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S   0# `` I  Y*    0T# `  I  [*    0# `  I  [*  H  0޽h ? ̙33 $Blank Presentation H@0 (  ,    0t#` \Momentum & Impulse(2    0#@ <Momentum (p) - the product of an objects mass and velocity. (= 2 1 =   0T#(,$D  fp = mv(2      0#  ,$D @Newton s 1st law of physics states that a force must be used to overcome the inertia of an object. Inertia deals with the mass of an object and the velocity, so a force must be used to change the momentum of an object. T 2    T  <Pd޽h ? ̙33    @0 $ ( ̙D(+h+ n  03,$D  fThe force used to change the momentum of an object for a certain amount of time is called an impulse. Fg 2] g   0d3`O lMomentum & Impulse,(2  l Z Z,$D   6$3 _ S F = = 2       63\h  nDp< 2  lB   <DԔ     63A  PDt 2  j@ ]  v S Z   6D3`  N pmDv< 2    63] .v RDt (2  lB  <DԔ @5@'  03> ,$D  -From this the following formulas can be made:8. 2( . 4  0ĉ3 U ,$D  FDt = mDv` (2  b  03V  ,$D  FDt = mv2 + mv1(2  T  <Pd޽h ? ̙33 vnP( jB)   0D3`O nMomentum & Impulse.(2  v  03 J "FDt = m1v2 + m1v1(2    0d31 \When looking at the forces between two objects, the above formula can be written as follows:&] 2\ ]   0Č3 lH,$  This formula describes the Law of Conservation of Momentum. This says that the momentum before a collision is the same after a collision.> 2O    03  emAv1 + mBv1 = mAv2 + mBv2(2  T  <Pd޽h ? ̙33 yq`( 13   0D3`O nMomentum & Impulse.(2  &  0t 4ym hLaw of Conservation of Momentum- the total momentum after a collision is the same as before a collision.0i 2I i   0 4*,$  `Practice Problem: A 10,000-kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the cars lock together when they collide, what is their speed after the collision? Hint: make sure you use +/- for direction.D 2f*    0 4 emAv1 + mBv1 = mAv2 + mBv2(2  T  <Pd޽h ? ̙33SG GFpR0 sF(    0T 4`O nMomentum & Impulse.(2    041y emAv1 + mBv1 = mAv2 + mBv2(2    0t4?@ CPractice Problem: A 1.0 x 104-kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the cars lock together when they collide, what is their speed after the collision? N 2f      044 - ,$D  v$We can use right as the + direction.$% 2$ % <8    T    #   BCDEDFN     $(@` 3   BIC"DEF"!"I""@`% :n\p  @BLCDEtF~ ~ ,|ib`^^^^ \ "gLiLsAsAm*m*s* ~<@@`1 S}h  8B0C*DEpFz..(##  ####!!#$#$((#0!0$**:<@`O  BC"DEF&""@`t !C  BCDE8FB'      @`Qf  BCqDE@FJ'q l jhbZZ^ hfdV4"$@`9 `N   BCDEF' @` , A  BC9DEF999 @` j   BC7DEF777 @`j<  #  * 8<  #  * 8<  #  j <  # j  `BCCEDEF!!!E(E.C4?:<=6A2C*C%CA?< 60*# !&.4 : =AC!EDH@`]     `BECCDEF!!!C(C.A6=::?6C0E(E#ECA= 82,$  !&.4 8;?A!CDH@` 6   `BCCCDEFa!! C(C.A4=99=6A0C(C"CA?; 50*"  !&.4 8 =?A CDH@`e     `BECEDEFa!! E(C.A5?9;?6A0E*E$EC?; 72,$  !