Chapter 9 Impulse and Momentum Physics, 6th Edition
Chapter 9. Impulse and Momentum Momentum 9-1. A 0.5-! "rench is dropped #rom a hei!ht o# 10 m. $hat is its momentum momentum %ust &e#ore it stries the #loor' ( First First find the velocity from from conservation of energy.) energy.) mgh = ½mv2;
v ) gh )(9.* m+s ,(10 m, )
p = mv = (0.5 !(1 m+s
v = 1.0 m+s
p = /.00 ! m+s, do"n
9-). Compute the momentum and inetic ener!y ener!y o# a )00-l& car moin! north at 55 mi+h. m
W g
)00 l& 2) #t+s )
m /5 slu!s v = 55 mi+h *0./ #t+s
p = mv = (/5 slu!s(*0./ #t+s
p = 6050 slu! #t+s
K = ½mv2 = ½(/5 ½(/5 slu!s(*0.66 #t+s )
K = ),000 #t l&
9-2. A )500-! truc truc traelin! at 0 m+h stries a &ric "all and comes to a stop in 0.) s. (a $hat is the chan!e in momentum' (& $hat is the impulse' impulse' (c $hat $hat is the aera!e #orce on the "all durin! the crash' Take + to e to!ard the !all . ( 0 m+h 11.1 m+s p
= mv f " " mvo = # $ ()500 !(11.1 m+s
p
- )/,*00 ! m+s
%mp&lse = p; F t = $)/,*00 $)/,*00 ! m+s 3orce 4 truc
F
)/,*00
0.) s
3orce on "all is opposite, so
3 -129,000
3 7 129,000
9-. $hat is the momentum momentum o# a 2-! &ullet moin! at 600 m+s in a direction direction 200 a&oe the 600 m+s hori8ontal' $hat are the hori8ontal and ertical ertical components o# this momentum' momentum' 200
p = mv = (2 !(600 m+s
p p 1*00 ! m+s, 200
p 1*00 cos 200 and p y = 1*00 sin 200
101
p ' = (*# ! (*# ! m+s p y 900 ! m+s
Chapter 9 Impulse and Momentum Physics, 6th Edition :9-5. A 0.)-! &ase&all traelin! traelin! to the le#t at )0 m+s is drien in the opposite direction direction at 25 m+s "hen it is hit &y a &at. ;he aera!e #orce on the &all is 600 .
25 m+s
( %mp&lse = change in moment&m. )
7
F t = mv f " " mvo = (0.) !(25 m+s = (0.) !(-)0 m+s t
(600 t 11 ! m+s
t
-)0 m+s
1./) ms
:9-6. A &at eerts an aera!e #orce o# )* l& on 0.6-l& &all #or 0.01 s. ;he incomin! elocity elocity o# the &all "as #t+s. I# it leaes in the opposite direction "hat is its its elocity' hoose positive + as direction a!ay from the at, making incoming all velocity negative F t = mv f " " mvo ;
F t = mv f " " mvo ;
m
0.6 l& 2) #t+s )
0.01*/5 slu!s
()0 l&(0.01 s (0.01*/5 slu!sv slu!sv f $ (0.01*/5 slu!s(- #t+s
0.01*/5 v f = = ). l& s = 0.*)5
v f = = *.0 #t+s
:9-/. A 500-! &all traels #rom le#t to ri!ht at )0 m+s. m+s. A &at dries the &all in the opposite direction "ith a elocity o# 26 m+s. ;he time o# contact "as 0.002 s. $hat "as the aera!e #orce on the &all'
( m = #. !, vo = 7)0 m+s, v f = = -26 m+s,
t
0.002 s
F t = mv f " " mvo ; F (0.002 (0.002 s (0.5 !(-26 m+s = (0.5 !()0 m+s F
1* ! m+s - 10 ! m+s
0.002 s
;
F = 9220
:9-*. A 00-! ru&&er &all is is dropped a ertical distance o# 1) m onto the paement. It is in contact "ith the paement #or 0.01 s and re&ounds to a hei!ht o# 10 m. $hat is the total chan!e in momentum' $hat aera!e #orce is is eerted on the the &all'
10)
Chapter 9 Impulse and Momentum Physics, 6th Edition To apply the imp&lse$moment&m theorem, !e need to first find the velocities &st efore and &st after impact !ith the gro&nd. mgho = ½mvo2;
/0 p ) 1eginning = /0 k ) ro&nd ;
1) m
v0 ) gh0 )(9.* m+s ) ,(1) m, >mv = mgh f ;
v f )(9.* m+s (10 m
F t = mv f " mvo;
h#
vo - 15.2 m+s )
2 f
v f 10 m
vo
v f = + 1 m+s
F (0.01 s = (0. !(1 m+s = (0. !(-15.2 m+s F 11/0
:9-9. A cue stic stries an ei!ht-&all "ith an aera!e #orce o# *0 oer a time o# 1) ms. I# the mass o# the &all is )00 !, "hat "ill &e its elocity' F t = mv f " mvo;
(*0 (0.01) s = (0.) !v f = 0
v = .*0 m+s
9-10. A !ol#er hits a 6 ! !ol# &all "ith an initial elocity o# 50 m+s at 200. $hat are the '$ and y-components o# the momentum imparted to the &all' 0
50 m+s
v y 0
v ' (50 cos 20 2.2 m+s v y (50 sin 20 )5.0 m+s p ' = (0.06 !(2.2 m+s p y (0.06 !()5 m+s
200
v '
p ' = 1.99 ! m+s p y 1.15 m+s
:9-11. ;he #ace o# the clu& in Pro&lem 9-10 is in contact "ith the &all #or 1.5 ms. $hat are the hori8ontal and ertical components o# the aera!e #orce on the &all' We need to treat hori3ontal and vertical imp&lses and momenta separately From previo&s prolem- po = #, p f = 1.99 ! m+s, p fy = 1.15 ! m+s F ' t = p f' " po' =1.99 ! m+s
F '
1.99 ! m+s
F ' t = p f' " po' =1.15 ! m+s
F y
1.15 ! m+s
0.0015 s
0.0015 s
102
F ' 1220
F y /6/
Chapter 9 Impulse and Momentum Physics, 6th Edition
Conservation of Momentum 9-1). A sprin! is ti!htly compressed &et"een a 6-! &loc and a )=! &loc and then tied "ith a strin!. $hen the strin! &reas, the )-! &loc moes to the ri!ht "ith a elocity o# 9 m+s. $hat m( = 6 !
