I_I
Launching
Device
~
Projectile
(13)201IMI
Mech. I.
A projectile is fired horizontally from a launching device, exiting with a speed Ivx. While the projectile is in the
launching device, the impulse imp3l1ed to it is J p, and the average force on it is Fm'g. Assume the force becomes zero
just as the projectile reaches the end of the launching device. Express your answers to pal1s (a) and (b) in terms of Vx,
.61" Fm'g, and fundamental constants, as appropriate.
(a) Determine an expression for the time required for the projectile to travel the length of the launching device.
r ~J
0i
(b' D""m;" ." ~P"";'"f", ,h, m." "f •• W,;~,ilo.
Yc1 .
\fp~ 6p -:: m(V,,-
\
The projectile is fired horizontally into a bl~Od
that is clamped to a tabletop so that it cannot move. The
projectile travels a distance d into thV,>l6ck before it stops. Express all algebraic answers to the following in terms of
d and the given quantities previously indicated, as appropriate.
(c) Derive an expression for the .,(ork done in stopping the projectile.
V\J~6¥- /
1-
_ 1. ~
_-I
1.
- "2 '(\{\ V'/.
2.
::0
"fi
(Y'I. \) -:::.
(d) Derive an expression for the average force Fb exel1ed on the projectile as it comes to rest in the block.
\ d. - -
.JpV)G
2
t"b
?
""
ow a new projectile and block are used, identical to the first ones, but the block is not clamped to the table. The
projectile. is again fired into the block of wood and travels a new distance d" into the block while the block slides
across the table a sh0l1 distance D. Assume the following: the projectile enters the block with speed x, the average
force Fb between the projectile and the block has the same value as determined in part (d), the average force of friction
between the table and the block isfr, and the collision is instantaneous so the frictional force is negligible during the
collision.
v,.
cA
(,'D~:W;':':~"~
fyt~",m, d,Dfr~: F: ~::P~;';p--..~-=-v--~-J L~
t::
"f
~. - f,.P - j\.c\.V) -=-.0
~bc\ -
5-..• 0 - +'0.:\1\
_---
~-{,,'-=/""'iM)V
I'll
Xl
V
\:
-=.
l
V\J(F~ ~\
~
.••
.l.MtM
v:
:::
1...
1
111
of the projectile, and the mass M of the block.
t"I\
\"V\-tt-A
')
2.("'tM)
V~~••.J =-1""TL:>
•.•.
~y\\)
~~-tf1"tl
r/
(rnJ!b):;; fw ~~ )lUd =+~-~
r
"'r. -:::JrD
V'A"f- .
~~
JI
+I?
-=-0
(I) Derive an expression for d" in terms of d, the mass
p; =-rm
0
\
(
3-(fbd)"'1;-D
~
-tlf/\) ~
= .rbd - .(:bdf\
d
-::[rn~d.-q
l\
~-l"Y"f'.
14
"
pe~Aulm:l
t.,.,H l;:J lj' ;)Jr.[;ing
Th
C
;c"") d\; I.U;",
to atr~kc
I"
-Jr
'>
---.>
1
N
>I
ve~ticill1)'
'$
'
PlL 11'",~
m.~"s m on a string
,Ie l't';;t 1n contact
of
:3 block
"lith
,nn
,,= "ng.lp_ "" a'lcl rc,l('~.se<.l.
"'."('"
_,,_~
to
1 i:;
Itl'lgth
oE !J'.'ss N.
a11ol,'ifl7,
the: h'l11
bl[)ck~
the
:5
+
II
.Sl
':1'
{.3)
Uhae
....
i&
the
>
is
P:e
(b) WI"t
~
'JllC'"J:fy ~c.jf(![l in
t,ote.ntlal
~'hi:'n. f. hf' r"~nti,,lum
-
'"
ball
t,) l.'h.l.ch the
(:mali
an".••
'e~
?; -:::?f
\
j '2' t...l- ..Q.l d> '::
V
_
ca -t ~'.tJ
(f
instc'.'lt!
(VI.
the
$,'('c,:l
Cftl
iif:i'H',
1'(Y'\ '1i'- --=
'].
\'yw"~b(l. -r \'V\ Vfob\
yY\v; '2.::
l"Y\(
'2.
'2.. ~
'(Y\I/, -::. V\'\ VI
-
of
-~'4Io~"J.
2.MMv,
ill'
sf"",,<1
lit
and
th~
result
\'V\ VI ~
••
{\A
Vfbl -t m
".
"fbI")..
'4"
VI-bl
c I.€s -
of
the
+ \VI V~bl.'2.
;1llp,l, ..
m;'l;;f:",~!s
expr.~F;n
1.he
impac.t.)~_~
L-.
hlod.:.
