PHYS 1443 – Section 001 Lecture #13

PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu

1Tuesday, June 28, 2011

PHYS 1443 – Section 001Lecture #13

Tuesday, June 28, 2011Dr. Jaehoon Yu

• Collisions• Center of Mass• Rotational Motion• Rotational Kinematics• Relationship Between Angular and

Linear quantities

Tuesday, June 28, 2011 2

Announcements• Reading Assignment

– CH9.10• Quiz #3 tomorrow, Wednesday, June 29

– Beginning of the class– Covers CH8.1 through CH9.9

• Planetarium Show extra credit– Must obtain the signature of the “Star Instructor” AFTER

watching the show on the ticket stub– Tape one side of the ticket stubs on a sheet of paper with

your name on it– Submit it on the last class Thursday, July 7

• Late submissions will not be accepted!!!PHYS 1443-001, Summer 2011 Dr.

Jaehoon Yu

Tuesday, June 28, 2011 PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu

3

Valid Planetarium Shows• Regular shows

– TX star gazing; Nanocam; Ice Worlds• Private shows for a group of 15 or more

– Bad Astronomy; Black Holes; IBEX; Magnificent Sun– Microcosm; Stars of the Pharaohs; Time Space– Two Small Pieces of Glass; SOFIA– Violent Universe; Wonders of the Universe

• Please watch the show and obtain the signature on the back of the ticket stub


4

Extra-Credit Special Project• Derive the formula for the final velocity of two objects which

underwent an elastic collision as a function of known quantities m1, m2, v01 and v02 in page 8 of this lecture note in a far greater detail than in the note.– 20 points extra credit

• Show mathematically what happens to the final velocities if m1=m2 and explain in detail in words the resulting motion.– 5 point extra credit

• NO Credit will be given if the process is too close to the note!• Due: Start of the class Tuesday, July 5

PHYS 1443-001, Summer 2011 Dr. Jaehoon Yu

5

Extra Credit: Two Dimensional Collisions

• Proton #1 with a speed 5.0x106 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and proton #2 deflects at an angle φ to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, φ. This must be done in much more detail than the book or on page 13 of this lecture note.• 10 points• Due beginning of the class Wednesday, July 6

Tuesday, June 28, 2011


6

Collisions

Consider a case of a collision between a proton on a helium ion.

The collisions of these ions never involve physical contact because the electromagnetic repulsive force between these two become great as they get closer causing a collision.

Generalized collisions must cover not only the physical contact but also the collisions without physical contact such as that of electromagnetic ones on a microscopic scale.

1 21dp F dtrr

t

F F12

F21

Assuming no external forces, the force exerted on particle 1 by particle 2, F21, changes the momentum of particle 1 by

Likewise for particle 2 by particle 1 2 12dp F dt

rr

Using Newton’s 3rd law we obtain So the momentum change of the system in a collision is 0, and the momentum is conserved

dpur

2

dpur12F dt

r21F dtr

−d pur1

d pur1 + d p

ur2

pursystem p

ur1 + p

ur2

constant0


7

Elastic and Inelastic Collisions

Collisions are classified as elastic or inelastic based on whether the total kinetic energy is conserved, meaning whether it is the same before and after the collision. A collision in which the total kinetic

energy and momentum are the same before and after the collision.

Momentum is conserved in any collisions as long as external forces are negligible.

Elastic Collision

Two types of inelastic collisions:Perfectly inelastic and inelastic

Perfectly Inelastic: Two objects stick together after the collision, moving together with the same velocity.Inelastic: Colliding objects do not stick together after the collision but some kinetic energy is lost.

Inelastic Collision

A collision in which the momentum is the same before and after the collision but not the total kinetic energy .

Note: Momentum is constant in all collisions but kinetic energy is only in elastic collisions.


8

Elastic and Perfectly Inelastic Collisions In perfectly inelastic collisions, the objects stick together after the collision, moving together. Momentum is conserved in this collision, so the final velocity of the stuck system isHow about elastic collisions?