is the elocity o# the 6-! &loc' ;otal moment&m is 3ero efore and after the event. 0 7 0 m(v( + m2v2 ;
v1
m) v) m1
() !,(9 m+s, (6 !,
m2 = ) !
v( !
v2
v( = $ 2.00 m+s
9-12. ;"o masses, one three times that o# the other, are compressed a!ainst a sprin! and then tied to!ether on a #rictionless sur#ace as sho"n in 3i!. 9-*. ;he connectin! strin! &reas and sends the smaller mass to the le#t "ith a elocity o# 10 m+s. $hat "as the elocity o# the lar!er mass' 4oment&m 3ero efore and after- 0 7 0 m(v( + m2v2 v)
m1v1 m)
m(10 m+s, (2m,
10 m+s v(
m
5m v2
v( = $ 2.22 m+s
9-1. A /0-! person standin! on a #rictionless sur#ace thro"s a #oot&all #or"ard "ith a elocity o# 1) m+s. I# the person moes &ac"ard at 2 cm+s, "hat "as the mass o# the #oot&all' 4oment&m 3ero efore and after- 0 7 0 m(v( + m2v2 m)
m1v1 v)
(/0 !,(0.2 m+s, (-1) m+s,
m2 = 1.9* m+s
9-15. A )0-! child is at rest in a "a!on. ;he child %umps #or"ard at ) m+s, sendin! the "a!on &ac"ard at 1) m+s. $hat is the mass o# the "a!on'
10
Chapter 9 Impulse and Momentum Physics, 6th Edition 0 m(v( + m2v2 ;
m1
m) v) v1
()0 !,() m+s, (-1) m+s,
m( = $ 2.22 !
9-16. ;"o children, "ei!hin! *0 and 50 l&, are at rest on roller sates. ;he lar!er child pushes so that the smaller moes a"ay at 6 mi+h. $hat is the elocity o# the lar!er child' 0 m(v( + m2v2 ;
v1
m) v) m1
(50 l&,(6 #t+s, (*0 l&,
v( = $ 2./5 #t+s
/6ere !ere ale to &se the !eight eca&se it is proportional to the mass)
9-1/. A 60-! #irecracer eplodes, sendin! a 5-! piece to the le#t and another to the ri!ht "ith a elocity o# 0 m+s. $hat is the elocity o# the 5-! piece' The t!o pieces add to 60 !- m( + m2 = *# !. 0 m(v( + m2v2 ;
v1
m) v) m1
;hus, m( = 7 g, m2 = ( g
(15 !(0 m+s
(5 !
