Physics 111physics.valpo.edu/courses/p111/lectures/lecture13.pdf · 2004. 10. 11. · momentum, and...

Physics 111

Title page

Thursday,October 07, 2004

Physics 111 Lecture 13

• Ch 9: MomentumConservation of Momentum

• Ch 7: Work

Kinetic Energy

Help this week:Wednesday, 8 - 9 pm in NSC 118/119

Sunday, 6:30 - 8 pm in CCLIR 468

Help sessions

AnnouncementsThursOct

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Don’t forget to read over the lab andprepare for the short quiz.

labs

AnnouncementsThursOct

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In fact, we can relateforce to momentumchanges directly usingNewton’s 2nd Law:

Fnet =

ΔpΔt

F

net= ma = m

ΔvΔt

This form is completelygeneral, as it allows forchanges in mass as wellas velocity with time.

con’t

Ch 9: Momentum and CollisionsThursOct

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p mv=

By how much will the momentummomentum change?

That depends upon the amount of time overwhich the force is applied to the object.

v

f= v

i+ a(Δt)

mv

f= mv

i+ ma(Δt)

mv

f− mv

i=F

netΔt

Here we’ll derive the casefor a constant force.

con’t


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p mv=

But that’s just impulse. Sowe now have a relationshipbetween momentum andimpulse!

We define impulse:

I ≡

Fnet Δt = Δp Impulse-Momentum

Theorem

Impulse-Momentum Theorem


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p

f− p

i=F

netΔt

Worksheet #1

CQ1: impulse


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Worksheet #1

CQ1 – impulse ?

Consider two carts, of masses m and 2m, at rest on an airtrack. If you push first one cart for 3 s and then the other forthe same length of time, exerting equal force on each, themomentummomentum of the light cart is

PI, Mazur (1997)

≥≥

1. four times2. twice3. equal to4. one-half5. one-quarter

the momentum of the heavy cart.


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Motion Diagram: Problem #1

? (P1) Bonds - motion

A 0.25 kg baseball arrives at the plate with aspeed of 50 m/s. Barry Bonds swings and theball leaves his bat at 40 m/s headed directlyback toward the mound along the same pathon which it arrived. Assuming that the ball isin contact with the bat for 10-3 s, what is themagnitude of the average force the bat exertson the ball? Ignore gravity in this problem.

LAtimes.com


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a

v

? (P1) Bonds - picture

Ch 9: Momentum & CollisionsThursOct

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Pictorial Representation:

a

+x

vxi, tiKnowns:vx,i = - 50 m/svx,f = + 40 m/sΔti = 0.001 sm = 0.250 kg

Unknowns: <Fx,net>

vyf, tf

p

x ,i= mv

x ,i= (0.25 kg)(−50 m/s) = −12.5 kg m/s

p

x , f= mv

x , f= (0.25 kg)(40 m/s) = 10 kg m/s

I = Δp = p

f− p

i= (10 − (−12.5)) kg m/s

= 22.5 kg m/s

I =

F

netΔt

F

net=IΔt

= (22.5 kg m/s)

10−3s = 22,500 N rightward

? (P1) Bonds - math


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Mathematical Representation:

The total momentum of an isolated system ofobjects is conserved, regardless of the nature ofthe force between the objects in that system.

Here, we speak of thelinear momentum of the system.

Thus, if our system consists of two objects...

v1im1 m2

v2i

p1i= m11v1i p2i= m2v2i

Momentum…collisions


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The total initial momentum of the system is

After these two balls hit, the situation changes...

p2f= m2v2fp1f= m1v1f

v1fm1 m2

v2f

psys i = p1i + p2i = m1v1i + m22v2i

collisions


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v1im1 m2

v2i

p1i= m11v1i p2i= m2v2i

The total final momentum of the system is

And by conservation of momentum...

psys f = p1f + p2f = m1v1f + m2v2f

psys f = psys i

p1f + p2f = p1i + p2i

m1v1f + m2v2f = m1v1i + m2v2i

con’t


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p2f= m2v2fp1f= m1v1f

v1fm1 m2

v2f

• NOTE: our formulation above is derivedfor a system of two POINT MASS objects.

• With no external forces acting upon theseobjects, when the two objects collide, thetotal linear momentum before and afterthe collision are the same.

m1v1f + m2v2f = m1v1i + m2v2i

Conservation of Momentum


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Notice that this statement can also bederived through Newton’s Laws.

Let’s say the collision lasts for a time Δt.

FBDs:

con’t


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m1

F2→1

m2

F1→2

con’t


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m1

F2→1

m2

F1→2

aleft =Fnet ,leftm1

=F2→1m1

ΔvleftΔt

=F2→1m1

m1Δvleft =

F2→1Δt

aright =Fnet ,rightm2

=F1→2

m2

ΔvrightΔt

=F1→2

m2

m2Δvright =

F1→2Δt

Newton’s 3rd Law: F

2→1= −

F1→2

con’t


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m1

F2→1

m2

F1→2

m1Δvleft =

F2→1Δt m2Δ

vright =F1→2Δt

Δp1 = −

F1→2Δt Δ

p2 =F1→2Δt

Δp1 = −Δp2

Or stated slightlydifferently:

con’t


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Δp1 = −Δp2

p1, f −

p1,i = −( p2, f −p2,i )

p1, f +

p2, f = (p1,i +

p2,i )

ptot , f =

ptot ,i

Worksheet #2

CQ2: momentum


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Worksheet #2

CQ2 – momentum ?

