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Unformatted text preview: 1. A 58—g tennis ball has a horizontal speed of 10 m/sec when it is struck by a tennis racket as shown. After the impact, the ball's velocity is 25 m/sec and makes an angle of 15° with initial direction. if the time of contact is 0.05 sec, determine the average force at impact (direction and magnitude)
of thetennis racket on the ball. +nglt : My? [05800)]: + Fﬂsﬁos) :: .058(25)[r¥ (1056; + 511415;] 56057“: [~58 —1.‘t]x + 375'.)
r3 —.. —.5‘i6:+ 7.573" (N) +~ N _ 1041'
P =— 40.3 1. A .06kg bullet is moving to the right with a speed of 450 m/sec when it
strikes and is then embedded into a 20—kg wooden box that is also moving
to the right at a Speed of 5 m/sec. The coefﬁcient of kinetic friction between
the box and the ﬂoor pk = 0.3. Use the conservations of momentum and
energy to determine a) the initial speed of the bullet—box system when the
bullet becomes embedded in the box, and b) the distance that the unit moves
before coming to reSt. ' Sml's ' v=0 ___.. __ g IV'M
m 1,. 3 “2%” 05X!» 33) — 4‘” 9 90'
T “‘0 U “ Fat: ~(2‘3'c‘09'8w‘
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.. 59.04 0‘
7: 4" + U52? : 7:? "L V4"
.. O .+ O 2. Twoblocks A and B each with a mass of_4kg slide on a smooth.
horizontal bar with the velocities shown, Determine their velocities
after they collide a) if they stick and slide together after impact, and
b) if they. rebound with a coefﬁcient of restitution e = 0.8. ‘ i MAVA + Mfg/(5 : (MA 40o) 4.. 4(~s) :. 3v
Vie 'ZS M/Slcr‘ \ \
4(a)) + ribs) .2 AMA + 6% _,; 20
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V6~VA 3 : 5’VA "v4 \4‘ (:1,
‘JW/s 2. A baseball bat strikes a baseball as shown. Before the impact, the base—
ball has a velocity vb = 132 (cos 45°i + cos 45°j) while the bat has a velocity ' VB 2 60 (—cos 45° i — cos 45°j ), both in ft/sec. The coefﬁcient of restitution
as a consequence of the impact e = 0.2. Determine the ball’s velocity after
impact. Assume that the ball and bat are moving horizontally. ASSUmc no dim/toe ill 1M4 Velocl'lq
0.52 +qu 'bd‘l (“QQeSSGI—rl w/£3 masses) baseball TL“ '75 M CLﬁﬂqe :1. ll»: wow? '1‘
0+ +lq loan Ila +1“ Y organe/ Vb ‘ 132 atoms" a 933:4 {l/m loqll .. x 4min” 1‘ l ‘ disc‘er
V " "bows‘KlT v42~4§"p‘4a bill "‘ x '
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a: 0'2:_ v' ‘ 43.3%”. .
7‘ ' ' Vb“: " 5,; I
= :v 9.53 V '
\ll,’q _ w .Vb; 2 ISZCoS‘K. == Cissy Vs... or \ib‘F :: .. 4,9,;ng 4. 93.373. 3. A satellite is in an elliptical orbit about the earth as shown. its velocity
at perigee A is 8640 m/sec. The radius of the earth is 6370 km. The mass
of the earth Me = 5.97600“) kg and the universal gravitational constant
G = 6.673 (10“) m3/(kg 5.2). Use the conservation of angular momentum
to determine the magnitude of the satellite’s velocity at C (the apogee),
and at point B in its orbit. Ell/MW) 1 {cam/c) 2000 ($940) = 74000 Va
Va 3 2380 Pvt/sec VETa : V5 ssh {30.08 I “"5014 V6 34;. 60.08” = TAO/MW) moss V’s (. see?) .2 $5000 (gave) V6: 4972.4, Via/5;“ ...
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This note was uploaded on 02/17/2011 for the course MAE 2323 taught by Professor B.p.wang during the Spring '08 term at UT Arlington.
 Spring '08
 B.P.WANG

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