v( = $ 12.2 m+s
:9-1*. A )-! &ullet is #ired "ith a mu88le elocity o# 900 m+s #rom a 5-! ri#le. 3ind the recoil elocity o# the ri#le and the ratio o# the inetic ener!y o# the &ullet to that o# the ri#le' 0 m(v( + m2v2 ;
v1
m) v) m1
() !,(900 m+s, ) �m v) 0kr �mr vr) (5000 !,(.2) m+s, )
0k
() !,(900 m+s, (5000 !,
v( = $ .2) m+s
?atio )0*
:9-19. A 6-! &o"lin! &all collides head on "ith 1.*-! pin. ;he pin moes #or"ard at 2 m+s and the &all slo"s to 1.6 m+s. $hat "as the initial elocity o# the &o"lin! &all' m& + # = mv + m pv p; (6 !& = (6 !(1.6 m+s 7 (1.* !(2 m+s *& = 8.* m+s 7 5. m+s
& = ).50 m+s
105
Chapter 9 Impulse and Momentum Physics, 6th Edition
:9-)0. A 60-! man on a lae o# ice catches a )-! &all. ;he &all and man each moe at * cm+s a#ter the &all is cau!ht. $hat "as the elocity o# the &all &e#ore it "as cau!ht' $hat ener!y "as lost in the process' ( 9 completely inelastic collision- vc = vm = v * cm+s m& + mm&m = /m + mm )vc ; 2& = .96 m+s
() !& + # = () ! 7 60 !(0.0* m+s
& ).* m+s
½m&2 + # =/m + mm )vc2; ½() !().* m+s) >(6) !(0.0* m+s) 7 @oss :oss = 6.15 = 0.19*
:oss = 5.95
:9-)1. A )00-! roc traelin! south at 10 m+s stries a 2-! &loc initially at rest. (a I# the t"o stic to!ether on collision, "hat "ill &e their common elocity' (& $hat ener!y "as lost in the collision' mr &r + m& = /mr + m )vc ; ) m+s 2.) vc
(0.) !(10 m+s + # = (0.) ! 7 2 !vc
vc 0.6)5 m+s
½mr &r 2 + # =/mr + m )vc2; ½(0.) !(10 m+s) > (2.) !(0.6)5 m+s) 7 @oss :oss =10.0 = 0.6)5
:oss = 9.2*
Elastic and Inelastic Collisions 9-)). A car traelin! at * m+s crashes into a car o# identical mass stopped at a tra##ic li!ht. $hat is the elocity o# the "reca!e immediately a#ter the crash, assumin! the cars stic to!ether' &( = *.00 m+s &2 = #, m( = m2 = m) 0 m&( + m&2 = /m + m)v c ; m&( = 2mvc
106
(
Chapter 9 Impulse and Momentum Physics, 6th Edition vc
&1 )
* m+s )
vc = .00 m+s
9-)2. A )000-! truc traelin! at 10 m+s crashes into a 1)00-! car initially at rest. $hat is the common elocity a#ter the collision i# they stic to!ether' $hat is the loss in ener!y' m(&( + m2&2 = /m( + m2 )vc ; )0,000 m+s 2)00 vc
()000 !(10 m+s + # = ()000 ! 7 1)00 !vc
vc 6.)5 m+s
½m(&(2 + # =/m( + m2 )vc2; ½()000 !(10 m+s) >(2)00 !(6.)5 m+s) 7 @oss :oss = 100,000 = 6),500
:oss = 2/,500
9-). A 20-! child stands on a #rictionless sur#ace. ;he #ather thro"s a 0.*-! #oot&all "ith a elocity o# 15 m+s. $hat elocity "ill the child hae a#ter catchin! the #oot&all' m(&( + 0 = m(v( + m2v2; (0.* !(15 m+s = (20 ! 7 0.* !vc (20.* !vc = 1) m+s
vc =0.290 m+s
:9-)5. A )0-! o&%ect traelin! to the le#t at * m+s collides head on "ith a 10-! o&%ect traelin! to the ri!ht at 5 m+s. $hat is their com&ined elocity a#ter impact' m(&( + m2&2 = /m( + m2 )vc ; -110 m+s 20 vc
()0 !(-* m+s + (10 !(5 m+s = ()0 ! 7 10 !vc
vc -2.6/ m+s, to le#t
:9-)6. 3ind the percent loss o# ener!y #or the collision in Pro&lem 9-)5. onservation of 0nergy- ½m(&(2 + ½m2&22 =½/m( + m2 )vc2 + :oss >()0 !(-* m+s) 7 ½(10 !(5 m+s) >(20 !(-2.6/ m+s) 7 @oss /65 )0) 7 :oss;
:oss = *5 ;
<:oss =
10/
562 /65
= /2.6B
Chapter 9 Impulse and Momentum Physics, 6th Edition
:9-)/. A )-! &loc o# clay is tied to the end o# a strin! as sho"n in 3i!. 9-9. A 500-! steel &all em&eds itsel# into the clay causin! &oth h
to rise a hei!ht o# )0 cm. 3ind the entrance elocity o# the &all'
1efore applying moment&m conservation, !e need to kno! the common velocity of the clay and all after the collision. 0nergy is conserved - ½/m( + m2 ) vc2 = /m( + m2 ) gh vc ) gh )(9.* m+s ) ,(0.)0 m, ; m(&( + 0 = /m( + m2 ) vc ; (0.5 !&( = .95 m+s
vc = 1.9* m+s