~
_
((\')+VV\q)-l(C\\~
)"2. ~
\if
__
•••••
Q:+--_-
find
just
aft,~r
t.Ii,.,
in t" r••s of vi")
'M V.f-b\
rvWfb'l. .•
M Vrb'
YY'Iv .••..
'(V\
to
,,,1:f.Cuf the
t'oe
tli"
(If
ehe maxlml.t"l
(eneI'By-co[lservinf().
t-p'Jcd
2
bl
--~
f'p<"Jd
~
i~: elastic
e):pce"" c-h"!
l M"+
parU
l-
the
and
point
Vr.,.'&=
+
\L_--'
j..~(rY\v'j
b,~LJ "E
(I!gain,
-r
-I
Lind
.(>l11~'lon
the ball
"(Y'o.V,
V\'\ -t W\
th"
V"fb~
lk7- ~':.fiO;;~- c,," 'I
{A~S='J the
ea led
t
Q
1.
/Y\ Vi::
'fY\V,
,lfte~ the
the collision
of
#- rr.
'2.
t
""
,,.
Legeth,~,
rll~nd,d"Jll ,;;',;in
f.cnns
ill
'Jt:;.ck
just
thf':' Cali be:
so
'.',"" "
h ~ L-A.cC{,e
.3nd hled:
cornh! "I,d "'asses
fL,~1d
7,r,witatio[lal
,;?
~~b~'
the
ene
~~(JJ-..t ~e)
W~
If
(1:)
angle
.9(;
r.;:;~'
:'
;;-
m
of
/
II
II
'.li " hall
C(J'lSi:;Li."g
-::.L -.1(01 e
~ "" ")..V ':
(rv\-tYlA)
'2." '1.
9
;:Vo"f¥- _ (\ _'
rn "1.V '2..
I
.
\ lrM-W:tt
iffb
(bS8')
Jl
-;;e-
15
;r.n!iC: 3D
;:];\'c:rtictll
.L~1JJ:Jt:i.l:~llh'.•
the b.:::.ll
O:i
ret.urn
ttl
!tt~~
w.al1.
t;;\..;]ndr.
:l{
r.lf!t
trictior.l"'~:Q
f.jn J\
ir.p
rid::
~lc:' thk~lo,Ig .tit 2 kt?;b:1l11 hnri~l£mtil]1y
!:.h~ bi"J~'
r;.nlchoCrt
'(11('
\1;~lk.
ltJ;:""lt'\r,~
gT.fl.'"d,ty -
nr.i";u:ll.'!
~r~' ..'.cl:1 rl't}r-:lmnt !ll 1)' I
5
-'5
?i -c;:.?-f
D
s
2:;tyJ P -7- \f '0
-===
"t)OVp == 2. (\
v
,..---'- ....• ..,.. ..•..•..•.•
a)
I "ball
2::E
~
"OU
_ (,pI
t<)
5
I
"boy -
-
-
1:..0 ~
t~~:~.:~~~:-._-~__
ft ft 1
_••_
_ ___ \_"_~~
_V_
Vi ~ Pf'
z.,(\
lO~
0')
+
'LO
+
-:::32..v~
"bt> (.
(p 1)
2.0-\
= 3 2.. v~
4D. \ _ Vs;.
'{:,2.-
-l-Il
~
I
16
A hall of oass ml
spring.
2 kg is launched on a frictionless
=
The velocity
launcher.
of
A second
distance
ball
1 m from the
collision
with
the
ball
of mass
launcher.
the second
is
VI
:a
5 m/s when it leaves
~2 ~ 1 kg was placed
The first
ball
(m2)
plane by a
hall
and they
(ml)
the
at rest
then
both move
at a
makes a head-on
forward
along
the
same direction.
2 kg
5 m/s
,
I kg
62
~~
Fig.
(a)
Find
v
(b)
lhe
velocity
of
the
center
of mass of. the
I
two ball
system
(Vern).
'5l'2-) + \(0)
C '"""'
"3
If an observer
1s moving
velocity
of
(vIc)
the
at the same velocity as
first
ball
observed
what Is the
Vern'
by him before
the
collision?
(c)
Find the velocities
of the two balls VI; and v2; after the collision
wi til respect
launcher.
V.
I
to
the
Assume that
the collision
Is
V"'2 -::::- (VI -\/2.')
5 -= - \}
V1. -= ';)
I
-+ \1'2.
-T
VI
=
.
(.lal
?j ::::Pf
2.("'=»
-= '2 ( VI')
V-z. ')
"2V, of Vi
lo -
-z.. VI + S. 1" VI
\/1..'"
-:? VI
VI -:: 5/3
0(" \.
~ -T (I.Lil)
-
-5
- 5
'----
.j. \ (,
10
'0 -
elasti!=_
\11
V. VI
""'.5
"
'f