Last Name: First Name ID s1

Physics 101. / Hour Exam I / Summer 2015

Last Name: First Name Net ID:______

Discussion Section: Discussion TA Name:

Instructions—

Turn off your cell phone and put it away.

Calculators cannot be shared. Please keep yours on your own desk.

This is a closed book exam. You have sixty (60) minutes to complete it.

1. Use a #2 pencil; do not use a mechanical pencil or a pen. Fill in completely (until there is no white space visible) the circle for each intended input – both on the identification side of your answer sheet and on the side on which you mark your answers. If you decide to change an answer, erase vigorously; the scanner sometimes registers incompletely erased marks as intended answers; this can adversely affect your grade. Light marks or marks extending outside the circle may be read improperly by the scanner.

2. Print your last name in the YOUR LAST NAME boxes on your answer sheet and print the first letter of your first name in the FIRST NAME INI box. Mark (as described above) the corresponding circle below each of these letters.

3. Print your NetID in the NETWORK ID boxes, and then mark the corresponding circle below each of the letters or numerals. Note that there are different circles for the letter “I” and the numeral “1” and for the letter “O” and the numeral “0”. Do not mark the hyphen circle at the bottom of any of these columns.

4. Print your UIN# in the STUDENT NUMBER designated spaces and mark the corresponding circles. You need not write in or mark the circles in the SECTION box..

5. This Exam Booklet is Version A. Mark the A circle in the TEST FORM box at the bottom of the front side of your answer sheet.

6. On the SECTION line, print your DISCUSSION SECTION. (You need not fill in the COURSE or INSTRUCTOR lines.)

7. Stop now and double-check that you have bubbled-in all the information requested in 2 through 6 above and that your marks meet the criteria in 1 above. Check that you do not have more than one circle marked in any of the columns.

8. Sign (DO NOT PRINT) your name on the STUDENT SIGNATURE line.

Before starting work, check to make sure that your test booklet is complete. You should have XX numbered pages including X Formula Sheet (s).

Academic Integrity—Giving assistance to or receiving assistance from another student or using unauthorized materials during a University Examination can be grounds for disciplinary action, up to and including expulsion.

Page 1 of 11

(22 problems)

Physics 101. / Hour Exam 3 / Summer 2015

Exam Grading Policy—

The exam is worth a total of 123 points, and is composed of three types of questions.

MC5: multiple-choice-five-answer questions, each worth 6 points.

Partial credit will be granted as follows.

(a) If you mark only one answer and it is the correct answer,
you earn 6 points.

(b) If you mark two answers, one of which is the correct answer,
you earn 3 points.

you earn 2 points.

(d) If you mark no answers, or more than three, you earn 0 points.

MC3: multiple-choice-three-answer questions, each worth 3 points.

No partial credit.

(a) If you mark only one answer and it is the correct answer,
you earn 3 points.

(b) If you mark a wrong answer or no answers, you earn 0 points.

TF: true-false questions, each worth 2 points.

No partial credit.

(a) If you mark only one answer and it is the correct answer,
you earn 2 points.

(b) If you mark the wrong answer or neither answer, you earn 0 points.

1. A uniform plank of mass 40kg and

length 5m is held up by two supports as

shown. The left support is attached to the

left end of the plank, the right support

is 3m from the left end of the plank.

What is the force exerted by the left support on the plank?

(A) F = 49.0 N

(B) F = 65.3 N

(D) F = 81.3 N

(E) F = 103.5 N

The following two problems refer to this situation:

A uniform rod of mass 3kg and length of 4m is supported by a rope as shown. On the far left end of the rod a block of mass 12kg is suspended. The system is in equilibrium.

2. What is the tension, T, in the rope supporting the rod?

(A) F = 147.0 N

(B) F = 152.4 N

(D) F = 161.3 N

(E) F = 196.0 N

3. What is the distance, d, from the left end of the rod to where the rope is attached?

(A) .15 m

(B) .21 m

(D) .65 m

(E) .80 m

The following two problems refer to this situation:

4. Curtis is painting his house one summer day. He sets a board of mass = 30kg and

length = 4m on two supports so he can reach higher on the wall. If Curtis has a mass of 70kg and is standing directly above the right support, what is the magnitude of the force that the right support applies to the board?