(0.5 ! &( = (0.5 ! 7 ) !(1.9* m+s &( = 9.90 m+s
:9-)*. In Pro&lem 9-)/, suppose the 500-! &all passes entirely throu!h the clay an emer!es "ith a elocity o# 10 m+s. "hat must &e the ne" entrance elocity i# the &loc is to raise to the same hei!ht o# )0 cm' We m&st find the velocity v2 of the clay /m2 ) after collision-
10 m+s
½/m( + m2 ) v22 = /m( + m2 ) gh vc ) gh )(9.* m+s ) ,(0.)0 m, ;
h
v2 = 1.9* m+s
4oment&m is conserved- m(&( + # = m(v( + m2v2; (0.5 !&( = (0.5 !(10 m+s 7 () !(1.9* m+s
&( = 1/.9 m+s
:9-)9. A 9-! &ullet is em&edded into a ).0 ! &allistic pendulum (see 3i!. *-12. $hat "as the initial elocity o# the &ullet i# the com&ined masses rise to a hei!ht o# 9 cm' ½/m( + m2 ) vc2 = /m( + m2 ) gh vc ) gh )(9.* m+s ) ,(0.09 m, ; m(&( + 0 = /m( + m2 ) vc ;
vc = 1.22 m+s
(0.009 ! &( = (0.009 ! 7 ) !(1.22 m+s 10*
Chapter 9 Impulse and Momentum Physics, 6th Edition (0.009 !&( = ).6* m+s
&( = )9/ m+s
:9-20. A &illiard &all moin! to the le#t at 20 cm+s collides head on "ith another &all moin! to the ri!ht at )0 cm+s. ;he masses o# the &alls are identical. I# the collision is per#ectly elastic, "hat is the elocity o# each &all a#ter impact' /onsider right as +.) Momentum m(&( + m2&2 = m(v( + m2v2
iven- m( = m2 = m, v( = $20 cm+s, v2 = #
m(-20 cm+s 7 0 mv( + mv2 ; v( + v2 (-20 cm+s 7 ()0 cm+s v( + v2 -10 cm+s 0nergy /e = ()- v2 " v( = &( " &2 = /$20 cm+s = ()0 cm+s v2 " v( = $ 50 cm+s From second e&ation v2 = v( = 50 cm+s >&stit&ting this for v2, !e otainv( + (v1 - 50 cm+s = $ 10 cm+s
and
v( = )0 cm+s, to right
9nd, v2 = v( " 50 cm+s ()0 cm+s = 50 cm+s
v2 = $20 cm+s, to left
9-21. ;he coe##icient o# restitution #or steel is 0.90. I# a steel &all is h(
dropped #rom a hei!ht o# / m, ho" hi!h "ill it re&ound' e
h) h1
e )
h) h1
h) h1e (/ m,(0.9, )
)
/m
h2
h2 =5.6/ m
:9-2). $hat is the time &et"een the #irst contact "ith the sur#ace and the second contact #or Pro&lem 9-21'
(We need to kno! vo to rise to 5.6/ m, then find t.)
�mv0) mgh 0 ) gh )(9.* m+s ) ,(5.6/ m, ; s
v0 v f )
t = 1.0/ s
t t
)s vo 0
; )t
)(5.6/ m, 10.5 m+s
; ).15 s
109
vo = 10.5 m+s
Chapter 9 Impulse and Momentum Physics, 6th Edition
:9-22. A &all dropped #rom rest onto a #ied hori8ontal plate re&ounds to a hei!ht that is *1 percent o# its ori!inal hei!ht. $hat is the coe##icient o# restitution' $hat is the reuired elocity on the #irst impact to cause the &all to re&ound to a hei!ht o# * m. e
e
v) v1 &1 &)
v2 = &2 = #;
h) h1
0.*1
e
v1 &1
e = 0.900
&1
v1 e
v1 (0.9,
�mv1) mgh 1 ) gh )(9.* m+s ) ,(* m, ; &( = $1.11v(;
&( = -1.11(-1).5 m+s
h(
;
7 h2
&( = $1.11v(
v( = $1).5 m+s
&( = 12.9 m+s
:9-2. A 200-! &loc moin! north at 50 cm+s collides "ith a )00-! &loc moin! south at 100 cm+s. I# the collision is completely inelastic, "hat is their common elocity a#ter sticin! to!ether' $hat is the loss in ener!y' /onsider north as positive)
4oment&m m(&( + m2&2 = m(v( + m2v2 ;
v( = v2 = vc for inelastic collision
(200 !(50 cm+s 7 ()00 !(-100 cm+s (200 ! 7 )00 !vc 15,000 ! cm+s = )0,000 ! cm+s (500 !vc;
vc = -10 cm+s, so&th
/?ote- When !orking !ith energy, it is necessary to &se ! for the mass &nit.) onservation of 0nergy- ½m(&(2 + ½m2&22 =½/m( + m2 )vc2 + :oss >(0.2 !(-* m+s) 7 ½(0.) !(5 m+s) >(0.2 ! 7 0.) !(-2.6/ m+s) 7 @oss >olving for @lossA, !e otain-
:oss = 0.125
:9-25. Duppose the collision in Pro&. 9-2 is per#ectly elastic. 3ind the elocities a#ter impact. m(&( + m2&2 = m(v( + m2v2;
m( = 5## !, m2 = )00 !, &( = 50 cm+s, &2 - 100 cm+s
(200 !(50 cm+s 7 ()00 !(-100 cm+s (200 !v( + ()00 !v2 110
Chapter 9 Impulse and Momentum Physics, 6th Edition Bividing each term y (## !
