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Unformatted text preview: Lab 7 The Ballistic Pendulum l Lab Report (100 pts) Name: PHYS 1114 327114 Section: QQQ Day and Time: 3—?» r ‘ " 17. L30 Station Check: USE LONG RANGE ONLY! ments of the Pendulum from Vertical 25 ts mc= 0.0893 kg L = 0.374 m
63 Average 8
(deg) (deg) nitude of the Anular Dis 1 lace Average R
(m) Questions and Exercises
1. Are there any nonconservative forces that do not do work on the block during the period of time when the pendulum (block and tubing) is swinging upward? Explain.(3 pts) /.’ z \ v Q l; N. .1 j elf,» 5: AJ‘A \DQFQ‘U} a t, Ma»...— "luﬁ Ivy" 3‘\ sank 6/ I Are there external non—conservative force(s), besides air resistance, that do work on the
ball/pendulum system? Explain.(3 pts) r‘ i \ d» ix . 1 . I
H Lawn ‘0 ~11.. Qv» »<::f m 69 Lab 7 The Ballistic Pendulum 2. The gravitational potential energy, GPE, of a particle of mass m located a distance of h above the reference position is deﬁned by GPE = m g h. H £1.19 L
I Show that the height change of the block is given by h = Lfl Xi‘hiL 2} “Uhr“ =73 tinLang : if,
t'aﬁ'ﬁ‘fﬁ}: x/L é x: Laose 1/!” . Use Your data recorded in Table 1 to calculate the h eight Change Ofthe bIOCk.
M Z L( g “@2353 A s
f? 7 Weigh,‘ a : “if?” ‘m’ . «5231:; Ann.“ .5 I What is the change in gravitational potential energy of the ball/block? Show all work,
units, and equations. (9951:. yth = @055 ku ﬂaw/,5? * gQﬁB‘W, = 432337“, 4 ~ .3 3. Use the principle of conservation of mechanical energy (Equation 1) to determine the common speed v of the ball and block immediately after the collision. Show all work, units,
and equations. (5 pts) r N 1
(Vi/W)" 1”“c}q‘v\ : If? Emat”: jV
I J 263%. = V Q \ $13»! ; \ : .etw’s / 4. Why is it permissible to apply conservation of linear momentum during the brief period of
time immediately before to immediately after the collision, but not during the longer period
of time when the block is swinging upward? (5 pts) v. ‘ ‘I‘ ‘ f ‘ ‘ rl/ I, “N
[:4 '3 5/“ \1 var M9459“ "~. ‘ , ‘ 7O Lab 7 The Ballistic Pendulum 5. What is the total horizontal momentum of the ball/block system immediately after the collision? Show all work, units, and equations. (5 pts)
(Wu/new => (.o‘ awe r «9293 mg . <3? = .0702 “mg / 6. Use the principle of conservation of the total horizontal momentum (Equation 2) of the
ball/block system to determine the initial speed V0 of the ball. Show all work, units, and equations. (4 pts) mam, : (\mgstV \ V0 : LMQ'kVWC‘j‘k/ : : 0%" {(6.5
m6 \ A J 7. What is the kinetic energy of the ball/block system before and immediately after the
collision? Show all work, units, and equations. (9 pts) 1/ m ‘ a, ‘ ‘3 ._ 1 3 / \‘P‘ A q
lit/M w Viarw/ ~ 'éCmemclV = Aromas xggttb’t’i'tu : 02339 a i y ' , {Ser‘A '2  ‘ “,
Egiozlwve; Verna V? = Vat ﬁrmsttlwm “9 * \36‘473’ /” o Is the kinetic energy conserved during the collision? If not, account for the decrease in
kinetic energy. (3 pts) .. f r" 
KEo' 24:4: : ‘l6l473ﬂoggg'y  ‘tgoqj‘
EtaL‘s“: 0;? ‘ii§§%33fk¥l¢§i ﬂ “wimp, o Is the ball/block collision an elastic or inelastic collision? Explain.(2 pts) MAE/63”»; Ct, cJ 1M: ‘30 ,/
1 ‘ j x a I’E'Ti far» 03 *3“ «i? ”
{Ur sow"? 5’1
”" Tn afar ’1 gsiaeee w <2 52 r‘ r “ e z ,, “a I \ 71 Lab 7 The Ballistic Pendulum 8. Altemately, the launch speed V0 of the steelball can be determined from the kinematic
equations for projectile motion. Figure 2 shows a oneinch diameter steel ball launched
horizontally from a springloaded projectile launcher a distance H above the ground, which
then lands a distance R from the point on the ground directly below the launch position of the
ball. The kinematic equations for projectile motion are KG) 2 X0 + ont (3)
and
Y(t)=yo +Vyot%gt2, '(4) Use Equations 3 and 4, the kinematic equations for projectile motion, to show that the initial
speed v0 of a steel ball that is launched horizontally is given by Equation 5, 2
magi, (5) where H is the vertical distance between the ﬂoor and the launch position and R is the
horizontal range . Show all work. (11 pts)
(USE SYMBOLS OR ZERO TO FILL IN BLANKS) 0 s (when balls leaves launcher) t (when ball hits the ﬂoor) X0: X: R
on= \/0 y= 0 Substitute the above values into Equation 3
KW: >4¢+VXQ+ =3 R. = V.) "r  Substitutetlre above values into Equation 4 ym: \t o i“ {413% “Va 39$} => 0 t ~ ifaafq // 0 Use the new equation 3 and 4, to show that the equation 5 is correct. mg: :3. '4," : “1;.ﬂ/U
2 ’ i f, \22 H: {Rim 3 1:4: .lQSK‘B’iw/f /// f \ / '1 3..
I m 6 Q
2p — ER 5:» o a
V01 L; 72 Lab 7 The Ballistic Pendulum  0 Use your data in Table 2 and Equation 5 to determine the launch speed of the ball, that is,
the speed of the ball before the collision with the block. (3 pts) 9. In Exercises 2 — 6 conservation of the mechanical energy of the ball and block after the
collision and conservation of linear momentum of the ball and block during the collision are used to calculate the speed of the ball immediately before the collision with the block, while
in Exercises 8 and 9 the kinematic equations for projectile motion are used to calculate the
launch speed. (12 pts) ‘ I Calculate the percent difference between the two speeds. tiesﬁﬂwagl ’00 I /
____.__________. K x: o
awe#4335 /// 25‘ S Q ’1
A ' How well do the conservation principles for linear momentum and for mechanical energy
apply to the ball/block system? a 55 I .3 x
No: my I List two physical factors and their effects that have not been taken into account that
could explain the difference in the two estimates of the launch speed of the ball. (Do not
list air resistance, or variations in the launch speed due to the spring loaded release
mechanism.) Human LJ ‘ '° (
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 Spring '08
 Nandi

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