Phys 2101 Gabriela González

•  There is potential energy associated with the work of conservative forces (gravity: U=mgh; springs: U= ½ kx2):

ΔU = Uf – Ui = -W

•  If the only forces doing work are conservative the sum of kinetic and potential energy is conserved:

ΔU + ΔK = 0 Uf + Kf = Ui + K

•  If there are non-conservative forces, the change in the total energy (potential + kinetic) is equal to the work done by those forces:

ΔU + ΔK = Wother

2

3

U(h)=mgh

h

x

U(x) = ½ kx2

Where is F=0? (equilibrium position)

4

€

ΔU = −W = − F(x)dx∫ ⇒ F(x) = −dU(x)dx

•  Force = slope of U(x) curve •  Equilibrium: F=0 = dU/dx. d2U/dx2>0: stable eq; •  If E=KE+U, when U(x)=E, KE =0 and v=0: turning point •  If E=KE+U, since KE is +ve, U<E: only certain positions are allowed.

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Figure 8-48 shows a plot of potential energy U versus position x of a 0.90 kg particle that can travel only along an x axis. (Nonconservative forces are not involved.) The particle is released at x = 4.5 m with an initial speed of 7.0 m/s, headed in the negative x direction. (a) If the particle can reach x = 1.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (b) magnitude and (c) direction of the force on the particle as it begins to move to the left of x = 4.0 m? Suppose, instead, the particle is headed in the positive x direction when it is released at x = 4.5 m at speed 7.0 m/s. (d) If the particle can reach x = 7.0 m, what is its speed there, and if it cannot, what is its turning point? What are the (e) magnitude and (f) direction of the force on the particle as it begins to move to the right of x = 5.0 m?

15J

35J

45J

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The center of mass of a body or systems of bodies is the point that moves as though ALL the mass were concentrated there and ALL external applied forces were applied there.

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......1

321

332211

1 +++

+++== ∑

= mmmrmrmrmrm

Mr i

N

iicom

Three masses, m1=1.2 kg, m2=2.5 kg, m3=3.4 kg, in an equilateral triangle with side equal to 140cm. Where is the center of mass?

r1 = 0 ;r2 = 140cm i ;

r3 = 140cm cos60o i + sin600 j( ) = 70cmi +121cmjrcom =

0 + 2.5kg1.40mi + 3.4kg 0.70mi +1.21m j( )1.2kg + 2.5kg + 3.4kg

=5.88kgm i + 4.11kgm j

7.1kg= 0.83mi + 0.58mj

Where is the center of mass if all masses are equal? Where is it of one mass is much heavier than the others?

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•  For a single particle,

•  For each particle in a system of particles,

•  For a system of particles,

amF =

Fi = mi

ai

Fext = M

acomNet forces!

Internal forces that cancel each other: •  tension; • explosions; • electromagnetic attractions or repulsions,…

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•  Definition of linear momentum for a single particle:

•  Newton’s law (again!?)

vmp =

dtpdFam

dtvdm

dtpd

=⇒==

•  Where is the force the largest? and smallest? and zero?

•  When is the particle slowing down?

more vectors

!

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•  Single particle:

•  Several particles:

vmp = am

dtpdF

==

...321 +++= pppP

( ) ( )

( )

( ) netcomcom FaMxMdtd

xmxmxmdtd

dtxdm

dtxdm

dtxdm

dtd

vmvmvmdtdppp

dtd

dtPd

===

+++=

+++=

++=+++=

2

2

3322112

2

33

22

11

332211321

...

...

......

If Fnet =0, momentum is conserved

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A man of mass m clings to a rope ladder suspended below a balloon of mass M. The balloon is stationary with respect to the ground.

a)  If the man begins to climb the ladder at speed v with respect to the ladder, in what direction will the balloon move?

b)  What is the state of motion after the man stops climbing?

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Ricardo, mass 95 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 35 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moved 40 cm relative to a submerged log during the exchange and calculates Carmelita's weight, which she has not told him. What is it?

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Change in momentum = “impulse”

JdttFpdtpdF

f

i

t

t

==Δ⇒= ∫ )(

integral = area under the curve

tFJdttFJ avg

t

t

f

i

Δ=⇒= ∫ )(

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The National Transportation Safety Board is testing the crash-worthiness of a new car. The 2300 kg vehicle, moving at 15 m/s, is allowed to collide with a bridge abutment, which stops it in 0.56 s. What is the magnitude of the average force that acts on the car during the impact?

National Highway Traffic Safety Administration: http://www.nhtsa.dot.gov/ http://www.safercar.gov/

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It is well known that bullets and other missiles fired at Superman simply bounce off his chest. Suppose that a gangster sprays Superman's chest with 10 g bullets at the rate of 100 bullets/min, the speed of each bullet being 700 m/s. Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest from the stream of bullets?

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"Rail guns are hyper-velocity weapons that shoot aluminum or clay rounds at just below the speed of light. In our film, we've taken existing stealth technology one step further and given them an X-ray scope sighting system," notes director Russell.

Warner Bros., production notes, 1996. http://movies.warnerbros.com/eraser/cmp/prodnotes.html#tech

Also: INSULTINGLY STUPID MOVIE PHYSICS: http://www.intuitor.com/moviephysics/

Conservation of momentum !!?? P0 = 0 = mSVS + mPVP

If mP~ 0.1kg, Vp~0.03 c~107 m/s and mS~100kg, then

Vs ~-mPvP/mS ~ 104 m/s ~22,000 mph

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In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 56 m/s (terminal speed), that his mass (including gear) was 85 kg, and that the magnitude of the force on him from the snow was at the survivable limit of 1.2x105 N. What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

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A pitched 140 g baseball, in horizontal flight with a speed vi of 39.0 m/s, is struck by a bat. After leaving the bat, the ball has a speed vf of 45.0 m/s at an upward angle of 30.0° . (a) What impulse J acts on the ball while it is in contact with the bat during the collision? (b) The impact time Δt for the baseball–bat collision is 1.20 ms. What average force acts on the baseball?

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In a collision, the system has a certain kinetic energy and a certain linear momentum before the collision. During the collision, the kinetic energy and linear momentum of each body are changed by the impulse from the other body. We will only consider collisions in systems that are closed (no mass enters or leaves them) and isolated (no net external forces act on the bodies within the system).

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•  Elastic collisions: kinetic energy is conserved •  Inelastic collisions: kinetic energy is not conserved •  The total linear momentum is always conserved! •  Before and after the collision, problems use

conservation of energy or work-energy theorem as before.

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m1v1i + m2

v2i = m1v1 f + m2

v2 f

Momentum is conserved:

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Momentum AND Energy is conserved:

m1v1i + m2

v2i = m1v1 f + m2

v2 fm1v1i

2 + m2v2i2 = m1v1 f

2 + m2v2 f2Collisions in a horizontal plane

(energy=kinetic energy):

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mv = (m + M )V12(m + M )V 2 = (m + M )gh⇒V = 2gh

v =(m + M )V

m=(m + M )

m2gh

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Initially, block 2 is at rest, spring is unstretched. Block 1 collides with Block 2, and they stick together. When the blocks momentarily stop, by what distance is the spring compressed?

m1vi = (m1 + m2 )vfv f = vmax12(m1 + m2 )vmax

2 =12kxmax

2

⇒ xmax =(m1 + m2 )

kvmax =

(m1 + m2 )k

m1

(m1 + m2 )vi =

m12

k(m1 + m2 )vi

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v0

If the collision is elastic, and we know v0, we can find out the velocities of each ball after the collision.