General Physics I (aka PHYS 2013)d.umn.edu/~vvanchur/2015PHYS2013/Slides2.pdfCHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8 REVIEW General Physics I (aka PHYS 2013) PROF.VANCHURIN (AKA VITALY)University

CHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8 REVIEW

General Physics I(aka PHYS 2013)

PROF. VANCHURIN(AKA VITALY)

University of Minnesota, Duluth(aka UMD)


OUTLINE

CHAPTER 5

CHAPTER 6

CHAPTER 7

CHAPTER 8


SECTION 5.1: USING NEWTON’S FIRST LAW

First Law. A body acted on by no net force, i.e.∑i

~Fi = 0

has a constant velocity (which may be zero) and zero acceleration.



Example 5.1. A gymnast with mass mG = 50 kg suspends herselffrom the lower end of a hanging rope of negligible mass. The upperend of the rope is attached to the gymnasium ceiling.(a) What is the gymnasts’s weight?(b) What force (magnitude and direction) does the rope exert on her?(c) What is the tension at the top of the rope?



Example 5.2. Find the tension at each end of rope in Example 5.1 ifthe weight of the rope is 120 N.



Example 5.4. A car of weight w rests on a slanted ramp attached to atrailer. (See figure below. Angle α is given. ) Only a cable runningfrom the trailer to the car prevents the car from rolling off the ramp.(The car brakes are off and its transmission is neutral.) Find thetension in the cable and the force that the ramp exerts on the car’s

tires.


SECTION 5.2: USING NEWTON’S SECOND LAW

Second Law. If a net external force acts on a body, the bodyaccelerates. The direction of acceleration is the same as the direction ofthe net force. The mass of the body times the acceleration vector of thebody equals to the net force vector, i.e.∑

i

~Fi = m~a (1)



Example 5.6. An iceboat is at rest on a frictionless horizontal surface.Due to the blowing wind, 4.0 s after the iceboat is released, it ismoving to the right at 6.0 m/s. What constant horizontal force FWdoes the wind exert on the iceboat? The combined mass of iceboat andrider is 200 kg.



Example 5.10. A toboggan loaded with physics students (totalweight w) slides down a snow-covered hill that slopes at a constantangle α. The toboggan is well waxed, so there is virtually no friction.(a) What is its acceleration? (b) What is normal force?


SECTION 5.3: FRICTION FORCESFriction force. There two types of contact forces betweenmacroscopic objects:

~n − normal force always perpendicular to the contact surface~f − friction force is always parallel to the contact surface .

Both forces arise due to microscopic (electromagnetic) interaction betweenmolecules, but we shall only study their macroscopic properties.

Kinetic Friction. If there is a motion along the surface of contact, thenthese two forces are related to each other by the so-called coefficientof kinetic friction

µk =fkn

(2)

orfk = µkn. (3)

Note that the friction coefficients are dimensionless, i.e. have no units.


SECTION 5.3: FRICTION FORCESStatic Friction If there is no motion along the surface of contact, thenthe friction force is bounded from above

fs ≤ (fs)max = µsn. (4)

It turns out that

µs > µk and thus fk < f maxs . (5)


SECTION 5.3: FRICTION FORCES

Table of Friction Coefficients Exact values of both coefficientsdepend on the materials in constant, e.g.

Material µs µk

steel on steel 0.74 0.57ice on steel 0.03 0.015

dry rubber on dry concrete 1.0 0.8wet rubber on wet concrete 0.3 0.25

There is also rolling friction which is typically much smaller. For steelwheels on steel rails it is ∼ 0.002− 0.003.



Example 5.13. You want to move a 500 N crate across a level floor.To start the crate moving, you have to pull with a 230 N horizontalforce. Once the crate starts to move, you can keep it moving atconstant velocity with only 200 N force. What are the coefficient ofstatic and kinetic friction?



Example 5.16. A toboggan loaded with physics students (fromExample 5.10) slides down a snow-covered hill. The wax has wornoff, so there is a nonzero coefficient of kinetic friction µk. The slope hasjust the right angle α to make the toboggan slide with constantvelocity. Find the angle in terms of w and µk.