Suppose rain falls vertically into an open cartrolling along a straight horizontal track withnegligible friction. As a result of the accumulatingwater, the speed of the cart _________.

PI, Mazur (1997)

≥≥

1. Increases2. Remains the same3. Decreases


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Work & Kinetic Energy

After developing an understanding of motionusing velocity, acceleration, forces,momentum, and impulse (all vector quantities),we’re now going to develop scalar tools thatwill help make some problems much simpler!

Ch 7: Work & Energy

Ch 7: Work and Kinetic EnergyThursOct

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We come to physics with at least a well-defined notion of what “work” is.However, in physics, the term has aVERY SPECIFIC meaning.

Work is the mechanical transfer of energybetween a system and its environment bypushes and pulls (forces) at the boundarywhere the forces are applied.

Work definition #1


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SystemMotional energy (kinetic, “K”)Stored energy (potential, “U”)Mechanical energy= K + U

Environment

Work: System & Environment


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Let’s try to provide acontext for understandingwhy “work” is such animportant concept inphysics.

Describe the following demonstrations:

(W3 & W4) Work demos


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Worksheet #31) Dropped ball

Worksheet #42) Tugged cart

Remember to identify the “system” and forces.

We come to physics with at least a well-defined notion of what “work” is.However, in physics, the term has aVERY SPECIFIC meaning.

The work done by a constant force on an objectis equal to the magnitude of the component ofthe force in the direction of motion of the objecttimes the displacement of the object.

Work definition in words


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Mathematically, thisstatement of worktranslates to

Where Δs is the distance the object travelsas measured along the path it follows (i.e.,its trajectory) and Fs is the component ofthe force in the direction of motion.

This statement is correct so long as theapplied force Fs is a constant.

Work: math definition


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W = FsΔs

[W ] = [Fs][Δs]

[W ] = N m Joule (J)

[W ] = kg m2

s2

Work: units

W = FsΔs


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We’re about to relate work to anew quantity by using an oldexpression...

s

f= s

i+ v

i(Δt) + 1

2a(Δt)2

v

f= v

i+ a(Δt)Now solve for Δt

(Δt) =

vf− v

i

a

s

f− s

i= Δs = v

i(Δt) + 1

2a(Δt)2

Plug back in above...

Relating the kinematic eqns to…


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Δs = vi

vf− v

i

a

⎛

⎝⎜

⎞

⎠⎟ +

1

2a

vf− v

i

a

⎛

⎝⎜

⎞

⎠⎟

2

Δs = (v

iv

f/ a) − (v

i2 / a) + 1

2(v

f2 − 2v

iv

f+ v

i2 ) / a

Δs = 1

2(v

f2 − v

i2 ) / a

aΔs = 1

2(v

f2 − v

i2 )

This last quantity looks something like work,right? All we need to do is multiply both sidesby mass, and we have force times distance!

something like work….


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ma

s(Δs) = F

s ,netΔs = 1

2m(v

s, f2 − v

s ,i2 )

Each term on the right-hand-side of this equation is called a“kinetic energy” term.

K mv= 12

2

And now we can rewrite work asa change in kinetic energy:

Wnet = ΔK = K f − Ki =12 mvf

2 − 12 mvi

2

kinetic energy


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[K ] = 1

2

⎡

⎣⎢

⎤

⎦⎥[m][v2]

[K ] = kg m2 / s2 = Nm = J

That’s good! It has the same units as work! WHEW!

Units: energy


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K mv= 12

2

Although we’ve derived this relationshipbetween work and kinetic energy for thespecial case of a constant applied force(i.e., constant acceleration), it turns outto be generally true, regardless of whetheror not the force is constant!

The NET WORK done on an object is equal to its change in KINETIC ENERGY.

generalization


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A man raises a pail of mass 5 kg at aconstant velocity to a height of 0.5 m.How much work has he done?

Problem #2

? (P2) Bucket

F


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1) - 49 J2) - 24.5 J3) 0 J4) 24.5 J5) 49.0 J

A man raises a pail of mass 5 kg at a constant velocityto a height of 0.5 m. How much work has he done?

Motion Diagram:

? (P2) Bucket - motion


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v

a= 0


FBD:

? (P2) Bucket - FBD


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Fman = Force of manon bucket

W = weight

3rd Law Pair????

NO!!


Pictorial Representation:

? (P2) Bucket - picture


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yf, tf, vf

yi, ti, vi

+y

Knowns:yi = 0; yf = 0.5 ma = 0; vf = viFy,man = W = m gm = 5.00 kg

Unknowns: W, v


Fman = W = (5.00 kg)(9.81 m/s2) = 49.1 N

Since the force is parallel to the displacement,the work done by the man is given by

Wman = Fman,s Δs = (49.1 N) (0.5 m) = 24.5 J

? (P2) Bucket - math


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Mathematical Representation:

Physics 111physics.valpo.edu/courses/p111/lectures/lecture13.pdf · 2004. 10. 11. · momentum, and...

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Transcript of Physics 111physics.valpo.edu/courses/p111/lectures/lecture13.pdf · 2004. 10. 11. · momentum, and...