&.4 9=AC EDH@` >   BCDEF&  @`    B C DE F*    @`    BCDEF"  @`    BC DEF&    @`v    B C DEF"  @`n z   BCDEF&  @`r    BCDEF&  @`.  B C DEF"  @`)4  BC DEF"  @`.  B CDEF"     @`   BC DEF"   @`  BCDEF&  @`  BC&DEHFR&oo A A&&&(@`i ),rB  BDo   64.w ^ R24.0 m/s 2     64F  y 1.0 x 104 -kg> 2    64t  KA(2f  BT    #    Z BCDEDFN     $(@` 3  m BIC"DEF"!"I""@`% :n\j n :BLCDEtF~ ~ ,|ib`^^^^ \ "gLiLsAsAm*m*s* ~<@@`1 S}b o 2B0C*DEpFz..(##  ####!!#$#$((#0!0$**:<@`O p BC"DEF&""@`t !C q BCDE8FB'      @`Qf r BCqDE@FJ'q l jhbZZ^ hfdV4"$@`9 `N  s zBCDEF' @` , A t zBC9DEF999 @` j  u zBC7DEF777 @`j< v #  * 8< w #  * 8< x #  j < y # j z ZBCCEDEF!!!E(E.C4?:<=6A2C*C%CA?< 60*# !&.4 : =AC!EDH@`]    { ZBECCDEF!!!C(C.A6=::?6C0E(E#ECA= 82,$  !&.4 8;?A!CDH@` 6  | ZBCCCDEFa!! C(C.A4=99=6A0C(C"CA?; 50*"  !&.4 8 =?A CDH@`e    } ZBECEDEFa!! E(C.A5?9;?6A0E*E$EC?; 72,$  !&.4 9=AC EDH@` >  ~ BCDEF&  @`    B C DE F*    @`    BCDEF"  @`    BC DEF&    @`v    B C DEF"  @`n z   BCDEF&  @`r    BCDEF&  @`.  B C DEF"  @`)4  BC DEF"  @`.  B CDEF"     @`   BC DEF"   @`  BCDEF&  @`  BC&DEHFR&oo A A&&&(@`i ),  64E   y 1.0 x 104 -kg> 2    644 KB(2f    64   mA = 1.0 x 104 kgV 2       64% E  mB = 1.0 x 104 kgV 2       6t4    s v1 = 24.0 m/s8 2      644:  Z  r v1 = 0.0 m/s8 2    jB @ BDԔx  x jB @ BDԔ  T  <Pd޽h ? ̙33MD CCLM C( 4 9    04`O nMomentum & Impulse.(2    04. rmA v1 + mBv1 = mAv2 + mBv2(2    04? QPractice Problem: A 1.0 x 104-kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the cars lock together when they collide, what is their speed after the collision? \ 2f   F           BCDEDFN     $(@` 3    BIC"DEF"!"I""@`% :n\p   @BLCDEtF~ ~ ,|ib`^^^^ \ "gLiLsAsAm*m*s* ~<@@`1 S}h   8B0C*DEpFz..(##  ####!!#$#$((#0!0$**:<@`O   BC"DEF&""@`t !C   BCDE8FB'      @`Qf   BCqDE@FJ'q l jhbZZ^ hfdV4"$@`9 `N    BCDEF' @` , A   BC9DEF999 @` j    BC7DEF777 @`j<   #  * 8<   #  * 8<   #  j <   # j   `BCCEDEF!!!E(E.C4?:<=6A2C*C%CA?< 60*# !&.4 : =AC!EDH@`]      `BECCDEF!!!C(C.A6=::?6C0E(E#ECA= 82,$  !&.4 8;?A!CDH@` 6    `BCCCDEFa!! C(C.A4=99=6A0C(C"CA?; 50*"  !&.4 8 =?A CDH@`e      `BECEDEFa!! E(C.A5?9;?6A0E*E$EC?; 72,$  !&.4 9=AC EDH@` >    BCDEF&  @`     B C DE F*    @`     BCDEF"  @`     BC DEF&    @`v     B C DEF"  @`n z    BCDEF&  @`r     BCDEF&  @`.   B C DEF"  @`)4   BC DEF"  @`. !  B CDEF"     @`  "  BC DEF"   @` #  BCDEF&  @` $  BC&DEHFR&oo A A&&&(@`i ),F    %    &  BCDEDFN     $(@` 3  '  BIC"DEF"!"I""@`% :n\p (  @BLCDEtF~ ~ ,|ib`^^^^ \ "gLiLsAsAm*m*s* ~<@@`1 S}h )  8B0C*DEpFz..(##  ####!!#$#$((#0!0$**:<@`O *  BC"DEF&""@`t !C +  BCDE8FB'      @`Qf ,  BCqDE@FJ'q l jhbZZ^ hfdV4"$@`9 `N  -  BCDEF' @` , A .  BC9DEF999 @` j  /  BC7DEF777 @`j< 0  #  * 8< 1  #  * 8< 2  #  j < 3  # j 4  `BCCEDEF!!!E(E.C4?:<=6A2C*C%CA?