2 v( + 2 v2 = $50 cm+s
0nergy /e = ()- v2 " v( = &( " &2 = (50 cm+s = (-100 cm+s v2 " v( = 150 cm+s >&stit&te v2 = v( + 150 cm+s into the earlier e&ation and solve for v(2 v( + ) (v( 7 150 cm+s - 50 cm+s v2 = (-*0 cm+s 150 cm+s
v( -*0 cm+s, to left
v2 /0 cm+s, to right
:9-26. A 5-l& and a 1)-l& o&%ect approach each other "ith eual elocities o# )5 #t+s. $hat "ill &e their elocities a#ter impact i# the collision is (a completely inelastic or (& per#ectly elastic' >ince !eight is proportional to mass, !e !ill &se the !eights instead.
4oment&m W (&( + W 2&2 = W (v( + W 2v2 ;
(5 l&()5 #t+s 7 (1) l&(-)5 #t+s (5 l& 7 1) l&vc;
v( = v2 = vc for inelastic collision
vc = $10.2 #t+s
0lastic case- (5 l&()5 #t+s 7 (1) l&(-)5 #t+s (5 l&v( + (1) l&v2 ; Bividing each term y l& v( + ). v2 = $25 #t+s 0nergy /e = ()- v2 " v( = &( " &2 = ()5 #t+s = (=)5 #t+s v2 " v( = 50 #t+s >&stit&te v( = v2 $ 50 #t+s into the earlier e&ation and solve for v((v2 " # #t+s 7 ). v2 - 25 #t+s
v2 .1 #t+s
v( = v2 = 50 #t+s .1 #t+s = 50 #t+s
v( -5.6 #t+s
Challenge Problems 9-2/. An aera!e #orce o# 000 acts on a 00-! piece o# metal causin! it to moe #rom rest to a elocity o# 20 m+s. $hat "as the time o# contact #or this #orce' F t = mv f " mvo = (0. !(20 m+s = 0 (000 t 1) ! m+s
t
2.00 ms 111
Chapter 9 Impulse and Momentum Physics, 6th Edition :9-2*. An 600-! o&%ect "hose elocity is initially 1) m+s, collides "ith a "all and re&ounds "ith hal# o# its ori!inal inetic ener!y. $hat impulse "as applied &y the "all' )
2
2 f
>mv# = 2/½mv ) ;
v f
v0 )
(1) m+s, ) )
v f *.9 m+s
F t = mv f " mvo = (0.6 !( =).5 m+s = (0.6 !(1) m+s F t = -1).2 m
:9-29. A 10-! &loc at rest on a hori8ontal sur#ace is struc &y a )0-! &ullet moin! at )00 m+s. ;he &ullet passes entirely throu!h the &loc and eits "ith a elocity o# 10 m+s. $hat is the v2 = C
elocity o# the &loc'
v( = 10 m+s
m(&( + # = m(v( + m2v2; m( = 0.0) ! (0.0) !()00 m+s (0.0) !(10 m+s 7 (10 !v2 ;
v2 = 0.2*0 m+s
9-0. In Pro&lem 9-29, ho" much inetic ener!y "as lost' onservation of 0nergy- ½m(&(2 + # = m(v(2 + m2v22 + :oss >(0.) !()00 m+s) ½(0.) !(10 m+s) + >(10 !(0.2*0 m+s) 7 @oss >olving for @lossA, !e otain-
:oss =2990
:9-1. A 60-! &ody hain! an initial elocity o# 100 cm+s to the ri!ht collides "ith a 150-! &ody moin! to the le#t at 20 cm+s. ;he coe##icient o# restitution is 0.*. $hat are the elocities a#ter impact. $hat percent o# the ener!y is lost in collision' m(&( + m2&2 = m(v( + m2v2;
m( = *# !, m2 = 150 !, &( = 100 cm+s, &2 - 20 cm+s
(60 !(100 cm+s 7 (150 !(-20 cm+s (60 !v( + (150 !v2 Bivide each term y 60 ! and simplify
v( + 2. v2 = )5 cm+s
v2 " v( = e/&( " &2 ); v2 " v( = 0.*100 cm+s = (-20 cm+sF v2 " v( = 10 cm+s >olve for v(-
v( = v2 = 10 cm+s
?o! s&stit&te to find v2.
11)
Chapter 9 Impulse and Momentum Physics, 6th Edition :9-1. (Cont.
(v2 = 10 cm+s 7 ).5 v2 = )5 cm+s
v( = v2 = 10 cm+s (26.9 cm+s = 10 cm+s
v2 = 26.9 cm+s, to right v( = -6/.1 cm+s , to left
onservation of 0nergy- ½m(&(2 + ½m2&22 = m(v(2 + m2v22 + :oss For energy !e m&st &se >% &nits !ith mass in @!G and velocity in Am+s.A 0 ok = >(0.06 !(1 m+s) 7 >(0.15 !(-0.2 m+s) 0 ok 0.026/5 0 fk = ½(0.06 !(-0.6/1 m+s) + >(0.15 !(0.269 m+s) 0 fk 0.0)2/
0ok 0 fk 0.026/5 A - 0.0)2/ A 100 0.026/5 A 0 ok
B :oss 100
<:oss = 25.5B
:9-). ;he &loc in 3i!. 9-6 "ei!hs 6 l&.
m W
vc = 5.6)5 #t+s
?o!, !e can find h &sing conservation of energy and the common initial velocity vc½/m( + m2 ) v22 = /m( + m2 ) gh ; The mass divides o&t. h
v0) ) g
(5.6)5 #t+s, ) )(2) #t+s ) ,
;
h 0.9 #t
h 5.92 in.