1 21 2i im v m v+r r

In elastic collisions, both the momentum and the kinetic energy are conserved. Therefore, the final speeds in an elastic collision can be obtained in terms of initial speeds as


21

211 fi vvm

m1 v1i −v1 f m 2 v2i −v2 f

v1 f m 1 −m 2

m 1 + m 2

⎛⎝⎜

⎞⎠⎟v1i +

2m 2

m 1 + m 2

⎛⎝⎜

⎞⎠⎟v2i

1 2( ) fm m v +r

1 21 2

1 2( )i i

fm v m vv

m m+

+

r rr

1 21 2f fm v m v +r r

222

211 2

121

ii vmvm + 222

211 2

121

ff vmvm +

22

222 fi vvm

m1 v1i −v1 f v1i +v1 f m2 v2i − v2 f( ) v2i + v2 f( )

From momentum conservation above

v2 f 2m 1

m 1 + m 2

⎛⎝⎜

⎞⎠⎟v1i +

m 1 −m 2

m 1 + m 2

⎛⎝⎜

⎞⎠⎟v2i

What happens when the two masses are the same?


9

Example for CollisionsA car of mass 1800kg stopped at a traffic light is rear-ended by a 900kg car, and the two become entangled. If the lighter car was moving at 20.0m/s before the collision what is the velocity of the entangled cars after the collision?

pi

The momenta before and after the collision are

What can we learn from these equations on the direction and magnitude of the velocity before and after the collision?

m1

20.0m/sm2

vfm1

m2

Since momentum of the system must be conserved

pi pf

The cars are moving in the same direction as the lighter car’s original direction to conserve momentum. The magnitude is inversely proportional to its own mass.

Before collision

After collision

m1v1i + m 2v2i 0 + m 2v2i

p f

m1v1 f + m 2v2 f m1 + m 2 vf

m1 + m 2 vf m 2v2i

v f

m 2v2im1 + m 2

900 × 20.0900 +1800

6.67m / s


10

The mass of a block of wood is 2.50-kg and the mass of the bullet is 0.0100-kg. The block swings to a maximum height of 0.650 m above the initial position. Find the initial speed of the bullet.

Ex.9 – 11: A Ballistic Pendulum

m1v f 1 + m 2vf2

m1 + m 2

v01

What kind of collision?No net external force momentum conserved

Perfectly inelastic collision

m1v01 + m 2v02

m1v01fv

Solve for V01 1 2

1

fm m vm+

What do we not know? The final speed!!How can we get it? Using the mechanical

energy conservation!


11

Ex. A Ballistic Pendulum, cnt’d

v01

Using the solution obtained previously, we obtain

m1 + m 2 vf

m1

Now using the mechanical energy conservation

mgh

m1 + m 2 ghf

gh f

fv

12 mv2

12 m1 + m 2 vf

2

12 v f

2

2gh f 22 9.80m s 0.650 m

Solve for Vf

m1 + m 2 2ghf

m1

0.0100 kg+ 2.50 kg0.0100 kg

⎛⎝⎜

⎞⎠⎟ 2 9.80m s2 0.650 m

+896m s


12

Two dimensional Collisions In two dimension, one needs to use components of momentum and apply momentum conservation to solve physical problems.

1 21 2i im v m v+ r r

21 12

1i

vm

m2m1

v1i

m1

v1f

θ

m2v2f

φ

Consider a system of two particle collisions and scatters in two dimension as shown in the picture. (This is the case at fixed target accelerator experiments.) The momentum conservation tells us:


And for the elastic collisions, the kinetic energy is conserved:

What do you think we can learn from these relationships?

fxfx vmvm 2211 + fθ coscos 2211 ff vmvm +

iyvm 11 0 fyfy vmvm 2211 + m1v1 f sinθ − m2v2 f sinφ

11 im vr

ixvm 11

222

211 2

121

ff vmvm +

1 1 2 2ix ixm v m v+

1 1 2 2iy iym v m v+

x-comp.

y-comp.

1 21 2f fm v m v+r r

1 1 2 2fx fxm v m v+

1 1 2 2fy fym v m v+


13

Ex. 9 – 13: Two Dimensional CollisionsProton #1 with a speed 3.50x105 m/s collides elastically with proton #2 initially at rest. After the collision, proton #1 moves at an angle of 37o to the horizontal axis and proton #2 deflects at an angle φ to the same axis. Find the final speeds of the two protons and the scattering angle of proton #2, φ.

Since both the particles are protons m1=m2=mp.Using momentum conservation, one obtainsm2

m1v1i

m1

v1f

θ

m2v2f

φ

ipvm 1

Canceling mp and putting in all known quantities, one obtains

smv f /1080.2 51

From kinetic energy conservation:

3.50 ×105 2v1 f

2 +v2 f2 3

Solving Eqs. 1-3 equations, one gets

(1) 1050.3cos37cos 521 + fff vv

Do this at home

fθ coscos 21 fpfp vmvm +

fθ sinsin 21 fpfp vmvm 0

x-comp.

y-comp.

