Physics 110

Spring 2010

Exam #1

April 16, 2010

Name______________

In keeping with the Union College policy on academic honesty, it is assumed that you will neither accept nor provide unauthorized assistance in the completion of this work.

Part Multiple Choice / 10

Problem #1 / 27 Problem #2 / 27 Problem #3 / 36

Total / 100

Part I: Free Response Problems Please show all work in order to receive partial credit. If your solutions are illegible

no credit will be given. Please use the back of the page if necessary, but number the problem you are working on. Each subpart of a problem is worth 9 points.

1. A Boeing F/A-18 Super Hornet with a mass of 30,000kg lands on the deck of an

aircraft carrier at a speed of 140 mi/hr (~63 m/s) and is brought to rest by an arresting cable (shown at the very back of the aircraft) that is stretched across the deck of the carrier as shown in the figure below.

a. If the jet needs to be brought to rest

in 1.5s, what are the minimum force and acceleration exerted by the arresting cable?

zxfzdfasdfasdf

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v fx = vix + axt→ 0 = 63 ms − ax 1.5s( )→ ax = 42 m

s2

Fx = max = 30,000kg× 42 ms2

=1.26 ×106N, both opposite the landing

aircraft.

b. From the time the jet hits the carrier deck, how far does it travel before coming to rest?

€

Δx = vixt − 12 axt

2 = 63 ms ×1.5s( ) − 1

2 42 ms2( ) 1.5s( )2 = 47.3m

c. Of course in order to land the jet, it must have taken off from the carrier. A steam catapult located on the front of the flight deck launches the planes. Suppose that the plane needs to be launched from rest to 63 m/s. If the flight deck is 100m long, what is the time it takes to launch this plane and what are the magnitudes of the acceleration and force needed to launch this plane?

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v fx2 = vix

2 + 2axΔx→ ax =v fx2

2Δx=

63 ms( )2

2 ×100m=19.9 m

s2

Fx = max = 30000kg ×19.9 ms2

= 5.95 ×105N

Δx = 12 axt

2 → t =2Δxax

=2 ×100m19.9 m

s2= 3.2s

in the direction of launch.

Arresting cable

2. In a local bar, a customer slides an empty beer mug down the counter for a refill. The bartender is momentarily distracted and does not see the mug, which slides off the counter and strikes the floor 1.40m from the base of the counter. Suppose that the height of the counter is 0.86m above the floor.

a. What was the speed of the mug when it leaves the counter?

b. What is the impact velocity of the mug just before the floor?

c. How long does it take the mug to reach the floor?

FNC

FNL

FTR FTL

FWC

FWL

FTR

FWR

FTL

3. Suppose that you are given the arrangement of masses shown below where the left mass mL = 1kg, center mass mC = ½ kg and the right mass mR = 4kg. The masses are on frictionless surfaces and are released from rest. You may assume that the center mass moves to the right.

a. Draw a carefully labeled force diagram showing all of the forces that act on each

mass. You can draw these on the diagram below, but be sure to specify a coordinate system for each mass. The triangular mass on the left is on an incline that is oriented at 51o with respect to the horizontal.

b. What is the magnitude of the acceleration of the system?

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mC : Fx : FTR − FTL = mca∑mR : Fx : FTR −mRg = −mRa∑ → FTR = mRg −mRa

mL : Fx : FTL −mLgsinθ = mLa∑ → FTL = mLgsinθ + mLa

FTR − FTL = mRg −mRa −mLgsinθ −mLa = mca

∴a =mRg −mLgsinθmL + mC + mR

=4kg × 9.8 m

s2( ) − 1kg × 9.8 ms2 sin51( )

5.5kg= 5.74 m

s2

c. What are the magnitudes of the tension forces that act on the left (FTL) and right (FTR) sides of the center mass?