:9-2. A sin!le railroad car moin! north at 10 m+s stries t"o identical, coupled cars initially moin! south at ) m+s. I# all three couple to!ether a#ter collidin!, "hat is their common elocity'
m1 m) m2 m
&( = 10 m+s
&) &5 = $) m+s
v( = v) v5 =vc
4oment&m is conserved- m&( + m/&2 + &5 ) = /5m)vc /4ass divides o&t.) 10 m+s = ) m+s = ) m+s 2 vc;
112
vc = ).00 m+s, north
h
Chapter 9 Impulse and Momentum Physics, 6th Edition :9-. An atomic particle o# mass ).00 10-)/ ! moes "ith a elocity o# .00 106 and collides head on "ith a particle o# mass 1.)0 10-)/ ! initially at rest. Assumin! a per#ectly elastic collision, "hat is the elocity o# the incident particle a#ter the collision' m( = 2 ' (#$2D !, m2 = 1.) 10-)/ !, &( = 106 m+s
m(&( + # = m(v( + m2v2;
(2 ' (#$2D !( 106 m+s () 10 -)/ !v( + (1.) 10-)/ ! v2 Bividing each term y ) 10-)/ !
v( + 0.6 v2 = 10 6 m+s
0nergy /e = ()- v2 " v( = &( " &2 = 106 m+s = 0 v2 " v( = 10 6 m+s >&stit&te v2 = / v( + 106 m+s into the earlier e&ation and solve for v(v( + 0.6 (v( 7 106 m+s 106 m+s
v( 1.00 106 m+s
:9-5. A &at stries a 00-! so#t&all moin! hori8ontally to the le#t at )0 m+s. It leaes the &at "ith a elocity o# 60 m+s at an an!le o# 20 0 "ith the hori8ontal. $hat are the hori8ontal and ertical components o# the impulse imparted to the &all'
60 m+s
v2y
First find components of velocity- v(' = $ )0 m+s v(y = # v2' = (60 cos 200 5).0 m+s v2y = (60 sin 200 20 m+s
200
v('
v2' -)0 m+s
F ' t = = mv 2' " mv(' ;
' = (0. !(5).0 m+s = (0. !(-)0 m+s ' = )*.* s
F y t = = mv 2y " mv(y ;
y = (0. !(20 m+s = 0 y = 1).0 s
:9-6. I# the &at in Pro&lem 9-5 "as in contact "ith the &all #or 5 ms, "hat "as the ma!nitude o# the aera!e #orce on the so#t&all' F
F
)*.* s 0.005 s F ')
F y)
5/60
F y
(5/60 )
1).0 s 0.005 s
()00 )
)00
F = 6)0
11
Chapter 9 Impulse and Momentum Physics, 6th Edition :9-/. A cart A has a mass o# 200 ! and moes on a #rictionless air trac at 1. m+s "hen it hits a second cart H at rest. ;he collision is per#ectly elastic and the 200-! carts elocity is reduced to 0.6)0 m+s a#ter the collision. $hat "as the mass o# the other cart and "hat "as its elocity a#ter the collision'
m( = 200 !
&( = 1. m+s v( = 0.6)0 m+s
m(&( + # = m(v( + m2v2; (200 !(1. m+s (200 !(0.6)0 m+s 7 m2v2 m2v2 = )2 ! m+s 0lastic collision- v2 " v( = &( " &2 = (1. m+s = 0 v2 = v( + 1. m+s 0.6)0 m+s 7 1. m+s m) v) )2 ! m+s
m)
v2 = ).0) m+s
)2 ! m+s
m2 = 116 !
).0) m+s
:9-*. I# the collision in 3i!. 9-10, assume that the collision o# the t"o masses is completely inelastic. $hat is the common elocity a#ter the collision and "hat is the ratio o# the #inal 15 m+s
inetic ener!y to the initial inetic ener!y' m(&( + # = /m( + m2 )vc ; &( = 15 m+s (1 !(15 m+s + 0 = (1 ! 7 ) !vc
�( m1 m) vc) 0k 1 �m1&1)
0k )
(2 !(5 m+s ) (1 !(15 m+s )
1 !