SECTION 5.3: FRICTION FORCESFluid (air) resistance. The (magnitude of) force of fluid resistancedepends on the velocity,

f =

{kv for “small” velocitiesDv2 for “large” velocities

(6)

where the coefficients depend on: type of fluid, shape of object, etc.Terminal velocity. For small velocities the second law implies

mg− (kv) = ma (7)

This acceleration must stop when the terminal velocity is reached

vt =mgk

(8)

Similarly in the regimes of large velocities the second law implies

mg− (Dv2) = ma. (9)

and thus the terminal velocity is

vt =

√mgD. (10)


SECTION 5.3: FRICTION FORCESExample 5.18. For a human body falling through air in a spread-eagleposition, the numerical value of the constant D in Eq. (5.6) is about 0.25kg/m. Find the terminal speed for a 50 kg skydiver.


SECTION 5.4: DYNAMICS OF CIRCULAR MOTIONIn Sec. 3.4 we derived the following eqs. for uniform circular motion

a⊥ =v2

R(11)

and

a⊥ =4π2R

T2 (12)

where

a⊥ − magnitude of accelerationv − constant speedR − radius of circular pathT − period of motion. (13)

An object in such a motion experiences a constant (in magnitude)acceleration and thus according to second law

Fnet = ma⊥ = mv2

R= m

4π2RT2 . (14)


SECTION 5.4: DYNAMICS OF CIRCULAR MOTION

Example 5.19. A sled with a mass of 25.0 kg rests on a horizontal sheet ofessentially frictionless ice. It is attached by a 5.00 m rope to a post set in theice. Once given a push, the sled revolves uniformly in a circle around thepost . If the sled makes five complete revolutions every minute, find the forceF exerted on it by the rope.


SECTION 6.1: WORK

Constant speedSpeed of an object is unchanged if

I the sum of all forces is acting on it is zero.

I the sum of all forces is acting in the direction orthogonal to thedirection of motion.

An important observation is that in both cases any function of speedwould also not change, e.g.

f (v(t)) ∝ v(t)2 = const. (15)

Changing speedSpeed of an object is changing on if the net force acting on the bodyhas a component parallel to the direction of motion. Then for example

f (v(t)) ∝ v(t)2 6= const. (16)


SECTION 6.1: WORKWorkLet us define a physical quantity, we shall call work , which willquantify whether there is a component of constant force ~F parallel toa straight-line displacement,~s,

W = ~F ·~s = Fs cosφ. (17)Although the work is done by each and every force acting on a body,according to above equation it may be positive, negative or even zero.Units of work

[Work] = [Force]× [Distance]. (18)In SI units

1 J = 1 N×1 m. (19)Note that the dimensions of work can be also written as

[Work] = [Mass]× [Acceleration]× [Distance] (20)

or[Work] = [Mass]× [Velocity]2. (21)


SECTION 6.1: WORKExample 6.2. A farmer hitches her tractor to a sled loaded withfirewood and pulls it a distance of 20 m along ground. The totalweight of sled and load is 147000 N. The tractor exerts a constant5000 N force at an angle of 36.9◦ above the horizontal. A 3500 Nfriction force opposes the sled’s motion. Find the work done by eachforce acting on the sled and the total work done by the forces.


SECTION 6.2: KINETIC ENERGY

Back in second chapter we argued that 1D motion with constantacceleration is described by following equations

x(t) = x0 + v0xt +12

axt2

v(t) = v0x + axta(t) = ax. (22)

Then we derived a useful relation

2ax (x(t)− x0) = vx(t)2 − v20x (23)

which can be written as

max (x− x0) =12

mv2x −

12

mv20x. (24)

Note that this equation does not depend on time explicitly, but only throughtime-dependence of x and vx.


SECTION 6.2: KINETIC ENERGYKinetic energy. In 3D the equation can be written as

m~a ·~s =12

mv2 − 12

mv20 (25)

where the left hand side takes the form of the equation for work.What about the right hand side of (25)? If we define kinetic energy by

K(v) =12

mv2 (26)

then equation (25) tells us that the work acting on a system changesits kinetic energy by the amount equal to work

W = K(t)− K(t0) = ∆K. (27)

In fact what is relevant is the total work of all forces, i.e.

Wtot = ∆K. (28)

This is known as the work-energy theorem or as we will see later anequation representing conservation of energy.



Example 6.3. Let’s come beck to Example 6.2. What is the speed of the sledafter it moves 20 m? Suppose sled’s initial speed v1 = 2.0 m/s.



Example 6.4. The 200− kg steel hammerhead of a pile driver is lifted3.00 m above the top of a vertical I-beam being driven into the ground. Thehammerhead is then dropped, driving the I-beam 7.4 cm deeper into theground. The vertical guide rails exert a constant 60−N friction force on thehammerhead. Use the work-energy theorem to find (a) the speed of thehammerhead just as it heats the I-beam and (b) the average force thehammerhead exerts on the I-beam. Ignore effects of air.



