!"#$%"$#&$'(()*$+"+&,-.+$/",0&12'3",4$

56! 7&/8%&$#98/9$":;&/-0$'1&$8,/).%&%$8,$-9&$

0*0-&+6$

<6! =&)'32&$-"$-9&$0*0-&+>$8%&,3?*$-9&$8,-&1,')$

',%$&@-&1,')$?"1/&06$

A6! B&18?*$-9'-$-9&$0*0-&+$80$80")'-&%6$

C6! D&-$-9&$E,')$+"+&,-.+$"?$-9&$0*0-&+$&F.')$

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+"+&,-.+$80$'$2&/-"1>$86&6>$8?$'$/'1$+'G&0$'$HI

-.1,$-9&,$-1'2&)$#8-9$0'+&$0(&&%$'0$-9'-$(18"1$-"$

-.1,8,J>$-9&$+"+&,-.+$80$,"-$-9&$0'+&6$$

!"!#$%&!''(#)%"*+'

K! !,-./01-'23/'4.'3/3567.8'46'19:/;'<.20,=9'

K! %>.'-,-./01-',?'3'23='322:8./0':9'.@135'0,'0>.'<.20,='91-'

,?'0>.'-,-./01-',?'.32>'23='A'B'C'4.?,=.'0>.'2,55:9:,/D'

A'

C'

L",0&12'3",$"?$+"+&,-.+$

L",0&12'3",$"?$+"+&,-.+$

?"1$8,%828%.')$$/"+(",&,-0$

M@>$*>$NO$

P)'03/$',%$Q,&)'03/$L"))808",0$$

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

R$/"))808",$8,$#98/9$-9&$-"-')$G8,&3/$&,&1J*$',%$+"+&,-.+$

'1&$.,/9',J&%$:&?"1&$',%$'S&1$-9&$/"))808",6$$$

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

Elastic Collision!

Two types of inelastic collision: Perfectly inelastic and inelastic

E.=?.2056'F/.539G2H'T#"$":;&/-0$03/G$-"J&-9&1$'S&1$-9&$/"))808",>$+"28,J$

-"J&-9&1$'-$'$/&1-'8,$2&)"/8-*6'

F/.539G2H'L"))8%8,J$":;&/-0$%"$,"-$03/G$-"J&-9&1$'S&1$-9&$/"))808",$:.-$0"+&$G8,&3/$

&,&1J*$80$)"0-6'

Inelatic Collision!

R$/"))808",$8,$#98/9$-9&$-"-')$G8,&3/$&,&1J*$80$,"-$-9&$0'+&$

:&?"1&$',%$'S&1$-9&$/"))808",>$:.-$+"+&,-.+$806$

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

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)"IIF+F"$+'

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-./$0&(&"(*%"&1$)(2(((

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.539G2',=':/.539G2'2,55:9:,/P'.539G2'

56$P)'03/$/"))808", $ $I $$$$$$$U$V$/",0-',-$

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K! TV'L;UTW'-X9.2U'Y'TZ'L;UT[DZ'-X9.2U''R''TV'L;'Y'Z'L;UT"3S.=U'''

K! ''''''''''''''''''''''''VD\'L;D-X9.2''R''T]'L;UT"3S.=U'''

K! ''''''''''''''''''''''''''''''''''''"3?=VD\'X]X']'L;'''''''''''''''''''

$"3S.='R''[D]''-X9.2'

$"3S.=R'$

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9D''_:/8'0>.'<.5,2:06',?'0>.'^9>':--.8:30.56'3S.='`51/2>aD''

The mass of the 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.!

P@6$R$Z'))803/$[&,%.).+$

What kind of collision?

No net external force ! momentum conserved

Perfectly inelastic

Solve for V01

What do we not know? The final speed!!

How can we get it? Using the mechanical

energy conservation!

Using the solution obtained previously, we obtain

Now using the mechanical energy conservation

Solve for Vf

Continued: A Ballistic Pendulum!

['13/)&$5X$

['13/)&$<X$

['13/)&$AX$

['13/)&$CX$$

['13/)&$\X$$

Momentum Conservation (N-particles)!