) !
vc = 5.00 m+s
Eatio = 0.222
:9-9. Assume the collision in Pro&lem 9-* is per#ectly elastic. $hat is the elocity o# each mass a#ter the collision' 0lastic- m(&( + # = m(v( + m2v2 and v2 " v ( = &( " &2 (1 !(15 m+s (1 !v( + () !v2 ;
v( + 2v2 = 15 m+s
v( = 15 m+s = ) v2
v2 " v ( = &( " &2 = (15 m+s = 0 v2 = 15 m+s 7 v( v2 = 15 m+s 7 (15 m+s = )v2 ) v( = $5 m+s
and
v2 = 10 m+s v( = 15 m+s = )(10 m+s -5 m+s
v2 = 10 m+s
115
Chapter 9 Impulse and Momentum Physics, 6th Edition :9-50. A )-! mass moes to ri!ht at ) m+s and collides "ith a 6-! mass moin! to the le#t at m+s. (a I# the collision is completely inelastic, "hat is their common elocity a#ter collidin!, and ho" much ener!y is lost in the collision' 4oment&m m(&( + m2&2 = m(v( + m2v2 ;
v( = v2 = vc for inelastic collision
() !() m+s 7 (6 !(- m+s () ! 7 6 !vc ! m+s = ) ! m+s (* !vc;
vc = -).50 m+s
onservation of 0nergy- ½m(&(2 + ½m2&22 = ½/m( + m2 )vc2 + :oss >() !() m+s) 7 ½(6 !(- m+s ) >() ! 7 6 !(-).50 m+s) 7 @oss >olving for @lossA, !e otain-
:oss = )/.0
::9-51. In Pro&lem 9-50, assume the collision is per#ectly elastic. $hat are the elocities a#ter the collision' m(&( + m2&2 = m(v( + m2v2;
m( = 2 !, m2 = 6 !, &( = ) m+s, &2 - m+s
() !() m+s 7 (6 !(- m+s () !v( + 6 !v2 Bividing each term y 2 !
v( + 2 v2 = $10 m+s
0nergy /e = ()- v2 " v( = &( " &2 = () m+s = (- m+s v2 " v( = 6 m+s >&stit&te v2 = v( + 6 m+s into the earlier e&ation and solve for v(v( + 2(v( 7 6 m+s - 10 m+s v2 = (-/.00 m+s 6 m+s
v( -/.00 m+s
v2 -1.00 m+s
116
Chapter 9 Impulse and Momentum Physics, 6th Edition
Critical Thinking uestions :9-5). An astronaut in or&it outside a capsule uses a reoler to control motion. ;he astronaut "ith !ear "ei!hs )00 l& on the earth. I# the reoler #ires 0.05-l& &ullets at )/00 #t+s, and 10 shots are #ired, "hat is the #inal elocity o# the astronaut' Compare the #inal inetic ener!y o# the ten &ullets "ith that o# the astronaut. $hy is the di##erence so !reat' 0 7 0 W ava + W v; va
W v W a
(0.05 l&,()/00 #t+s, )00 l&
0ach shot changes va y "#.*D m+s v f = 10(0.6/5 m+s We need masses- m
0.05 l& 2) #t+s
)
0.00156 slu!s ; m
; va = $0.6/5 #t+s v f = $ 6./5 #t+s
)00 l& 2) #t+s
)
6.)5 slu!s
0 k = 10 (>mv2 ) = (5(0.00156 slu!s()/00 #t+s) 0 k 56,950 #t l& 0 ka = ½mava2 = >(6.)5 slu!s(6./5 #t+s)
0 ka = 1) #t l&
The kinetic energy of the &llets is m&ch larger eca&se !hen finding the kinetic energy, one m&st deal !ith the s&are of velocity. The speeds dominate.
:9-52. In applyin! conseration o# momentum #or collidin! o&%ects to #ind #inal elocities, could "e use the "ei!ht o# the o&%ects instead o# the mass' $hy, or "hy not' Jeri#y your ans"er &y applyin! it to one o# the eamples in the tet. >ince !eight is proportional to mass- W = mg, and since mass appears in every term involving conservation of moment&m, the !eight can e &sed instead of the mass to calc&late either velocities or !eights of colliding oects. For e'ample, see ro. 8$5*. m(&( + m2&2 = m(v( + m2v2 W1 g
&1
W) g
&)
W1 g
v1
W) v) g
W (&( + W 2&2 = W (v( + W 2v2
11/
Chapter 9 Impulse and Momentum Physics, 6th Edition :9-5. A )0-! &ullet, moin! at )00 m+s, stries a 10-! "ooden &loc and passes entirely throu!h it, emer!in! "ith a elocity o# 10 m+s. $hat is the elocity o# the &loc a#ter impact'
v2 = 0.2*0 m+s
onservation of 0nergy- ½m(&(2 + # = ½/m( + m2 )vc2 + :oss >(0.0) !()00 m+s) >(10.0) !(0.2* m+s) 7 @oss
:oss = 299
:9-55. A 0.20-! &ase&all moin! hori8ontally at 0 m+s is struc &y a &at. ;he &all is in contact "ith the &at #or a time o# 5 ms, and it leaes "ith a elocity o# 60 m+s at an an!le o# 200, "hat are the hori8ontal and ertical components o# the aera!e #orce actin! on the &at' First find components of velocity- v(' = $ )0 m+s v(y = # 0
60 m+s
v2y 200
0
v2' = (60 cos 20 5).0 m+s v2y = (60 sin 20 20 m+s
v('
F ' t = mv2' " mv(' = (0.2 !(5).0 m+s = (0.2 !(-0 m+s
v2' -0 m+s
F '/#.## s )/.6 s F ' = 55)0 F y t = mv2y " mv(y = (0.2 !(20 m+s = 0 3y(0.005 s 9.00 s
3y 1*00
:9-56. $hen t"o masses collide they produce eual &ut opposite impulses. ;he masses do not chan!e in the collision, so the chan!e in momentum o# one should &e the ne!atie o# the chan!e #or the other. Is this true "hether the collision is elastic or inelastic. Jeri#y your ans"er &y usin! the data in Pro&lems 9-50 and 9-51. 4oment&m is conserved !hether energy is lost in collision or not. Therefore, e&al &t opposite imp&lses sho&ld al!ays prod&ce e&al and opposite changes in moment&m. ro. 8$#- The test is !hether- m(v( " m(&( = $/ m2&2 " m2v2 );
11*
v( = v2 -).5 m+s
Chapter 9 Impulse and Momentum Physics, 6th Edition :9-56. (Cont.