Example 6.5. Two iceboats hold a race on a frictionless horizontal lake.

The two iceboats have masses m and 2m. The iceboats have identical sails, sothe wind exerts the same constant force ~F on each iceboat. They start fromrest and cross the finish line a distance s away. Which iceboat crosses thefinish line with greater kinetic energy.


SECTION 6.3: VARYING FORCESUp to now we were dealing with work done by constant forces

W = Fx(xf − xi

). (29)

This can generalized in two different ways:

I we can consider a motion along a curve and/or

I we can consider forces which change with position.

In the latter case the motion can be broken into small segments

W1 = F1∆xW2 = F2∆x

...

Wn = Fn∆x (30)

and by adding all these small pieces of work we getn∑

i=1

Wi =

n∑i=1

Fi∆x (31)


SECTION 6.3: VARYING FORCESIn the limit of infinite n this sum reduces to integral

W ≡ limn→∞

n∑i=1

Fi∆x =

∫ xf

xi

Fx(x)dx (32)

where Fx(x) is x-component of ~F(x) which is a function of position.For example, force required to stretch an ideal spring is

Fx(x) = kx. (33)

where k is the so-called spring constant measured

[k] =[Force]

[Distance].

From (32) and (33) the work need to stretch an ideal spring is

W =

∫ X

0Fx(x)dx =

[12

kx2]X

0=

12

kX2. (34)


SECTION 6.3: VARYING FORCESWork done on an ideal spring to stretch it

W =

∫ X

0Fx(x)dx =

[12

kx2]X

0=

12

kX2. (35)

Note that the position for non-stretched spring was set to x = 0, but ifthe origin is displaced then the equations would have to be modified

Fx(x) = k(x− x0)

W =12

k (X − x0)2. (36)


SECTION 6.3: VARYING FORCES

Example 6.6. A woman weighing 600 N steps on a bathroom scale thatcontains a stiff spring. In equilibrium, the spring is compressed 1.0 cmunder her weight. Find the force constant of the spring and the total workdone on it during compression.


SECTION 6.3: VARYING FORCES

Example 6.7. An air-track glider of mass 0.100 kg is attached to the end of ahorizontal air track by a spring with force constant 20.0 N/m. Initially thespring is unstretched and the glider is moving at 1.50 m/s to the right. Findthe maximum distance d that the glider moves to the right (a) is the air trackis turned on, so that there is no friction, and (b) if the air is turned off, sothat there is a kinetic friction coefficient µk = 0.47


SECTION 6.4: POWER

I So far we did not pay attention to time it takes to do the work.I The physical quantity which represents the rate at which the

work is done is called power.I This suggest that dimensions of power should be

[Power] =[Work]

[Time]. (37)

I In SI units power is measures in watts (W) which is defined as

1 W =1 J1 s

(38)

I Average power is defined from total work done in the same wayas average velocity was defined from total displacement

vavg =∆x∆t

Pavg =∆W∆t

. (39)


SECTION 6.4: POWER

Similarly the (instantaneous) power is defined as (instantaneous)velocity by taking limit of interval to zero

v ≡ lim∆t→0

∆x∆t

=dxdt

P ≡ lim∆t→0

∆W∆t

=dWdt. (40)

Since work is given byW = ~F ·~s (41)

the power can also be expressed as

P =dWdt

=d(~F ·~s

)dt

=d~Fdt·~s + ~F · d~s

dt(42)


SECTION 6.4: POWER

But for constant (time-independent) forces acting on a particle

d~Fdt

= 0 (43)

the power is

P =d~Fdt·~s + ~F · d~s

dt= ~F · ~v (44)

where~v =

d~sdt. (45)

is the velocity of particle.

Example 6.9. Each of the four jet engines on an Airbus A380 airlinerdevelops a thrust (a forward force on the airliner) of 322000 N (72000 lb).When the airplane is flying at 250 m/s (900 km/h, or roughly 560 mi/h) whatis horsepower does each engine develop.


SECTION 7.1: GRAVITATIONAL POTENTIAL ENERGYPotential Energy

I So far the only form of energy that we defined was kinetic energywhich is a function of velocity

K(v) =12

mv2. (46)

I It turns out that it is also useful to define another form of energypotential energy - which is a function of position, e.g.