(A) 620 N

(B) 648 N

(D) 833 N

(E) 935 N

5. How far can Curtis walk to the right (from his initial position directly over the right support) before the board tips up?

(A) 0.43 m

(B) 0.50 m

(D) 0.77 m

(E) 0.81 m

6. My wristwatch is water resistant to 50m. How much pressure differential can my watch take before it will leak? (Assume that there is atmospheric pressure inside the watch and the density of water is 1000 kg/m3)

(A) 4.90*105 Pa

(B) 5.03*105 Pa

(D) 5.83*105 Pa

(E) 5.98*105 Pa

The next three questions pertain to the following situation:

A solid disk of mass m starts from rest at a height h and rolls without slipping down an incline plane. The linear velocity of the disk at the bottom of the ramp is v.

7. What is the total Kinetic Energy of the disk at the bottom of the incline?

(A) 1/2mv2

(B) 1/4mv2

8. If v is 5 m/s what is h, the initial height of the disk?

(A) 1.25 m

(B) 1.31 m

(D) 1.67 m

(E) 1.91 m

9. If v is 5 m/s and m = 4 kg what is the rotational Kinetic Energy of the disk at the bottom of the incline?

(A) 20 J

(B) 25 J

(D) 32 J

(E) 36 J

10. One way to measure a person’s percentage of body fat is to

fill a tub to the very top with water and have the person get

into the tub and become completely submerged. Of course

water will spill out of the tub. The water that spills out is

collected and weighed. If the weight of the overflowed water is

637 N and the weight of the person is 588N, what is the density of the person?

(A) ρ = 860 kg/m3

(B) ρ = 923 kg/m3

(D) ρ = 1017 kg/m3

(E) ρ = 1210 kg/m3

11. An open U shaped tube is initially partially filled with water (rw = 1000 kg/m3). A 10 cm high column of oil (roil = 880 kg/m3) is carefully poured into the right side of the tube (see picture). What is the height of the oil above the water, H.

(A) H = 1.2 cm

(B) H = 2.0 cm

(D) H = 3.1 cm

(E) H = 3.6 cm

12. A block with a mass, m1 = 2kg, is rotating in a circle of radius 1m with an initial angular velocity, ω = 10 rad/s. A string is attached to the block and passes over a pulley and a force, F, is keeping the block moving in a circle. The maximum tension the rope can apply before breaking is 300N. If the force is increased pulling the block inward, what is the minimum radius the block can have without the rope breaking? (Hint: Use Newton’s 2nd law, centripetal acceleration and conservation of momentum)

(A) 0.12 m

(B) 0.35 m

(D) 0.73 m

(E) 0.87 m

The next three problems pertain to the following situation:

Water (rw = 1000 kg/m3) flows though a cylindrical pipe of radius R1 = 10 cm with a speed of v1 = 0.25 m/s. After a narrowing of the pipe to a smaller radius R2, the speed of the water is measured to be v2 = 2 m/s.

13. What is the radius of the narrow part of the pipe?

(A) R2 = 3.5 cm

(B) R2 = 5.2 cm

(D) R2 = 8.8 cm

(E) R2 = 9.2 cm

14. What is the magnitude of the difference in pressure in the two segments of the pipe.

(A) |P2-P1| = 354 Pa

(B) |P2-P1| = 546 Pa

(D) |P2-P1| = 1553 Pa

(E) |P2-P1| = 1969 Pa

15. Compare the difference in pressure in the two segments of the pipe.

(A) P1 < P2

(B) P1 = P2

16. Your grandfather clock is running too fast. (The period is too short) How would you adjust the pendulum to correct it?

(A) Make the pendulum shorter

(B) Make the pendulum longer

The next four questions refer to the following situation:

A block of mass m = 5kg and initial velocity v = 2m/s runs into and compresses a spring of spring constant k = 70N/m. After compressing the spring, the block remains connected to the spring and continues to oscillate back and forth. The surface between the block and floor is frictionless.