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FTR = mR g − a( ) = 4kg 9.8 ms2− 5.74 m

s2( ) =16.2N

FTL = mL gsinθ + a( ) =1kg 9.8 ms2sin51+ 5.74 m

s2( ) =13.4N

d. How long does it take the center mass to move 0.5m and how fast is the right-

most block moving after it has traveled 1.2m?

€

Δx = vixt + 12 axt

2 → t =2Δxax

=2 × 0.5m5.74 m

s2= 0.42s

v fx2 = vix

2 + 2aΔx→ v fx = 2aΔx = 2 × 5.74 ms2×1.2m = 3.7 m

s

FA

m

Part II: Multiple-Choice Circle the best answer to each question. Any other marks will not be given credit. Each multiple-choice question is worth 2 points for a total of 10 points.

1. In the diagram below, the block of mass m is pulled to the right by an applied force

FA oriented at an angle θ above the horizontal and the block experiences an acceleration a. The magnitude of the frictional force is given by a. FAsinθ − ma b. FAcosθ − ma c. FAcosθ + ma d. FN

2. Suppose that you stand up quickly and accidentally bump your head on heavy object

located above your head. The greater force of impact will be

a. on the your head. b. on the heavy object. c. the same for both. d. unable to be determined with the information given.

3. Suppose that an object travels in three steps. In the first step the object covers 100m

in 10 seconds in the negative x-direction then the second step has the object decelerates to rest from velocity of 10 m/s in 10 seconds and third step has the object traveling 150m in the positive x-direction for 25s. For the entire trip, the average velocity of the object is

a. -4.8 m/s b. 0 m/s c. 9.8 m/s d. 12.7 m/s

4. Suppose that a 10,000kg block is sitting on a horizontal frictionless surface. A

0.01kg washer is connected to the block by a string passing over a pulley that is attached to the block. If the washer is released from rest, the acceleration of the block is a. equal to 0 m/s2 b. 0 m/s2< a < 9.8 m/s2 c. greater than 9.8 m/s2 d. unable to be determined.

5. Two marbles sitting on a tabletop. One is dropped from rest vertically

(marble A) and the other shot horizontally off the table (marble B). Compared to marble B, the acceleration of marble A’s is a. the same. b. greater. c. less. d. unable to be determined.

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ω = 2πf =2πT

TS = 2π mk

TP = 2π lg

v = ±kmA 1− x 2

A2

12

x t( ) = Asin 2πtT( )

v t( ) = A kmcos 2πt

T( )

a t( ) = −A kmsin 2πt

T( )

v = fλ =FTµ

fn = nf1 = n v2L

I = 2π 2 f 2ρvA2

€

g = 9.8m s2 G = 6.67×10−11 Nm 2

kg2

NA = 6.02×1023 atomsmole kB = 1.38×10−23 J Kσ = 5.67×10−8 W m 2K 4 vsound = 343m s

Useful formulas: Motion in the r = x, y or z-directions Uniform Circular Motion Geometry /Algebra

€

rf = r0 + v0rt + 12 art

2 ar =v 2

r

v fr = v0r + art Fr = mar = m v 2

r

v fr2 = v0r

2 + 2arΔr v =2πrT

FG =G m1m2

r2

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Circles Triangles SpheresC = 2πr A = 1

2 bh A = 4πr2

A = πr2 V = 43 πr

3

Quadratic equation : ax 2 + bx + c = 0,

whose solutions are given by : x =−b ± b2 − 4ac

2a

Vectors Useful Constants

Linear Momentum/Forces Work/Energy Heat

€

p→

= mv→

Kt = 12mv

2

p→

f = p→

i+ F→

Δt Kr = 12 Iω

2

F→

= ma→

Ug = mgh

Fs→

= −k x→

US = 12 kx

2

Ff = µFN WT = FdCosθ = ΔET

WR = τθ = ΔER

Wnet =WR +WT = ΔER + ΔET

ΔER + ΔET + ΔUg + ΔUS = 0ΔER + ΔET + ΔUg + ΔUS = −ΔEdiss

Rotational Motion Fluids Simple Harmonic Motion/Waves Sound