U(x) = mgx or U(x) =12

kx2. (47)

I The idea is that it might be possible to store energy in the formof potential energy by placing an object at certain position.

I The process of storing energy require work to be done on theobject, but later on the potential energy can be released.


SECTION 7.1: GRAVITATIONAL POTENTIAL ENERGYGravitational Potential Energy

I The most familiar example of potential energy is thegravitational potential energy, when the work must be done onan object to lift it up (or down).

I For example, if on object has mass m and is lifted from height x1to height x2, then the work done by gravitational force is

Wgrav =(−mg i

)((x2 − x1) i

)= mgx1 −mgx2 (48)

I Then it is convenient to define a gravitational potential energy

Ugrav ≡ mgx (49)

and then the work done by gravitational force can be written as

Wgrav = −∆Ugrav = Ugrav,1 −Ugrav,2. (50)


SECTION 7.1: GRAVITATIONAL POTENTIAL ENERGYGravitational Potential Energy

I Now if we assume that there are no other forces acting on abody, then according to work-energy theorem

Wgrav = Kf − Ki (51)

and by combining (50) and (51) we get

Ugrav,i −Ugrav,f = Kf − Ki (52)

orUgrav,i + Ki = Ugrav,f + Kf . (53)

I The latter equation suggest that the total energy is unchanged

E ≡ K + Ugrav = constant. (54)

I More generally there might be other forces acting on a givensystem and then the work energy theorem would imply

Wother = Ef − Ei (55)

I Conservation of energy is the first of many conservation laws.


SECTION 7.1: GRAVITATIONAL POTENTIAL ENERGY

Example 7.1. You throw a 0.145− kg baseball straight up, giving it aninitial velocity of magnitude 20.0 m/s. How high it goes?



Example 7.2. In Example 7.1 suppose your hand moves upward by 0.50 mwhile you are throwing the ball. The ball leaves your hand with an upwardvelocity of 20 m/s. (a) Find the magnitude of the force (assumed constant)that your hand exerts on the ball. (b) Find the speed of the ball at a point15.0 m above the point where it leaves your hand. Ignore air resistance.



Example 7.3. A batter hits two identical baseballs with the same initialspeed and from the same initial height but at different initial angles. Provethat both balls have the same speed at any height h is air resistance can beneglected.


SECTION 7.1: GRAVITATIONAL POTENTIAL ENERGYExample 7.6. We want to slide a 12− kg crate up a 2.5−m-long rampinclined at 30◦ angle. A worker, ignoring friction, calculates that he can dothis by giving it an initial speed of 5.0 m/s at the bottom and letting it go.But, friction is not negligible; the crate slides only 1.6 m up the ramp, stops,and slides back down. (a) Find the magnitude of friction acting on crate,assuming that it is constant. (b) How fast is the crate moving when itreaches the bottom of the ramp?


SECTION 7.2: ELASTIC POTENTIAL ENERGY

I Remember that the work done on a spring is given by

W =12

kx2f −

12

kx2i (56)

but so the work done by a spring is

Wel =12

kx2i −

12

kx2f (57)

Similarly to work done by gravity,

Wgrav = −∆Ugrav = mgyi −mgyf (58)

the work done by spring can be expressed as

Wel = −∆Uel =12

kx2i −

12

kx2f (59)



I The main difference is that

Ugrav = mgy (60)

butUel =

12

kx2 (61)

where the origin corresponds to unstretched spring.

I It is sometime useful to define the total potential energy

U = Ugrav + Uel = mgy +12

kx2 (62)

and the total mechanical energy as

E = U + K (63)

I We can also modify the work-energy theorem further

Wother = Ef − Ei. (64)



Example 7.7. A glider with mass m = 0.200 kg sits on a frictionlesshorizontal air track, connected to a spring with constant k = 5.00 N/m. Youpull on the glider, stretching the spring 0.100 m, and releasing it from rest.The glider moves back towards its equilibrium position (x = 0). What is itsx-velocity when x = 0.080 m?


SECTION 7.2: ELASTIC POTENTIAL ENERGYExample 7.9. A 2000− kg (or 19600−N) elevator with broken cables in atest rig is falling at 4.00 m/s when it contacts a cushioning spring at thebottom of the shaft. The spring is intended to stop the elevator, compressing2.00 m as it does so. During the motion a safety clamp applies a constant17000−N frictional force to the elevator. What is the necessary forceconstant k for the spring.


SECTION 7.3: CONSERVATIVE AND

NONCONSERVATIVE FORCES

I We were able to define potential energy associated with workdone by gravitational and elastic forces. All such forces arecalled conservative forces.