17. What is the period of the oscillation?

(A) .12 sec

(B) .35 sec

(D) 1.7 sec

(E) 2.0 sec

18. What is the amplitude of the oscillation?

(A) .12 m

(B) .35 m

(D) .63 m

(E) 1.2 m

19. What is the total energy of the block and spring system?

(A) 10 J

(B) 12 J

(D) 18 J

(E) 21J

20. What is the maximum acceleration of the block?

(A) 4.2 m/s2

(B) 5.5 m/s2

(D) 7.5 m/s2

(E) 8.3 m/s2

A hydraulic lift is used to lift a block of mass M = 2000 kg. The block rests on the right massless plunger. A force F1 is applied to the left massless plunger. The two plungers are at the same height. The left plunger is circular with diameter D1 = 0.02 m. The right plunger is circular with diameter D2 = 0.5 m. (Radius is D/2 and area = pr2)

21. What force F1 is required to lift the block?

(A) 1.2 N

(B) 3.9 N

(D) 15.3 N

(E) 31.4 N

22. A freely hanging block of mass 5 kg is attached to a massless string, which is wrapped several times around a pulley (solid cylinder I = ½ M R2) of mass 10 kg and radius 0.2 m. (see diagram). There is no friction in the pulley mechanism. Initially the block and pulley are at rest. The block is released. What is the velocity of the block after it has fallen a distance of 1.5m?

(A) 2.6 m/s

(B) 3.8 m/s

(D) 5.4 m/s

(E) 6.3 m/s

Physics 101 Formulas

Kinematics
vave = Dx/Dt aave = Dv/Dt
v = v0 + at x = x0 + v0t + 1/2at2 v2 = v02 + 2aDx
g=9.81m/s2 = 32.2ft/s2 (near Earth’s surface)

Dynamics
SF = ma Fg = Gm1m2 / R2 Fg = mg (near Earth’s surface)
fs,max = msFNGravitational constant, G = 6.7 x 10-11 N.m2/kg2
fk = mkFN ac = v2 / R = w2R

Work & Energy
WF = FScos(q) KE = 1/2mv2 WNET = DKE = KEf - KEi
Wnc = DE = Ef - Ei = (KEf + PEf) - (KEi + PEi)
Wgrav = -mgDy PEgrav = mgy

Impulse & Momentum
Impulse I = FaveDt = Dp FaveDt = Dp = mvf - mvi Fave = Dp/Dt
SFextDt= DPtotal = Ptotal,final- Ptotal,initial (momentum conserved if SFext = 0)
Xcm = (m1x1+ m2x2)/ (m1+m2)

Rotational Kinematics
w = w0 + at q = q0 + w0t + 1/2at2 w2 = w02 + 2aDq
vT = wR aT = aR ( so for rolling without slipping v = wR a = aR )

Rotational Statics & Dynamics
t = Fr sin q
St = 0 and SF=0 (static equilibrium)
St = Ia
I = Smr2 (for a collection of point particles)
I = 1/2MR2 (solid disk or cylinder)I = 2/5MR2 (solid sphere)I = 2/3MR2 (hollow sphere)
I = MR2 (hoop or hollow cylinder) I = 1/12 ML2 (uniform rod about center)
W = tq (work done by a torque)
L =Iw StextDt = DL (angular momentum conserved if Stext = 0)
KErot=1/2Iw2 =L2/2I KEtotal=KEtrans + KErot = 1/2mv2 + 1/2Iw 2

Simple Harmonic Motion
Hookes Law: Fs = -kx
Wspring = 1/2kxi2 - 1/2kxf2 PEspring = 1/2kx2
x(t) = Acos(wt) or x(t) = Asin(wt)
v(t) = -Awsin(wt) or v(t) = Awcos(wt)
a(t) = -Aw2cos(wt) or a(t) = -Aw2sin(wt)
w2 = k/m (mass and spring), T = 2p/w, f = 1/T
xmax= A vmax = wA amax= w2A
For a simple pendulum w 2 = g/L

Fluids
P2 = P1 + rg(y1-y2) change in pressure with depth
Buoyant force FB = rgVdis = weight of displaced fluid
Flow rate Q = v1A1 = v2A2 continuity equation (area of circle A = pr2)
P1 + 1/2rv12 + rgy1 = P2 + 1/2rv22 + rgy2 Bernoulli equation
rwater = 1000 kg/m3 1m3 = 1000 liters
r = M/V 1 atmos. = 1.01 x 105 Pa

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(22problems)