.025)(17)2

6.425J = 6.425J