I Work done by conservative forces:I can always be expressed as difference between initial and

final values of a suitably defined potential energyI it is reversible and is independent on the trajectory of the

body, but only on initial and final pointsI One might wonder if it is possible to do the same for all

macroscopic forces which would allow to rewrite thework-energy theorem as a simple law of conservation of energy.

I It turns out that there are other or non-conservative forces forwhich it is not possible to define potential energy.

I For example, frictional force or air resistance forces arenonconservative.


SECTION 7.2: CONSERVATIVE / NONCONSERVATIVE

Example 7.11. In a region of space the force of an electron is ~F = C x j,where C is a positive constant. The electron moves around a square loop inthe xy-plane. Calculate the work done on the electron by the force ~F during acounterclockwise trip around square.



I The non-conservative forces cannot be described in terms ofmechanical potential energies, but one can still associate withthe other energies such as internal energy.

I For example the frictional force is non-conservative, but whenfriction is applied to objects in contact the internal properties ofobjects change. In particular friction leads to increase intemperature or internal energy.

I On the microscopic level it corresponds to the change of kineticenergy of individual molecules. But from a macroscopic point ofview one can think of work done by all non-conservative forcesas a measure of change of internal energies

Wother = −∆Uint (65)

and the work energy theorem implies the conservation law

∆K + ∆U + ∆Uint = 0. (66)



I The main difference is that one cannot use internal energy to doany useful work.

I This follows from the so-called second law of thermodynamicsthat we will see later in the course.

I The conservation law of energy and the second law ofthermodynamics had passed a very large number of tests, butthis does not stop people from trying to build a perpetualmotion machine.

I Perpetual motion machines:

I first kind (do work without input of energy)I second kind (do work using internal energy).


SECTION 7.4: FORCE AND POTENTIAL ENERGY

I Work done on a particle when displacement is small

W = Fx(x)∆x (67)

which can also be expressed as a change in potential energy

W = −∆U. (68)

By equating (67) and (68) we get

Fx(x)∆x = −∆U (69)

or in the limit of small infinitesimal displacement

Fx(x) = − lim∆x→0

∆U∆x

= −dUdx. (70)

For example

U =12

kx2 ⇒ Fx(x) = −kx (71)

orU = mgx ⇒ Fx(x) = −mg. (72)



I This can be easily generalized to 3D

~F = −(∂U(x, y, z)

∂x,∂U(x, y, z)

∂y,∂U(x, y, z)

∂z

). (73)

It is convenient to define a gradient operator (pronounced nabla)

~∇ ≡(∂

∂x,∂

∂y,∂

∂z

)(74)

such that~F = −~∇U(x, y, z). (75)

For example the gravitational potential energy is

U = mgy (76)

and so

~F = −~∇(mgy) = −(

i∂

∂x+ j

∂

∂y+ k

∂

∂z

)(mgy) = −mgj. (77)



Example 7.14. A puck with coordinates x and y slides on a level,frictionless air-hockey table. It is acted on by a conservative force describedby the potential-energy function

U(x, y) =12

k(x2 + y2) . (78)

Find a vector expression for the force acting on the puck, and find anexpression for the magnitude of the force.


SECTION 7.5: ENERGY DIAGRAM

I Physical systems can be described in terms of energy diagrams.I For example harmonic oscillator whose potential energy is

U =12

kx2 (79)

and total energy isE = K + U. (80)

I The main point is to study the possible solutions qualitativelywithout having to solve equations of motion quantitatively.




SECTION 7.5: ENERGY DIAGRAMWhich statement correctly describes what happens to particlewhen it is at the maximum (of red line) between x2 and x3?

1. The particle’s acceleration is zero.2. The particle accelerates in the positive x-direction; the

magnitude of the acceleration is less than at any otherpoint between x2 and x3.

3. The particle accelerates in the positive x-direction; themagnitude of the acceleration is greater than at any otherpoint between x2 and x3.

4. The particle accelerates in the negative x-direction; themagnitude of the acceleration is less at any other pointbetween x2 and x3.

5. The particle accelerates in the negative x-direction; themagnitude of the acceleration is greater than at any otherpoint between x2 and x3.