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+"+&,-.+$

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0.1?'/&$',%$80$0-1./G$:*$'$0&/",%$(./G$R>$#98/9$

#'0$"18J8,'))*$-1'2&)8,J$'-$C]6]$+^0$',%$#98/9$80$

%&_&/-&%$A]6]`$?1"+$8-0$"18J8,')$%81&/3",6$$[./G$

Z$'/F.81&0$'$2&)"/8-*$'-$'$Ca6]`$',J)&$-"$-9&$

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-9&$0'+&$+'006$$3U''),-O10.'0>.'9O..8',?'.32>'O12L'

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D").3",X$M'O$T9&$@$',%$*$/"+(",&,-0$"?$-9&$+"+&,-.+$'1&$

0&('1'-&)*$/",0&12&%6$$=8J9-V("0832&$@>$H(V("0832&$*6$$D-'1-$#8-9$@$

%81&/3",$M0.:0/18(-$b5c$1&(1&0&,-0$Z&?"1&>b<c$1&(1&0&,-0$RS&1$

/"))808",O>$$'p1x = p2x!

!

m1vA1

= m1vA 2cos30 + m

2vB 2cos45

d'00$/',/&)$08,/&$:"-9$(./G0$9'2&$-9&$0'+&$+'00>$$$

!

vA1

= vA 2cos30 + v

B 2cos45

\"#$8,$-9&$*$%81&/3",>$$$5$

p1y = p2y!

!

0 = mvA 2 sin30 "m v

B 2 sin45 (2nd term is negative)

!

0 = vA 2sin30 " v

B 2sin45

!

vA 2

= 2 "2

2vB 2

= 2vB 2

<$

D.:03-.-&$PF6$<$8,-"$PF6$5X$

!

vA1

= 23

2vB 2

+ vB 2

2

2=1.93v

B 2

!

vB 2

= 20.70 m/s

!

vA 2

= 20.7 " 2 = 29.27 m/s

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/"))808",$8,$('1-$M'OX$

!

K1

=1

2mv

A1

2

!

K2

=1

2mv

A 2

2

+1

2mv

B 2

2

!

K2

K1

=1

2mv

A 2

2

+1

2mv

B 2

2"

# $

%

& ' ÷

1

2mv

A1

2"

# $

%

& '

!

=vA 2

2

+ vB 2

2

vA1

2

!

K2

K1

=vA 2

2

+ vB 2

2

vA1

2

!

=29.27

2+ 20.70

2

402

!

=29.27

2+ 20.70

2

402

!

= 0.80\"#$-9&$/9',J&$8,$-9&$

G8,&3/$&,&1J*X$

!

"K

K1

=K2#K

1

K1

= 0.8 #1= #0.2

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!

Coefficient of Restitution : e =v

2'-v

1'

v1

- v2

_,='0>.';./.=35'239.'9>,J/'34,<.H$

E.=?.2056'F/.539G2'),55:9:,/9'

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!

Coefficient of Restitution : e =v

2'-v

1'

v1

- v2

= 0

E.=?.2056'#539G2'),55:9:,/9'

!

Coefficient of Restitution : e =v

2'-v

1'

v1

- v2

=1

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+,51G,/H'

!

Coefficient of Restitution : e =v

2'-v

1'

v1- v

2

=v

1'

v1

=v'

v

!

Before bouncing : v 2 = v0

2- 2g(-h)

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!

v = 2gh

!

after bouncing : 0 = v '2 -2gh'

!

v' = 2gh'

!

e =v

1'

v1

=2gh'

2gh=

h'

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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.

Motion of the Center of Mass!T"$0&&$-9&$08J,8E/',/&$"?$-9&$/&,-&1$"?$+'00$M/6"6+6O$%8g&1&,3'-&$

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('13/)&0$+"2&X$

78g&1&,3'3,J$'J'8,$#&$J&-X$

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-"$-9&$-"-')$+"+&,-.+$"?$-9&$0*0-&+6c$

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?"1/&0$",$Rhh$"?$-9&$('13/)&0$8,$-9&$0*0-&+6$$R0$:&?"1&>$/)'008?*$

-9&0&$'0$&8-9&1$8,-&1,')$"1$&@-&1,')$-"$-9&$0*0-&+6$$Z.-$'0$#&$0'#$

\&#-",$QQQ$8+()8&0$$$$

P@'+()&$?"1$L&,-&1$"?$d'00$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

Starting from rest, two skaters push off against each other on ice where friction is negligible. One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of +2.5 m/s, the male skater moves with a velocity of -1.5 m/s. What is final v for center of mass?!

Q/&$DG'-&1$[1":)&+$