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Unformatted text preview: PHY 203 Exam llt Prof. John Rollino 1216108 1) Two blocks are sliding on a frictionless plane. A 20.0 kg block(A) is sliding to the right at 10.0 mils. while a 30 kg block(B) is sliding to the right at 3.00 mils. Block A collides with the back of block B. Block B is observed to leave the
collision with a velocity of 15.0 We to the right. a. Calculate the ﬁnal velocity of block A. VA .:  0 0 ("I [I
£9 [0 C 3 b
L“ “‘4’ iii ' “é C To (.51: 7') (.20 )(m .9 + (30)(3> 4205460305)
200 +330 T ZOJA + ‘{50
._ I60 : 20V!) b. Calculate the coefﬁcient of restitution.
é : VFD "' —" ?~ I S “Er”?! T 5—(0 : 3'19 c. Calculate the change in kinetic energy of the system as a percent of the original kinetic energy .. I_ L .L L
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2. £3 70 ﬂOrC C‘ d. How fast is the center of mass moving for the system described in part a? 200 ‘i’O'
VC '1 _.____:__. 5.80m/5 2.043’0 :— 9. Calculate the impulse applied to particle(A) in part g FM. WAVA— a”; : 206. 3») = _ 3m Ns f. Suppose we do the experiment over but the collision is 100% elastic: calculate the ﬁnal velocities. . l/A=(20.— so>/D + z(3:)(3) Vi}: 1“ 2036;”) n
57’ a 30 r5400 : .mm
400 + m” \f ‘ H h? A. B 55
5b )
= Ham/z; (wear gnaw—m =/0—? / 2) A 7.25 kg bowling bail is rolling without sliding down a 10.0 meter ramp inclined at 37°. Calculate the rotational
kinetic energy when it reaches the bottom of the ramp assuming that it statted from rest at the top of mi mp. ‘ ‘ ' t 2— 2
Assume the bowling ball is a solid sphere. % E {r : 2 (s m a £55 gpmsl, ,_. gmmpoawst‘) "lg/E”
2 2' 4 r
g  ‘39’2?.é I rEiCﬁTAc. J27 gfma ya 9;”
h = =,/
Elia .%Ha?é=r22 T 1;: = 3’; (1o)
0.. ____ 3) .A 150g plastic meter stick balances at the 50.0 cm mark. If we put a pivot (fulcrum) under the 35.0 mark, w ere
must we a h a 100 9 mass so the meter stick balances on the pivot? Acfemﬁf/ «or {‘01, a/ Mom AT 00 0’” WM“ “‘0’ loo 1» 293(33) ~— Hog—a) = 5
at T + W 192‘” ﬂ? IO/Vo’T 100
1000) = For") (A, w 12f?“
Y . 2L9.” 10
1009 u Mme» ﬁr lZSVM. ‘ ’
4) A large roll of wrapping paper, found often in a florist shop ( m = 15.0 kg ) which is 0.900 meters long has a diameter
of 30.0 cm. The friction the center axle has with the bearings is negligible. A ﬂorist begins to pull on the paper with
a constant force of 16.0N.  = (‘xQMR2
a. When the roll achieves an angular velocity of 10.0 radis, what length of paper has come off the roll 2T: Rl—‘elvs 1 9Q=___CQ_3_ QCJMw/JJl
¢z§)(w)r_—‘£(w)¢w) o< (:04 ‘ W3
ole M224 AG .~ 3.527142
5‘0 63/»: Sega“? (3.52 fMXJS‘pﬁ) = c b. Calculate the roll‘s kinetic energy. Q. —~ '1' l —~ .1 ‘
Er: 2 la) =31;— ‘sXJbJ (to) 1. «WI
W = 7—18 : 2VONd") (and) : QAHI / c. How much time elapsed during this event? ’L‘? rg‘)” “ MA— : ¢?c)3_s O< H. ?, Veal/51 (15) 5) 6) A horizontal wheel whose rotational inertia is 5.00 kgm2 is rotating about its vertical central axis at 300 revolutions per minute. It is stopped by friction by a brake which presses on the outer rim. The coefﬁcient of kinetic friction is 0.40. The radius of the wheel is 5.00 cm.
A. With what force must the brake be applied to the wheel in order to stop the wheel in 12.0 seconds? _ a.) : lo’TI‘V‘hMS
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5'1
5K :(S'OOJQ'953TP>_ to gas)
.. 26!.8‘
N ‘ E 7 ——__ z .4 ‘/
B. How many revolutions are need to stop the wheel
9": (U {OJ1 _é
1
f1
_ n  600M :— 3d.0 Vex/3
Q‘ . I02.— )2. .r (20) In another experiment. the wheel in problem ﬁve (5) is rotating freer at 200 rpm and a disc whose mass is 2.00 kg
and radius is 0.300m is dropped on the wheel so that both central axes form a straight line. The wheel and disc stick together and rotate as one.
A. Calculate the ﬁnal angular velocity. 1 T::: viacow)" =4)“ “‘1‘”
1::glh 51w Ham F”
(53000) = may 513)
:31on .20.; m! .— '
5m“ (~721me (R 5" 4
B. Calculate what percent of the kinetic energy remains.
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,— 2. “l 07
7i? ﬁ/lﬂ‘ I«‘ 20— hitﬂ “
6 {C '—""'""1 ' j '31:? (15) ...
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This note was uploaded on 01/29/2009 for the course PHYSICS 750:203 taught by Professor James during the Fall '08 term at Rutgers.
 Fall '08
 James
 Physics

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