I The total energy was defined as a sum

E(

x,dxdt

)=

12

m(

dxdt

)2

+12

kx2. (81)

I However, a more fundamental quantity (called Lagrangian) is

L(

x,dxdt

)=

12

m(

dxdt

)2

− 12

kx.2 (82)

I For any given trajectory x(t) Lagrangian L is a function of timeand thus can be integrated over some interval, i.e.∫ tf

ti

L(

x,dxdt

)dt. (83)

I Clearly, for any trajectory x(t) the integral would be some realnumber and one might wonder what trajectory would producethe smallest (or largest) number? Answer: Classical trajectories.

I Note, however, that all other trajectories are also possible!


SECTION 8.1: MOMENTUM AND IMPULSE

I Newton’s second law can be expressed as∑i

~Fi =d~pdt

(84)

where the new quantity (known as momentum) is defined as:

~p ≡ m~v = md~xdt. (85)

I In context of the so-called Hamiltonian mechanics it is defined as

~p =∂E(x, dx

dt

)∂( dx

dt ). (86)

I For example if

E(~x,

d~xdt

)=

12

m(

d~xdt

)2

+12

kx2 ⇒ ~p =∂E(~x, d~x

dt

)∂( d~x

dt )= m

d~xdt(87)



I The units of momentum are given by

[Momentum] = [Mass]× [Velocity] (88)

and for example in SI units

1 N · 1 s = 1 kg ·m/s. (89)

I Alternatively one can write units of momentum as

[Momentum] = [Mass]× [Velocity]

= [Mass]× [Acceleration]× [Time]= [Force]× [Time] (90)

I The latter form suggests that when some force is applied to asystem for some time then, it may effect the momentum.


SECTION 8.1: MOMENTUM AND IMPULSEI If we define impulse (of the time-independent net force) as

~J =∑

i

~Fi∆t (91)

then from (84) we get~J = ~pf − ~pi. (92)

I When the net force is time dependent the impulse is defined as

~J =

∫ tf

ti

∑i

~Fidt (93)

and from (84) we get

~J =

∫ tf

ti

∑i

~Fidt =

∫ tf

ti

d~pdt

dt = ~pf − ~pi (94)

I Note also that if the average force is defined as

~Favg ≡

∫ tfti

∑i~Fidt

tf − ti(95)

then the impulse is just~J = ~Favg(tf − ti) (96)



I Similarly to the energy conservation which is fundamentallydue to time-shift symmetry of physics laws, the momentumconservation is due to space-shift symmetry.

I Conservation of energy expresses changes in time

~J =∑

i

~Fidt = ~pf − ~pi (97)

I Conservation of momentum expresses changes in space

W =∑

i

~Fi · d~l = Kf − Ki. (98)

I The fact that the two expressions look so much alike might besurprising at first but this is what led people to eventuallydiscover a more fundamental and unified conserved quantitythe energy-momentum tensor as well as other conservedquantities such as electric charge.



Example 8.1. We can now go back to the example 6.5 where we considered arace of two iceboats on a frictionless frozen lake. The boats have masses mand 2m, and the wind exerts the same constant horizontal force ~F on eachboat. The boats start from rest and cross the finish like a distance s away.Which boat crosses the final line with greater momentum.



Example 8.2. You throw a ball with a mass of 0.40 kg against a brick wall.It hits the wall moving horizontally to the left at 30 m/s and reboundshorizontally to the right at 20 m/s. (a) Find the impulse of the net force onthe ball during its collision with the wall. (b) If the ball is in contact with thewall for 0.010 s, find the average horizontal force that the wall exerts on theball during impact.


SECTION 8.2: CONSERVATION OF MOMENTUM

I Similarly to the energy conservation which is fundamentallydue to time-shift symmetry of physics laws, the momentumconservation is due to space-shift symmetry.

I Conservation of energy expresses changes in time

~J =∑

i

~Fidt = ~pf − ~pi (99)

I Conservation of momentum expresses changes in space

W =∑

i

~Fi · d~l = Kf − Ki. (100)

I The fact that the two expressions look so much alike might besurprising at first but this is what led people to eventuallydiscover a more fundamental and unified conserved quantitythe energy-momentum tensor as well as other conservedquantities such as electric charge.


SECTION 8.2: CONSERVATION OF MOMENTUMIt is useful to distinguish two types of forces:

I internal forces (forces exerted by objects inside the system)

I external forces (forces exerted by objects outside of the system)

When there are no external forces the systems is said to be closed orisolated. For isolated systems one can write down theimpulse-momentum theorem for each object separately

~FA on B =d~pB

dt

~FB on A =d~pA

dt(101)

but because of the Newton’s third law

~FA on B = −~FB on A (102)

we getd~pB

dt= −d~pA

dt. (103)


SECTION 8.2: CONSERVATION OF MOMENTUMIf we now define the total momentum of all particles as

~P = ~pA + ~pB (104)

then we get the law of conservation of total momentum

d~Pdt

= 0. (105)

This is true for an arbitrary collection of particles, i.e.

