Buchanan_-_Phys_1A_-_Midterm_Exam_1_-_Solutions

Buchanan_-_Phys_1A_-_Midterm_Exam_1_-_Solutions - Cdb...

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Unformatted text preview: Cdb MidTerm Exam #1 Physics 1A One 3x5 formula card and calculator are allowed (and encouraged) Write units for all answers. NAME So ] u‘t ‘n o n 3 Student ID Number Problem #1A /8 Problem #IB /9 Problem #1C /8 Problem #2 /20 Problem #3 /30 Problem #4 /25 TOTAL / 100 Wes/’3 ‘v in 511$ ace #1A) 8 Points Person A holds a dollar bill, which is 15.6 cm long- Person B has his fingers at the bottom of the bill, ready to snap closed on it. Person A releases the bill. How quickly, after person A releases the bill, must Person B close his fingers together in order to catch the bill? ) Z Tapas-m = xm=igt t Z 3;}. 1%IM’L : (LDXSM, 3 CL? #lB) 9Points A block of mass M (weight =Mg) sits at rest on a horizontal table, as shown. The figure has three forces drawn in it: ° The force of gravity on the block. ' The Normal force up on the block from the table. ° The force from the block down on the table. Note: Each of these forces is equal to Mg (the block’s weight). a) 4 pts. What is the Newton’s 3pd Law partner of the Normal force? 1)) 5 pts. What is the Newton’s 3‘11 Law partner of the force of gravity on the block? é-—-—- w {~— ghtd’ll RWY“ bloeK W3 N .— N’s — E“*Q\‘.’ immkl‘i‘vk Ma ohbloclk’s Law VQYXMQTXS the. sofa. ei- grkv\\\,-l <3“ 8w Eav‘W imam-km black. #1C) 8 Points A block of mass M on a frictionless horizontal surface has a force F applied to it at angle of theta below the horizontal (as shown). H3 N1 ° a) 4pts. WritetheexpressionfortheNormal if]: N-Mcs‘ FWQ ’2: M43 "LO Force from the table on the block. ‘ =3 N —_ rig-t- Home b) 4 pts. Write the expression for the acceleration of the block. w: L, , 1': #2) 20 Points Two masses of 10kg and 5 kg are on frictionless slopes of 45° and 30°, respectively, as shown. A s \I M 3 system of massless, frictionless pulleys connect them Mzc.‘ ‘ such that the string in each case is parallel to the surface. _ __ ,‘L a) 14 pts. Calculate the acceleration of MI. E Fm : h‘ 63 Mgr l : H, Q \. “I; z p“: T- mauve; = me Add .. M‘ (5 MQFHICSMQIB 2 (H3911) 6x a _ C5_(n,sime,-“1W9e\ : cm [.ogwagtwmse h ‘ H1 1 | s’ - I b) 6 pts. Calculate the tension in the string. ’ l'qq “/ 5 —r.. nutagmezw)=§(q-%*%+1~qq)=3°|'” u I r ‘ mtg-“é ‘4- 1}; l M‘Ql . n), s l .. L __ f 1., ' [\«H'l-h‘v 31S3=)i$1a‘ : Siw‘filw‘ (:32 lhp‘ffiy, Livl"‘rr‘“:f)j 3323‘ Rm #3) 30 Points 4/ Shaquille O’Neal is going to attempt a freethrow. As V0 diagrammed, the ball begins “exactly” 7 feet (2.134111) “0 above the floor, the center of the basket is 15 ft (4572m) , away horizontally, and 10 ft (3.048m) abov e floor -— T ‘9) that is, 0.914m above the start height. g éfigem' lsecz. 7F.“ He shoots the ball at 55° above the horizaontal. 2.13%“ h " tsrt : 4.572 m ’9‘ a) 13 pts. Pick a coordinate system and write k the equations of motion x(t) and y(t) for the ball, plugging in as many numbers as you can. W V (t): v.m55°t 03$: =3 ‘+.§71=Q>.S33(9 Vutt, 9" *2. Wu —. 1.13% +vbz§w~55°i —%t"' Yo ‘5 (saws—mu») = voxbfsvn tb— that); W Q,°\l'-t Rm mm *51 LEB— : 237; b) 12pts. Find,tofour significant figures, ,smvo V0 the speed at which he must shoot for the 1 center of the balltoyass through the center , _ q 7m! ’5 “)fl ofthebasket. '- Ayah): 00”” ' 055‘ 1 q: '6‘ Van. W WV :3 , N’snrs : , Vt’ de> 7 %‘l5\“/§ c) 3 pts. Calculate the horizontal and vertical components of the hall’s velocity 'ustasit ase thro hth b kt. ._ .. = ” 'o-v . J p SS “1; e as e vy(t*)-v°y 63H 7.stis Hme, M bJBGl‘! 7' 'Ll'.3cll m/S d) 2 pts. Convert fliese results into the magnitude and direction of the velocity “3:”; 5 '15 m 5 as it passes through the basket. V :Vux * V)‘ c) ‘ I #4) 25 Points The Rainbow at Magic Mountain is an “interesting” ride. About 20 years ago, my children talked me into going on it with them. It’s a large circular steel apparatus, initially horizontal, with a 7 ft wall around it. People enter it and press their backs against the wall. It begins to revolve and, when it is up to speed, then it tips up so that is revolving vertically. That is, its passengers are now forced into Uniform Circular Motion in a vertical plane. Miraculously, the passengers remain in contact with the wall, even at the top where you might think they might fall out. Thinking of future 1A classes, I measured the time for one revolution and found it to be 6.00 seconds. When the ride was over, I stepped off the radius and found it to be 30.0 ft. a) ZPtS. Show that the radius is 9.15 meter. R 1 3 5'6 Q /3-1efl/‘“: c“ l 5'“ .\ s) b) 2 pix. Show that the speed at which the people V = iii—$3 : 12; are going around is 9.58 m/sec. ‘9 1 qS‘KN/s }' cl'sgl... L : x ». Cry/R: T ,le.0‘kM/s b mm is ) c) 4 pts. Calculate the acceleration. .1 g d) 4 pts. Define the person at the top as a system. Draw all the forces g the person. Label and explain them carefully. d) 5 pts. Pick a coordinate system (I would suggest you pick down as the +x direction) and set up Newton’s 2Ind Law for the person at the top. : Fx: M%*N:fla i)": )0: quscvx‘u lotions geramd lo») wat\\ to theluo‘to. «’1 me“- Una“ \WS how“ on "v Wu“ 'ts Quskx \\ ryuszm \mwm up wam =5 {Lt-7 SLV \‘nw'k‘ct * e) 8 pts. Interpret What is happening. Why do the passengers stay in contact With the wall at the top? i? (gt $3“ ...
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This note was uploaded on 05/29/2011 for the course PHYS 1A taught by Professor Musumeci during the Summer '08 term at UCLA.

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Buchanan_-_Phys_1A_-_Midterm_Exam_1_-_Solutions - Cdb...

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