(2) sin37sin 21 fff vv

smv f /1011.2 52

0.53f


14

Center of MassWe’ve been solving physical problems treating objects as sizeless points with masses, but in realistic situations objects have shapes with masses distributed throughout the body. Center of mass of a system is the average position of the system’s mass and

represents the motion of the system as if all the mass is on the point.

Consider a massless rod with two balls attached at either end.

CMx

The total external force exerted on the system of total mass M causes the center of mass to move at an acceleration given by as if all the mass of the system is concentrated on the center of mass.

ar F

ur∑ / M

What does above statement tell you concerning the forces being exerted on the system?

m1 m2x1 x2

The position of the center of mass of this system is the mass averaged position of the systemxCM CM is closer to the

heavier object1 1 2 2m x m x+

1 2

m m+


15

Motion of a Diver and the Center of Mass

Diver performs a simple dive.The motion of the center of mass follows a parabola since it is a projectile motion.

Diver performs a complicated dive.The motion of the center of mass still follows the same parabola since it still is a projectile motion.

The motion of the center of mass of the diver is always the same.


16

Example 9 – 14 Thee people of roughly equivalent mass M on a lightweight (air-filled) banana boat sit along the x axis at positions x1=1.0m, x2=5.0m, and x3=6.0m. Find the position of CM.

Using the formula for CM

ii

iii

CM m

xmx

1.0M × 12.03

MM

M M M+ +

4.0( )m5.0M+ × 6.0M+ ×


17

Ex9 – 15: Center of Mass in 2-DA system consists of three particles as shown in the figure. Find the position of the center of mass of this system.

Using the formula for CM for each position vector component

ii

iii

CM m

xmx

One obtains

CMx

CMrr

If kgmmkgm 1;2 321

3 4 0.754

CMi jr i j+

+

r rr r r

m1y=2 (0,2)

m2x=1

(1,0)m3x=2

(2,0)

(0.75,1)rCM

ii

iii

CM m

ymy

ii

iii

m

xm

321

332211

mmmxmxmxm

++++

321

32 2mmm

mm++

+

CMy

ii

iii

m

ym

321

332211

mmmymymym

++++

321

12mmm

m++

CMx ir 2 3 1

1 2 3

2 2m m i m jm m m+ +

+ +

r rCMy j+

r


18

Center of Mass of a Rigid ObjectThe formula for CM can be extended to a system of many particles or a Rigid Object

A rigid body – an object with shape and size with mass spread throughout the body, ordinary objects – can be considered as a group of particles with mass mi densely spread throughout the given shape of the object

CMx

ii

iii

CM m

ymy

The position vector of the center of mass of a many particle system is

CMrr

xCM ≈Δm ixi

i∑

M

CMx

1CMr rdm

M

r r

Δmi

rirCM

ii

iii

CM m

zmz

CM CM CMx i y j z k + +r r r i i i i i i

i i i

ii

m x i m y j m z k

m

+ +

r r r

iii

CM

m rr

M

rr

M

xmi

ii

mi

Δ

Δ 0lim

1M

x dm∫

1 1 2 2 n nm x m x m x+ +×××+ i ii

m x i

i

m1 2

nm m m

+ +×××+


19

Ex 9 – 16: CM of a thin rod

The formula for CM of a continuous object is

Lx

xCM xdmM

x0

1

Therefore

L

x dxdm=λdx

Since the density of the rod (λ) is constant;

CMx

Show that the center of mass of a rod of mass M and length L lies in midway between its ends, assuming the rod has a uniform mass per unit length.

Find the CM when the density of the rod non-uniform but varies linearly as a function of x, λαx

CMxM

dxdm LM /

The mass of a small segment

Lx

xxdx

M 0

1 Lx

x

xM

0

2

211

2

211 L

M

ML

M 211

2L

Lx

xdx

0

Lx

xxdx

0α

Lx

x

x

0

2

21α 2

21 Lα

Lx

xxdx

M 0

1

Lx

xdxx

M 0

21 αLx

x

xM

0

3

311 α

3

311 L

Mα

ML

M 321

32L

CMx

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PHYS 1443 – Section 001 Lecture #13

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