~P = ~pA + ~pB + ~pC + ... = mA~vA + mB~vB + mC~vC + ... (106)

given that there are no external (but only internal) forces.Note that since (105) is a vector equation we have three equatiions

dPx

dt= d(mAVAx+mBVBx+mCVCx+...)

dt = 0

dPy

dt=

d(mAVAy+mBVBy+mCVCy+...)

dt = 0

dPz

dt= d(mAVAz+mBVBz+mCVCz+...)

dt = 0 (107)


SECTION 8.2: CONSERVATION OF MOMENTUM

Example 8.4. A marksman holds a rifle of mass mR = 3.00 kg loosely, so itcan recoil freely. He fires a bullet of mass mB = 5.00 g horizontally with avelocity relative to the ground of vBx = 300 m/s. What is the recoil velocityvRx of the rifle? What are the final momentum and kinetic energy of thebullet and rifle?


SECTION 8.2: CONSERVATION OF MOMENTUMExample 8.6. Consider two batting robots on a frictionless surface. RobotA, with mass 20 kg, initially moves at 2.0 m/s parallel to the x-axis. Itcollides with robot B, which has mass 12 kg and is initially at rest. After thecollision, robot A moves at 1.0 m/s in a direction that makes an angleα = 30◦ with its initial direction. What is the final velocity of robot B.


SECTION 8.3: INELASTIC COLLISIONSIn this section we will discuss instantaneous events - we callcollisions - which suddenly change the kinetic energies of objects.

I If forces between colliding objects are conservative, then thetotal kinetic energy right before and right after collision is thesame and the the collision is called elastic.

∆K = 0 (108)

I If the forces are non-conservative, then the total kinetic energy isnot conserved and the collision is called inelastic (or completelyinelastic is objects stick together).

∆K 6= 0 (109)

The key point is that although the (kinetic) energy might not beconserved, the momentum is still conserved for both elastic andinelastic collisions:

∆~P = 0 (110)


SECTION 8.3: INELASTIC COLLISIONSConsider a completely inelastic two-bodies collision, i.e.

~vA2 = ~vB2 = ~v2 (111)

then conservation of momentum implies

mA~vA1 + mB~vB1 = (mA + mB)~v2 (112)

or~v2 =

mA~vA1 + mB~vB1

(mA + mB). (113)

For example if object B was originally at rest, then

v2 =mA

(mA + mB)vA1 (114)

and then after the (completely inelastic) collision

K2

K1=

mA

(mA + mB)< 1 (115)

or the kinetic energy after the collision is lower than before collision

K2 < K1. (116)


SECTION 8.3: INELASTIC COLLISIONSExample 8.8. Consider a ballistic pendulum, a simple system of measuringspeed of a bullet. A bullet of mass mB makes a completely inelastic collisionwith a block of wood of mass mW , which is suspended like a pendulum. Afterimpact, the block swings up to a maximum height y. In terms of y, mB, andmW , what is the initial speed v1 of the bullet.


SECTION 8.3: INELASTIC COLLISIONS

Example 8.9. A 1000− kg car traveling north at 15 m/s collides with a2000− kg truck traveling east at 10 m/s. The occupants, wearing seat belts,are uninjured, but the two vehicles move away from the impact point as one.The insurance adjustor asks you to find the velocity of the wreckage justafter the impact. What is your answer?


SECTION 8.4: ELASTIC COLLISIONS

In elastic collisions the total kinetic energy before and after collisionsis unchanged and so one can use both conservation of energy andmomenta

∆K = 0∆~P = 0 (117)


SECTION 8.4: ELASTIC COLLISIONS

Example 8.12. Consider an elastic collision of two pucks (massesmA = 0.500 kg and mB = 0.300 kg) on a frictionless air-hockey table. PuckA has an initial velocity of 4.00 m/s in the positive x-direction and a finalvelocity of 2.00 m/s in an unknown direction α. Puck B is initially at rest.Find the final speed vB2 of puck B and the angles α and β.


SECTION 8.5: CENTER OF MASSThe conservation of momentum can be restated in terms of center ofmass defined as the mass-weighted average position of particle

~rcm ≡m1~r1 + m2~r2 + m3~r3 + ...

m1 + m2 + m3 + ...=

∑i mi~ri∑i mi

.