< 60*# !&.4 : =AC!EDH@`]    5  `BECCDEF!!!C(C.A6=::?6C0E(E#ECA= 82,$  !&.4 8;?A!CDH@` 6  6  `BCCCDEFa!! C(C.A4=99=6A0C(C"CA?; 50*"  !&.4 8 =?A CDH@`e    7  `BECEDEFa!! E(C.A5?9;?6A0E*E$EC?; 72,$  !&.4 9=AC EDH@` >  8  BCDEF&  @`   9  B C DE F*    @`   :  BCDEF"  @`   ;  BC DEF&    @`v   <  B C DEF"  @`n z  =  BCDEF&  @`r   >  BCDEF&  @`. ?  B C DEF"  @`)4 @  BC DEF"  @`. A  B CDEF"     @`  B  BC DEF"   @` C  BCDEF&  @` D  BC&DEHFR&oo A A&&&(@`i ),dB E <Do   F 0D~:m. T  R24.0 m/s 2    G 0: K  y 1.0 x 104 -kg> 2   H 0: E   y 1.0 x 104 -kg> 2  K J 0$:I i ,$  3(1.0 x 104 kg)(24.0 m/s) +(1.0 x 104 kg)(0.0 m/s) =V4 2    4 { K 0:{ i ,$  '(1.0 x 104 kg)(v2) +(1.0 x 104 kg)(v2) ( 2       (  L 0D:;  SA$(2   M 0:  KB(2  T  <Pd޽h ? ̙33fG GGOR$F( ;(  $ $ 0d:`O nMomentum & Impulse.(2   $ 0$:. rmA v1 + mBv1 = mAv2 + mBv2(2   $ 0:? RPractice Problem: A 1.0 x 104 -kg railroad car traveling at a speed of 24.0 m/s strikes an identical car at rest. If the cars lock together when they collide, what is their speed after the collision? \ 2f   " I$ 0D:I i ,$D  ((240,000 kg m/s) =8 2   d K$ 0:Q q ,$D  (2 (1.0 x 104 kg v2)z 2     L$ 0ą:{ ,$D  F120,000 kg m/s = 1.0 x 104 kg v2$ 2   $  O$ 0:_4,$D   v2 = 12 m/s P 233 33  <8   R$ N    $    $ BCDEDFN     $(@` 3  $ BIC"DEF"!"I""@`% :n\p $ @BLCDEtF~ ~ ,|ib`^^^^ \ "gLiLsAsAm*m*s* ~<@@`1 S}h  $ 8B0C*DEpFz..(##  ####!!#$#$((#0!0$**:<@`O  $ BC"DEF&""@`t !C  $ BCDE8FB'      @`Qf  $ BCqDE@FJ'q l jhbZZ^ hfdV4"$@`9 `N   $ BCDEF' @` , A $ BC9DEF999 @` j  $ BC7DEF777 @`j< $ #  * 8< $ #  * 8< $ #  j < $ # j $ `BCCEDEF!!!E(E.C4?:<=6A2C*C%CA?< 60*# !&.4 : =AC!EDH@`]    $ `BECCDEF!!!C(C.A6=::?6C0E(E#ECA= 82,$  !&.4 8;?A!CDH@` 6  $ `BCCCDEFa!! C(C.A4=99=6A0C(C"CA?; 50*"  !&.4 8 =?A CDH@`e    $ `BECEDEFa!! E(C.A5?9;?6A0E*E$EC?; 72,$  !&.4 9=AC EDH@` >  $ BCDEF&  @`   $ B C DE F*    @`   $ BCDEF"  @`   $ BC DEF&    @`v   $ B C DEF"  @`n z  $ BCDEF&  @`r   $ BCDEF&  @`. $ B C DEF"  @`)4  $ BC DEF"  @`. !$ B CDEF"     @`  "$ BC DEF"   @` #$ BCDEF&  @` $$ BC&DEHFR&oo A A&&&(@`i ),N    %$   &$ BCDEDFN     $(@` 3  '$ BIC"DEF"!"I""@`% :n\p ($ @BLCDEtF~ ~ ,|ib`^^^^ \ "gLiLsAsAm*m*s* ~<@@`1 S}h )$ 8B0C*DEpFz..(##  ####!!#$#$((#0!0$**:<@`O *$ BC"DEF&""@`t !C +$ BCDE8FB'      @`Qf ,$ BCqDE@FJ'q l jhbZZ^ hfdV4"$@`9 `N  -$ BCDEF' @` , A .$ BC9DEF999 @` j  /$ BC7DEF777 @`j< 0$ #  * 8< 1$ #  * 8< 2$ #  j < 3$ # j 4$ `BCCEDEF!!!E(E.C4?:<=6A2C*C%CA?< 60*# !&.4 : =AC!EDH@`]    5$ `BECCDEF!!!C(C.A6=::?6C0E(E#ECA= 82,$  !&.4 8;?A!CDH@` 6  6$ `BCCCDEFa!! C(C.A4=99=6A0C(C"CA?; 50*"  !&.4 8 =?A CDH@`e    7$ `BECEDEFa!! 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