() !(-).50 m+s - () !() m+s -(6 ! (-).50 m+s - (6 !(- m+sF
- 9 ! m+s - 9 ! m+s %t !orks for inelastic collisions. ro. 8$(- >ame test- m(v( " m(&( = $/ m2v2 " m2&2 );
v( = -/ m+ s; v2 -1 m+s
() !(-/ m+s - () !() m+s -(6 ! (-1 m+s - (6 !(- m+sF - 1* ! m+s - 1* ! m+s %t also !orks for elastic collisions.
:9-5/. ;"o toy cars o# masses m and 2m approach each other, each traelin! at 5 m+s. I# they couple to!ether, "hat is their common speed a#ter"ard' $hat are the elocities o# each car i# the collision is per#ectly elastic' ( m( = m, m2 = 5m, &( = m+s, &2 = $ m+s m(&( + m2&2 = /m( + m2 ) vc ; -10 m+s vc;
vc = $).50 m+s
m(&( + m2&2 = m(v( + m2v2 ; v( + 2v2 $10 m+s;
m(5 m+s 7 2m(-5 m+s (m 7 2m vc For inelastic case
m(5 m+s 7 2m(-5 m+s mv( + 5mv2
?o! for elastic- v2 " v( = &( " &2
v2 " v( = 5 m+s = (=5 m+s 10 m+s
v( = v2 " 10 m+s
(v2 " 10 m+s 7 2 v2 = -10 m+s
v2 = 0
v( = /#) " 10 m+s - 10 m+s
v1 -10 m+s
:9-5*. An *-! &ullet is #ired hori8ontally at t"o &locs restin! on a #rictionless sur#ace. ;he #irst &loc has a mass o# 1-! and the second has a mass o# )-!. ;he &ullet passes completely throu!h the #irst &loc and lod!es into the second. A#ter the collisions, the 1-! &loc moes "ith a elocity o# 1 m+s and the )-! &loc moes "ith ) m+s. $hat is the elocity o# the &ullet &e#ore and a#ter emer!in! #rom the #irst &loc' $$Gserve the fig&re on the ne't page $$
119
Chapter 9 Impulse and Momentum Physics, 6th Edition ) !
1 !
*!
:9-5*. (Cont.
1 m+s
) m+s
Total moment&m at start = total moment&m at finish
(0.00* ! v( = (1 !(1 m+s 7 ().00* !() m+s
v( = 6)/ m+s
To find velocity emerging from 1-! mass, !e apply conservation to first lock only(0.00* !(6)/ m+s (0.00* !ve 7 (1 !(1 m+s
ve = 50) m+s
:9-59. A 1-! mass A is attached to a support &y a cord o# len!th *0 cm, and it is held hori8ontally as in 3i!. 9-11. A#ter release it s"in!s do"n"ard striin! the )-! mass H "hich is at rest on a #rictionless ta&letop. Assumin! that the collision is per#ectly elastic "hat are the elocities o# each mass immediately a#ter impact' First find & 9 from energy of fall- ½mv2 = mgh v ) gh
) )(9.* m+s ,(0.* m,
m 9& 9 + # = m 9v 9 + m 1v 1;
v = 2.96 m+s
@ *0 cm m 9 = 1 ! & 9 = 2.96 m+s
(1 !(2.96 m+s (1 !v 9 + () ! v 1
m 1 = ) !
v 9 7 ) v 1 2.96 m+s 0lastic- v 1 " v 9 = & 9 " & 1 = 2.96 m+s = 0 m 9 = ) ! v 1 " v 9 = 2.96 m+s v 9 = v 1 " 2.96 m+s >&stit&te for v 9 in the other e&ation. (v 1 " 2.96 m+s 7 ) v 1 = 2.96 m+s From !hichv 9 = v 1 " 2.96 m+s ).6 m+s = 2.96 m+s
v 1 = ).6 m+s
v 9 = -1.2) m+s
1)0