Then the total momentum can be written as

~P = (m1 + m2 + m3 + ...)d~rcm

dt(118)

or simply~P = M~vcm

whereM =

∑i

mi

is the total mass and~vcm =

d~rcm

dtis the velocity of the center of mass.


SECTION 8.5: CENTER OF MASS

Example 8.14. James (mJ = 90.0 kg) and Ramon (mR = 60.0 kg) are 20 mapart on a frozen pond. Midway between them is a mug of their favoritebeverage. They pull on the ends of a light rope stretched between them.When James has moved 6.0 m towards the mug, how far and in whatdirection has Ramon moved?


SECTION 8.5: CENTER OF MASS

If there are external forces acting on the objects, then the totalmomentum changes as

∑~Fext =

d~Pdt

=ddt

(M~vcm) = M~acm.

Example. Will the center of mass continue on the same parabolic trajectoryeven after one of the fragments hits the ground? Why or why not?


CHAPTER 0: MATHEMATICS

I AlgebraI Solving linear equations (numerically/algebraically)I Solving quadratic equations (numerically/algebraically)I Solving system of two equations with two unknowns

(numerically/algebraically)I Trigonometry

I Hypotenuse from two sides (numerically/algebraically)I Side from hypotenuse and another side

(numerically/algebraically)I Angle from hypothenuse and one side

(numerically/algebraically)I Angle from two sides, etc. (numerically/algebraically)

I CalculusI Differentiation of standard functionsI Product rule, quotient rule, chain rule.I Indefinite integrals (i.e. antiderivitives)I Definite integrals over interval


CHAPTER 5: APPLYING NEWTON LAWSI Coordinate system

I Choosing 1D, 2D or 3D coordinate systemI Drawing a free-body diagram for each object

I Using Newton’s First LawI Identifying equilibrium or motion without accelerationI Applying First Law to every object in equilibrium

I Using Newton’s Second LawI Identifying non-equilibrium or motion with accelerationI Applying Second Law to every object out of equilibrium

I Using Newton’s Third LawI Identifying action-reaction pairs for objects in contactI Applying Third Law to every action-reaction pair

I Frictional ForceI Static and kinetic frictional forces and frictional coefficientsI Relation between normal force and frictional forces

I Circular motionI Relation between radius, velocity and accelerationI Relation between radius, period and acceleration


CHAPTER 6: WORK AND KINETIC ENERGY

I Work

I Work done by constant force for straight-line displacementI Average acceleration and instantaneous acceleration

I Kinetic Energy

I Definition of kinetic energyI Applying the work-energy theorem to every object

I Varying forces

I Work done by varying forces and along straight lineI Work done by varying forces and along curved lineI Applying the work-energy theorem to varying forces

I Power

I Definition of average and instantaneous powerI Power from velocity of a particle and force acting on it


CHAPTER 7: POTENTIAL ENERGY

I Potential EnergyI Definition of gravitational potential energyI Definition of elastic potential energyI Understanding conservation of total mechanical energyI Applying the Work-energy theorem for every object

I Conservative and Nonconservative forcesI Distinguishing conservative and non-conservative forcesI Applying energy conservation for only conservative forcesI Applying work-energy theorem with nonconservative force

I Force and potential energyI Expressing force as a derivative of potential energy in 1DI Expressing force as a gradient of potential energy in 3D

I Energy diagramsI Drawing energy diagramsI Determining stable/unstable equilibriums


CHAPTER 8: MOMENTUM, IMPULSE, COLLISIONS

I Momentum and ImpulseI Definition of momentumI Definition of impulseI Applying the impulse-momentum theorem to every object

I Conservation of momentumI Conservation of the total momentum of all objectsI Applying the conservation of total momentum

I Inelastic collisionsI Distinguishing inelastic and elastic collisionsI Applying momentum conservation to inelastic collisions

I Elastic collisionsI Applying momentum conservation to elastic collisionsI Applying energy conservation to elastic collisions

I Center of massI Definition of the center of massI Applying second law for motion of center of mass

General Physics I (aka PHYS 2013)d.umn.edu/~vvanchur/2015PHYS2013/Slides2.pdfCHAPTER 5 CHAPTER 6 CHAPTER 7 CHAPTER 8 REVIEW General Physics I (aka PHYS 2013) PROF.VANCHURIN (AKA VITALY)University

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