chap 15 test

chap 15 test - 1. A 58—g tennis ball has a horizontal...

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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. +ngl-t : 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 .06-kg 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‘ 4 " x3: " " - ' .. 59.04 0‘ 7: 4" + U52? : 7:? "L V4" .. O .+ O 2. Twoblocks A and B each with a mass of_4-kg 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 'Z-S- M/Slcr‘ \ \ 4(a)) + ribs) .2 AMA + 6% _,-; 20 . ‘ \ VA -i—\/6 : € \ \- \ ‘ I V #V I_e:__\/5_ " " _§5 _: 6 “A? “5“0; 10“(">) '4 l H‘ \ ' s‘ \ 3 3.§w 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—r-l 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‘Kl-T v42~4§"p‘4a bill "‘ x ' Elli— : \/(g 4 '3‘ Vin" (“42'”) ' I M : I“ 4%“) 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 VET-a : 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.

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chap 15 test - 1. A 58—g tennis ball has a horizontal...

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